Complex numbers

Class 11 Mathematics, unit 1, book pages 5–29

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A complex number $x + i y$ extends the real numbers by the imaginary unit $i$ with $i^{2} = - 1$ , so that equations such as $x^{2} + 1 = 0$ can be solved. This unit builds the algebra of complex numbers (parts, equality, the four operations, conjugate and modulus), solves linear systems and quadratic and cubic polynomials with complex numbers, and then introduces polar coordinates and the polar form $r (cos θ + i sin θ)$ , in which multiplication and division become simple. It ends with applications to simple harmonic motion, electrical circuits and waves.

Learning outcomes (14)

Recall a complex number $z$ represented by an expression of the form $z = a + ib$ or of the form $(a, b)$ , where $a$ and $b$ are real numbers and $i = - 1$ .
Recognise $a$ as the real part of $z$ and $b$ as the imaginary part of $z$ .
Know the condition for equality of complex numbers.
Carry out basic operations on complex numbers.
Define $\overset{z}{ˉ} = a - ib$ as the complex conjugate of $z = a + ib$ .
Define $∣ z ∣ = a^{2} + b^{2}$ as the absolute value or modulus of a complex number $z = a + ib$ .
Solve simultaneous linear equations with complex coefficients. For example, $5 z - (3 + i) w = 7 - i$ and $(2 - i) z + 2 i w = - 1 + i$ .
Write a polynomial $P (z)$ as a product of linear factors. For example, $z^{2} + a^{2} = (z - ia) (z + ia)$ and $z^{3} - 3 z^{2} + z + 5 = (z + 1) (z - 2 - i) (z - 2 + i)$ .
Solve a quadratic equation of the form $p z^{2} + q z + r = 0$ by completing squares, where $p, q, r$ are real numbers and $z$ is a complex number. For example, $z^{2} - 2 z + 5 = 0 \Rightarrow (z - 1 - 2 i) (z - 1 + 2 i) = 0 \Rightarrow z = 1 + 2 i, 1 - 2 i$ .
Explain the polar coordinate system.
Describe the polar representation of a complex number.
Apply the operations with complex numbers in polar representation.
Demonstrate simple equations and inequations involving complex numbers in polar form.
Apply the concepts of complex numbers to real world problems (such as cryptography, wave phenomena, calculating voltage, current, circuits, and the velocity and pressure of a fluid).

1.1Complex numberOften examined

A complex number is an expression of the form $x + i y$ where $x, y \in R$ . It is denoted by $z$ , that is $z = x + i y$ , and the set of all complex numbers is denoted by $C$ . The number $i$ is the imaginary unit, defined by $i^{2} = - 1$ (Euler's notation), so $i = - 1$ .

The complex number $x + i y$ is also written as the ordered pair $(x, y)$ . This is justified because there is a one to one correspondence between $x + i y$ and $(x, y)$ . Clearly

i = 0 + i = (0, 1), 1 = 1 + 0 i = (1, 0)

Real and imaginary parts of a complex number (1.1.1)

Every complex number $z = x + i y$ has two parts, $x$ and $y$ . $x$ is called the real part and $y$ the imaginary part:

Re (z) = x, Im (z) = y

If the real part is zero ( $z = i y$ ), the number is called a pure imaginary number.
If the imaginary part is zero ( $z = x + i 0$ ), the number is a real number.
Every real number $x$ can be written as $x + i 0$ , so every real number is a complex number. But a complex number need not be real: only those with zero imaginary part are real. Hence the set of real numbers is a subset of the set of complex numbers, $R \subset C$ .

Example 1

Let

z = 2 + 3 i

. Name the real and imaginary parts of

z

Compare $z = 2 + 3 i$ with the standard form $x + i y$ , so $x = 2$ and $y = 3$ .
$Re (z) = 2, Im (z) = 3$
As an ordered pair this is the point with coordinates $(2, 3)$ on the complex plane (figure below).
$z = 2 + 3 i = (2, 3)$

−1

x

y

z = (2, 3)

2

3

The number

z = 2 + 3 i

drawn as the point

(2, 3)

: the real part

2

is measured along the

x

-axis and the imaginary part

3

along the

y

-axis.

Re (z) = 2, Im (z) = 3

Condition for the equality of two complex numbers (1.1.2)

Unlike real numbers, two complex numbers cannot be compared by size: we cannot say that one complex number is greater or less than another. Two complex numbers are equal if both have the same real part and the same imaginary part.

Equality of complex numbers: Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ . Then $z_{1} = z_{2}$ if and only if $x_{1} = x_{2}$ and $y_{1} = y_{2}$ .

Example 2

Let

z_{1} = x_{1} + i y_{1}

and

z_{2} = x_{2} + i y_{2}

be two complex numbers. When is

z_{1} = z_{2}

Write both numbers in the form real part plus $i$ times imaginary part.
$z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$
They are equal exactly when the real parts agree and the imaginary parts agree.
$z_{1} = z_{2} ⟺ x_{1} = x_{2} and y_{1} = y_{2}$

z_{1} = z_{2} ⟺ x_{1} = x_{2} and y_{1} = y_{2}

1.2Basic algebraic operations on complex numbersOften examined

Throughout, $z_{1} = x_{1} + i y_{1} = (x_{1}, y_{1})$ and $z_{2} = x_{2} + i y_{2} = (x_{2}, y_{2})$ . The four operations are built from the real operations together with $i^{2} = - 1$ .

Addition of two complex numbers (1.2.1)

Add real parts to real parts and imaginary parts to imaginary parts.

z_{1} + z_{2} = (x_{1}, y_{1}) + (x_{2}, y_{2}) = x_{1} + i y_{1} + x_{2} + i y_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2}) = (x_{1} + x_{2}, y_{1} + y_{2})

Example 3

Find the sum of

z_{1} = 2 + 3 i

and

z_{2} = 6 + 8 i

Write the sum and group the real parts and the imaginary parts.
$z_{1} + z_{2} = (2 + 3 i) + (6 + 8 i) = (2 + 6) + (3 + 8) i$
Add inside each bracket.
$z_{1} + z_{2} = 8 + 11 i = (8, 11)$

z_{1} + z_{2} = 8 + 11 i

Subtraction of two complex numbers (1.2.2)

Subtract real part from real part and imaginary part from imaginary part.

z_{1} - z_{2} = (x_{1}, y_{1}) - (x_{2}, y_{2}) = (x_{1} + i y_{1}) - (x_{2} + i y_{2}) = (x_{1} - x_{2}) + i (y_{1} - y_{2}) = (x_{1} - x_{2}, y_{1} - y_{2})

Example 4

z_{1} = 4 - 3 i

and

z_{2} = 7 + 6 i

, find

z_{1} - z_{2}

Write the difference, taking care that the minus sign acts on both parts of $z_{2}$ .
$z_{1} - z_{2} = (4 - 3 i) - (7 + 6 i) = (4 - 7) + (- 3 - 6) i$
Simplify each bracket.
$z_{1} - z_{2} = - 3 + (- 9 i) = - 3 - 9 i$

z_{1} - z_{2} = - 3 - 9 i

Product of two complex numbers (1.2.3)

Derivation Product of two complex numbers

Aim. Express $z_{1} z_{2}$ in the form $x + i y$ .

Multiply the two binomials term by term, as in ordinary algebra.
$z_{1} z_{2} = (x_{1} + i y_{1}) (x_{2} + i y_{2}) = x_{1} x_{2} + i x_{1} y_{2} + i x_{2} y_{1} + i^{2} y_{1} y_{2}$
Replace $i^{2}$ by $- 1$ .
$z_{1} z_{2} = x_{1} x_{2} + i (x_{1} y_{2} + x_{2} y_{1}) - y_{1} y_{2}$
Collect the real part and the imaginary part.
$z_{1} z_{2} = (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})$

Result

z_{1} z_{2} = (x_{1} x_{2} - y_{1} y_{2}, x_{1} y_{2} + x_{2} y_{1})

In practice you do not memorise this result: multiply the brackets out and replace $i^{2}$ by $- 1$ .

Example 5

Find the product of the complex numbers

z_{1} = (2, - 6)

and

z_{2} = (4, 9)

Convert the ordered pairs to the form $x + i y$ .
$z_{1} = 2 - 6 i, z_{2} = 4 + 9 i$
Multiply term by term.
$z_{1} z_{2} = (2 - 6 i) (4 + 9 i) = 8 + 18 i - 24 i - 54 i^{2}$
Use $i^{2} = - 1$ , so $- 54 i^{2} = + 54$ .
$z_{1} z_{2} = 8 - 6 i + 54$
Collect real and imaginary parts.
$z_{1} z_{2} = 62 - 6 i = (62, - 6)$

z_{1} z_{2} = 62 - 6 i = (62, - 6)

Division of complex numbers (1.2.4)

Division is not as simple, because the denominator has two independent parts. To make the denominator a single term we rationalise: multiply the numerator and the denominator by the conjugate of the denominator. The new denominator is then a single real number, and the division can be completed.

Derivation Quotient of two complex numbers

Aim. Express $\frac{z _{1}}{z _{2}}$ in the form $x + i y$ , where $z_{2} \neq = 0$ .

Write the quotient and multiply top and bottom by $x_{2} - i y_{2}$ , the conjugate of the denominator. This multiplies the quotient by 1, so its value does not change.
$\frac{z _{1}}{z _{2}} = \frac{x _{1} + i y _{1}}{x _{2} + i y _{2}} \times \frac{x _{2} - i y _{2}}{x _{2} - i y _{2}}$
Expand the numerator and the denominator.
$= \frac{x _{1} x _{2} - i x _{1} y _{2} + i x _{2} y _{1} - i ^{2} y _{1} y _{2}}{x _{2}^{2} - i x _{2} y _{2} + i x _{2} y _{2} - i ^{2} y _{2}^{2}}$
The two middle terms of the denominator cancel and $i^{2} = - 1$ . The denominator becomes the real number $x_{2}^{2} + y_{2}^{2}$ .
$= \frac{( x _{1} x _{2} + y _{1} y _{2} ) + i ( x _{2} y _{1} - x _{1} y _{2} )}{x _{2}^{2} + y _{2}^{2}}$
Split into real and imaginary parts.
$\frac{z _{1}}{z _{2}} = \frac{x _{1} x _{2} + y _{1} y _{2}}{x _{2}^{2} + y _{2}^{2}} + i \frac{x _{2} y _{1} - x _{1} y _{2}}{x _{2}^{2} + y _{2}^{2}}$

Result

\frac{z _{1}}{z _{2}} = (\frac{x _{1} x _{2} + y _{1} y _{2}}{x _{2}^{2} + y _{2}^{2}}, \frac{x _{2} y _{1} - x _{1} y _{2}}{x _{2}^{2} + y _{2}^{2}})

Rationalising always works because $(x_{2} + i y_{2}) (x_{2} - i y_{2}) = x_{2}^{2} + y_{2}^{2}$ is real and positive when $z_{2} \neq = 0$ .

Example 6

z_{1} = 3 + 7 i

and

z_{2} = - 4 + 6 i

, find the sum, difference, product and quotient of the two complex numbers.

Sum. Add real parts and imaginary parts.
$z_{1} + z_{2} = (3 + 7 i) + (- 4 + 6 i) = (3 - 4) + (7 + 6) i = - 1 + 13 i$
Difference. Subtract, changing the sign of both parts of $z_{2}$ .
$z_{1} - z_{2} = (3 + 7 i) - (- 4 + 6 i) = 3 + 7 i + 4 - 6 i = (3 + 4) + (7 - 6) i = 7 + i$
Product. Multiply term by term.
$z_{1} z_{2} = (3 + 7 i) (- 4 + 6 i) = - 12 + 18 i - 28 i + 42 i^{2}$
Replace $i^{2}$ by $- 1$ : $42 i^{2} = - 42$ .
$z_{1} z_{2} = - 12 - 10 i - 42 = - 54 - 10 i$
Quotient. Multiply the numerator and the denominator by $- 4 - 6 i$ , the conjugate of $- 4 + 6 i$ .
$\frac{z _{1}}{z _{2}} = \frac{3 + 7 i}{- 4 + 6 i} \times \frac{- 4 - 6 i}{- 4 - 6 i} = \frac{- 12 - 18 i - 28 i - 42 i ^{2}}{( - 4 ) ^{2} - ( 6 i ) ^{2}}$
Simplify: $- 42 i^{2} = + 42$ in the numerator and $- (6 i)^{2} = + 36$ in the denominator.
$= \frac{- 12 - 46 i + 42}{16 + 36} = \frac{30 - 46 i}{52}$
Divide each part by 52 and reduce.
$\frac{z _{1}}{z _{2}} = \frac{30}{52} - \frac{46}{52} i = \frac{15}{26} - \frac{23}{26} i$

z_{1} + z_{2} = - 1 + 13 i, z_{1} - z_{2} = 7 + i, z_{1} z_{2} = - 54 - 10 i, \frac{z _{1}}{z _{2}} = \frac{15}{26} - \frac{23}{26} i

Example 7

Write the complex number

\frac{( 2 + 3 i ) ( 2 + i )}{1 - i}

in the form

x + i y

First multiply the two brackets in the numerator.
$(2 + 3 i) (2 + i) = 4 + 2 i + 6 i + 3 i^{2} = 4 + 8 i - 3 = 1 + 8 i$
So the number is $\frac{1 + 8 i}{1 - i}$ . Multiply the top and bottom by $1 + i$ , the conjugate of $1 - i$ .
$\frac{1 + 8 i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{1 + i + 8 i + 8 i ^{2}}{1 ^{2} - i ^{2}}$
Use $i^{2} = - 1$ in the numerator and the denominator: $8 i^{2} = - 8$ and $1 - i^{2} = 2$ .
$= \frac{1 + 9 i - 8}{1 - ( - 1 )} = \frac{- 7 + 9 i}{2}$
Write the real and imaginary parts separately.
$= - \frac{7}{2} + \frac{9}{2} i$

- \frac{7}{2} + \frac{9}{2} i

Example 8

Find the values of

x

and

y

\frac{x}{2 + 3 i} - y (1 + 2 i) = 1 + i

Rationalise the fraction: multiply the top and bottom by $2 - 3 i$ .
$\frac{x}{2 + 3 i} = \frac{x ( 2 - 3 i )}{( 2 + 3 i ) ( 2 - 3 i )} = \frac{2 x - 3 x i}{4 - 9 i ^{2}} = \frac{2 x - 3 x i}{13}$
Substitute back, expand $y (1 + 2 i) = y + 2 i y$ and split the left side into real and imaginary parts.
$\frac{2 x}{13} - \frac{3 x}{13} i - y - 2 y i = 1 + i \Rightarrow (\frac{2 x}{13} - y) + (- \frac{3 x}{13} - 2 y) i = 1 + i$
Two complex numbers are equal when their real parts and imaginary parts are equal. Compare the real parts, then the imaginary parts.
$\frac{2 x}{13} - y = 1 (1) - \frac{3 x}{13} - 2 y = 1 (2)$
Eliminate $y$ : multiply (1) by 2 to get $\frac{4 x}{13} - 2 y = 2$ , then subtract (2) from it.
$(\frac{4 x}{13} + \frac{3 x}{13}) = 2 - 1 \Rightarrow \frac{7 x}{13} = 1 \Rightarrow x = \frac{13}{7}$
Put $x = \frac{13}{7}$ in (1) to find $y$ .
$\frac{2}{13} \times \frac{13}{7} - y = 1 \Rightarrow \frac{2}{7} - y = 1 \Rightarrow y = \frac{2}{7} - 1 = - \frac{5}{7}$
Check in (2): $- \frac{3}{13} \cdot \frac{13}{7} - 2 (- \frac{5}{7}) = - \frac{3}{7} + \frac{10}{7} = 1$ . It holds.

x = \frac{13}{7}, y = - \frac{5}{7}

1.3Conjugate of a complex numberOften examined

The conjugate of a complex number z=x+iy is denoted by \bar z and is defined as

\bar z=x-iy

Geometrically, the conjugate of a complex number is its **mirror image about the x-axis**. For example, if z=3+4i then \bar z=3-4i.

The number z=3+4i and its conjugate \bar z=3-4i are mirror images of each other in the x-axis.

Example 9

Find the conjugate of z=(1+i)(2-i).

First simplify z by multiplying the brackets.
z=(1+i)(2-i)=2-i+2i-i^2=2+i+1=3+i
Change the sign of the imaginary part.
\bar z=\overline{3+i}=3-i

\bar z=3-i

\overline{z_1z_2}=\bar z_1\,\bar z_2,\qquad \overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar z_1}{\bar z_2}

1.4Magnitude or modulus of a complex numberOften examined

Consider the complex number z=x+iy=(x,y). Its magnitude (or modulus) is denoted by |z|. It is found from a right-angled triangle on the complex plane.

Derivation Magnitude of a complex number

Aim. Find |z| for z=x+iy.

The point P(x,y) represents z. OAP is a right-angled triangle with OA=x, AP=y and hypotenuse OP=|z|.

Draw z=x+iy=(x,y) as the point P on the complex plane.
Draw the perpendicular from P on the real axis, meeting it at A. Then PAO is a right-angled triangle.
Apply Pythagoras' theorem to the triangle OAP.
|\overrightarrow{OP}|^2=|\overrightarrow{OA}|^2+|\overrightarrow{AP}|^2
Here |\overrightarrow{OP}|=|z|, |\overrightarrow{OA}|=x and |\overrightarrow{AP}|=y.
|z|^2=x^2+y^2
Take the positive square root, because a length is not negative.
|z|=\sqrt{x^2+y^2}

Result|z|=\sqrt{x^2+y^2}=\sqrt{\left(\operatorname{Re}(z)\right)^2+\left(\operatorname{Im}(z)\right)^2}

|z| is the distance of the point z=(x,y) from the origin.

Example 10

Find the conjugate and magnitude of z=\dfrac{(3+2i)(1-2i)}{4+3i}.

Conjugate. Take the conjugate of the numerator and the denominator, then simplify.

Conjugate of a quotient is the quotient of the conjugates, and the conjugate of a product is the product of the conjugates.
\bar z=\dfrac{\overline{(3+2i)(1-2i)}}{\overline{4+3i}}=\dfrac{(3-2i)(1+2i)}{4-3i}
Multiply the brackets in the numerator.
(3-2i)(1+2i)=3+6i-2i-4i^2=3+4+4i=7+4i
Rationalise by multiplying the top and bottom by 4+3i.
\bar z=\dfrac{7+4i}{4-3i}\times\dfrac{4+3i}{4+3i}=\dfrac{28+21i+16i+12i^2}{4^2-(3i)^2}
Use i^2=-1: the numerator is 28-12+37i and the denominator is 16+9=25.
\bar z=\dfrac{16+37i}{25}=\dfrac{16}{25}+\dfrac{37}{25}i

Magnitude. Use |z_1z_2|=|z_1||z_2| and \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}.

Find the modulus of each factor with |x+iy|=\sqrt{x^2+y^2}.
|z|=\dfrac{|3+2i|\,|1-2i|}{|4+3i|}=\dfrac{\sqrt{3^2+2^2}\,\sqrt{1^2+(-2)^2}}{\sqrt{4^2+3^2}}
Evaluate each root.
|z|=\dfrac{\sqrt{13}\,\sqrt5}{\sqrt{25}}=\dfrac{\sqrt{65}}{5}

\bar z=\dfrac{16}{25}+\dfrac{37}{25}i,\qquad |z|=\dfrac{\sqrt{65}}{5}

Worked example: moduli of a product and a sum

For z_1=3+4i and z_2=5-12i, verify that |z_1z_2|=|z_1||z_2| and show that |z_1+z_2|\neq|z_1|+|z_2|.

Find the two moduli.
|z_1|=\sqrt{9+16}=5,\qquad |z_2|=\sqrt{25+144}=13
Find the product by expanding and using i^2=-1.
z_1z_2=15-36i+20i-48i^2=63-16i
Its modulus.
|z_1z_2|=\sqrt{63^2+16^2}=\sqrt{3969+256}=\sqrt{4225}=65
Compare with the product of the moduli.
|z_1||z_2|=5\times13=65\;\checkmark
Now the sum.
z_1+z_2=8-8i,\qquad |z_1+z_2|=\sqrt{64+64}=8\sqrt2\approx11.3
The sum of the moduli is larger, so the rule for products does not carry over to sums.
|z_1|+|z_2|=5+13=18\neq8\sqrt2

|z_1z_2|=65=|z_1||z_2|,\quad |z_1+z_2|=8\sqrt2\neq18

|z_1z_2|=|z_1|\,|z_2|,\qquad \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}

Theorem: modulus is non-negative

Theorem. If z is a complex number then |z|\ge0, and |z|=0 if and only if z=0.

Derivation Proof that |z|\ge0 and |z|=0\iff z=0

Aim. Show |z|\ge0, and |z|=0 exactly when z=0.

Non-negative. Let z=x+iy, so |z|=\sqrt{x^2+y^2} with x,y real. The square of a real number is never negative, so x^2+y^2\ge0. The square root of a non-negative number is non-negative.
|z|=\sqrt{x^2+y^2}\ge0
**If |z|=0 then z=0.** Suppose |z|=0. Squaring gives a sum of two squares equal to zero.
\sqrt{x^2+y^2}=0\ \Rightarrow\ x^2+y^2=0
Each square is at least zero, so the sum is zero only if each square is zero.
x=0\ \text{and}\ y=0\ \Rightarrow\ z=x+iy=0+i0=0
**Conversely, if z=0 then |z|=0.** Put z=0+i0.
|z|=\sqrt{0^2+0^2}=0

Result|z|\ge0\quad\text{and}\quad |z|=0\iff z=0

1.5Real and imaginary parts of complex numbers of the typesOften examined

This section finds the real and imaginary parts of complex numbers of two types, for n=\pm1 (and, as the examples need, n=\pm2):

Type I: (x+iy)^n
Type II: \left(\dfrac{x_1+iy_1}{x_2+iy_2}\right)^{n}, where x_2+iy_2\neq0

Type I: (x+iy)^n

Write z=x+iy and |z|^2=x^2+y^2.

Derivation Real and imaginary parts of z^{-1}

Aim. Find \operatorname{Re}(z^{-1}) and \operatorname{Im}(z^{-1}) for z=x+iy.

**For n=1** nothing is needed: z=x+iy, so \operatorname{Re}(z)=x and \operatorname{Im}(z)=y.
**For n=-1** write z^{-1}=\dfrac1{x+iy} and rationalise with the conjugate x-iy.
z^{-1}=\dfrac{1}{x+iy}\times\dfrac{x-iy}{x-iy}=\dfrac{x-iy}{x^2-i^2y^2}=\dfrac{x-iy}{x^2+y^2}
Separate the real and imaginary parts.
z^{-1}=\dfrac{x}{x^2+y^2}-i\,\dfrac{y}{x^2+y^2}
Since x^2+y^2=|z|^2, x=\operatorname{Re}(z) and y=\operatorname{Im}(z), read off the parts.
\operatorname{Re}(z^{-1})=\dfrac{x}{x^2+y^2}=\dfrac{\operatorname{Re}(z)}{|z|^2},\qquad \operatorname{Im}(z^{-1})=\dfrac{-y}{x^2+y^2}=-\dfrac{\operatorname{Im}(z)}{|z|^2}

Result\operatorname{Re}(z^{-1})=\dfrac{\operatorname{Re}(z)}{|z|^2},\qquad \operatorname{Im}(z^{-1})=-\dfrac{\operatorname{Im}(z)}{|z|^2}

Derivation Real and imaginary parts of z^2 and z^{-2}

Aim. Find the parts of z^2 and z^{-2} (used in Example 11).

Square z=x+iy with i^2=-1.
z^2=x^2+2ixy+i^2y^2=(x^2-y^2)+i(2xy)
Read off the parts.
\operatorname{Re}(z^2)=(\operatorname{Re}z)^2-(\operatorname{Im}z)^2,\qquad \operatorname{Im}(z^2)=2\operatorname{Re}(z)\operatorname{Im}(z)
For z^{-2}=(z^{-1})^2 apply the same rule to z^{-1}=\dfrac{x}{|z|^2}-i\dfrac{y}{|z|^2}, whose real part is \dfrac{x}{|z|^2} and imaginary part -\dfrac{y}{|z|^2}.
z^{-2}=\dfrac{x^2-y^2}{|z|^4}+i\left(-\dfrac{2xy}{|z|^4}\right)
Read off the parts.
\operatorname{Re}(z^{-2})=\dfrac{(\operatorname{Re}z)^2-(\operatorname{Im}z)^2}{|z|^4},\qquad \operatorname{Im}(z^{-2})=\dfrac{-2\operatorname{Re}(z)\operatorname{Im}(z)}{|z|^4}

Result\operatorname{Re}(z^2)=x^2-y^2,\ \operatorname{Im}(z^2)=2xy;\quad \operatorname{Re}(z^{-2})=\dfrac{x^2-y^2}{|z|^4},\ \operatorname{Im}(z^{-2})=\dfrac{-2xy}{|z|^4}

Example 11

Find the real and imaginary parts of (i) 3+4i, (ii) (3+4i)^{-1}, (iii) (3+4i)^2, (iv) (3+4i)^{-2}.

Let z=3+4i. Then \operatorname{Re}(z)=3, \operatorname{Im}(z)=4 and |z|^2=3^2+4^2=25.

(i) z itself.
\operatorname{Re}(3+4i)=3,\qquad \operatorname{Im}(3+4i)=4
(ii) Use \operatorname{Re}(z^{-1})=\dfrac{\operatorname{Re}z}{|z|^2} and \operatorname{Im}(z^{-1})=-\dfrac{\operatorname{Im}z}{|z|^2}.
\operatorname{Re}(3+4i)^{-1}=\dfrac{3}{25},\qquad \operatorname{Im}(3+4i)^{-1}=\dfrac{-4}{25}
(iii) Use \operatorname{Re}(z^2)=(\operatorname{Re}z)^2-(\operatorname{Im}z)^2.
\operatorname{Re}(3+4i)^2=3^2-4^2=9-16=-7
and \operatorname{Im}(z^2)=2\operatorname{Re}(z)\operatorname{Im}(z).
\operatorname{Im}(3+4i)^2=2(3)(4)=24
(iv) Use |z|^4=25^2=625 (the book writes (\sqrt{3^2+4^2})^4=5^4).
\operatorname{Re}(3+4i)^{-2}=\dfrac{3^2-4^2}{5^4}=\dfrac{9-16}{625}=-\dfrac{7}{625}
and
\operatorname{Im}(3+4i)^{-2}=\dfrac{-2(3)(4)}{5^4}=-\dfrac{24}{625}

Check (iii): (3+4i)^2=9+24i+16i^2=-7+24i. Check (iv): \dfrac{1}{-7+24i}=\dfrac{-7-24i}{49+576}=-\dfrac{7}{625}-\dfrac{24}{625}i.

\begin{aligned}&3,\ 4;\quad \tfrac{3}{25},\ -\tfrac{4}{25};\quad -7,\ 24;\quad -\tfrac{7}{625},\ -\tfrac{24}{625}\end{aligned}

Type II: \left(\dfrac{x_1+iy_1}{x_2+iy_2}\right)^n

Let z_1=x_1+iy_1 and z_2=x_2+iy_2, so the number is \left(\dfrac{z_1}{z_2}\right)^n with z_2\neq0.

Derivation Real and imaginary parts of \dfrac{z_1}{z_2} and \left(\dfrac{z_1}{z_2}\right)^{-1}

Aim. Find the parts for n=1 and n=-1.

**For n=1:** rationalise with x_2-iy_2.
\dfrac{x_1+iy_1}{x_2+iy_2}\times\dfrac{x_2-iy_2}{x_2-iy_2}=\dfrac{x_1x_2-ix_1y_2+ix_2y_1-i^2y_1y_2}{x_2^2-i^2y_2^2}
Use i^2=-1 and separate the parts.
=\dfrac{(x_1x_2+y_1y_2)+i(x_2y_1-x_1y_2)}{x_2^2+y_2^2}
Read off the real and imaginary parts.
\operatorname{Re}\left(\dfrac{z_1}{z_2}\right)=\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2},\qquad \operatorname{Im}\left(\dfrac{z_1}{z_2}\right)=\dfrac{x_2y_1-x_1y_2}{x_2^2+y_2^2}
**For n=-1:** \left(\dfrac{z_1}{z_2}\right)^{-1}=\dfrac{z_2}{z_1}, which is the case n=1 with the roles of z_1 and z_2 exchanged. Rationalise with x_1-iy_1.
\dfrac{x_2+iy_2}{x_1+iy_1}=\dfrac{(x_1x_2+y_1y_2)+i(x_1y_2-x_2y_1)}{x_1^2+y_1^2}
Read off the parts.
\operatorname{Re}\left(\dfrac{z_1}{z_2}\right)^{-1}=\dfrac{x_1x_2+y_1y_2}{x_1^2+y_1^2},\qquad \operatorname{Im}\left(\dfrac{z_1}{z_2}\right)^{-1}=\dfrac{x_1y_2-x_2y_1}{x_1^2+y_1^2}

Result\begin{aligned}\operatorname{Re}\dfrac{z_1}{z_2}&=\dfrac{x_1x_2+y_1y_2}{|z_2|^2}\\[4pt]\operatorname{Im}\dfrac{z_1}{z_2}&=\dfrac{x_2y_1-x_1y_2}{|z_2|^2}\\[4pt]\operatorname{Re}\left(\dfrac{z_1}{z_2}\right)^{-1}&=\dfrac{x_1x_2+y_1y_2}{|z_1|^2}\\[4pt]\operatorname{Im}\left(\dfrac{z_1}{z_2}\right)^{-1}&=\dfrac{x_1y_2-x_2y_1}{|z_1|^2}\end{aligned}

In the book's notation the numerators are \operatorname{Re}(z_1)\operatorname{Re}(z_2)+\operatorname{Im}(z_1)\operatorname{Im}(z_2) and \operatorname{Im}(z_1)\operatorname{Re}(z_2)-\operatorname{Re}(z_1)\operatorname{Im}(z_2).

Derivation Real and imaginary parts of \left(\dfrac{z_1}{z_2}\right)^{2} and \left(\dfrac{z_1}{z_2}\right)^{-2}

Aim. Find the parts for n=2 and n=-2.

Let a=\dfrac{x_1x_2+y_1y_2}{|z_2|^2} and b=\dfrac{x_2y_1-x_1y_2}{|z_2|^2} be the real and imaginary parts of \dfrac{z_1}{z_2}. Squaring a+ib gives real part a^2-b^2 and imaginary part 2ab.
\left(\dfrac{z_1}{z_2}\right)^2=(a^2-b^2)+i\,(2ab)
Expand a^2-b^2 over the common denominator |z_2|^4=(x_2^2+y_2^2)^2 and group the terms.
(x_1x_2+y_1y_2)^2-(x_2y_1-x_1y_2)^2=(x_1^2-y_1^2)(x_2^2-y_2^2)+4x_1x_2y_1y_2
Expand 2ab in the same way.
2(x_1x_2+y_1y_2)(x_2y_1-x_1y_2)=2\left[x_1y_1(x_2^2-y_2^2)-x_2y_2(x_1^2-y_1^2)\right]
So for n=2 (the book's result):
\begin{aligned}\operatorname{Re}\left(\dfrac{z_1}{z_2}\right)^2&=\dfrac{(x_1^2-y_1^2)(x_2^2-y_2^2)+4x_1x_2y_1y_2}{(x_2^2+y_2^2)^2}\\ \operatorname{Im}\left(\dfrac{z_1}{z_2}\right)^2&=\dfrac{2\left[x_1y_1(x_2^2-y_2^2)-x_2y_2(x_1^2-y_1^2)\right]}{(x_2^2+y_2^2)^2}\end{aligned}
For n=-2 exchange z_1 and z_2, because \left(\dfrac{z_1}{z_2}\right)^{-2}=\left(\dfrac{z_2}{z_1}\right)^2. The real part keeps its form, and the imaginary part changes sign.
\begin{aligned}\operatorname{Re}\left(\dfrac{z_1}{z_2}\right)^{-2}&=\dfrac{(x_1^2-y_1^2)(x_2^2-y_2^2)+4x_1x_2y_1y_2}{(x_1^2+y_1^2)^2}\\ \operatorname{Im}\left(\dfrac{z_1}{z_2}\right)^{-2}&=\dfrac{-2\left[x_1y_1(x_2^2-y_2^2)-x_2y_2(x_1^2-y_1^2)\right]}{(x_1^2+y_1^2)^2}\end{aligned}

Result\text{The denominators are }|z_2|^4\text{ for }n=2\text{ and }|z_1|^4\text{ for }n=-2.

Example 12

If x_1+iy_1=12+5i and x_2+iy_2=3+2i, find the real and imaginary parts of (i) \dfrac{x_1+iy_1}{x_2+iy_2}, (ii) \left(\dfrac{x_1+iy_1}{x_2+iy_2}\right)^{-1}, (iii) \left(\dfrac{x_1+iy_1}{x_2+iy_2}\right)^{2}, (iv) \left(\dfrac{x_1+iy_1}{x_2+iy_2}\right)^{-2}.

Here x_1=12, y_1=5, x_2=3, y_2=2. So x_1^2+y_1^2=169 and x_2^2+y_2^2=13. Three quantities are used repeatedly:

\begin{aligned}x_1x_2+y_1y_2&=36+10=46\\ x_2y_1-x_1y_2&=15-24=-9\\ x_1y_1(x_2^2-y_2^2)-x_2y_2(x_1^2-y_1^2)&=60(5)-6(119)=-414\end{aligned}

(i) Use the n=1 formulas with denominator x_2^2+y_2^2=13.
\operatorname{Re}=\dfrac{46}{13},\qquad \operatorname{Im}=\dfrac{-9}{13}
(ii) Use the n=-1 formulas with denominator x_1^2+y_1^2=169. The imaginary numerator is x_1y_2-x_2y_1=24-15=9.
\operatorname{Re}=\dfrac{46}{169},\qquad \operatorname{Im}=\dfrac{9}{169}
(iii) Use the n=2 formulas with denominator (x_2^2+y_2^2)^2=13^2=169. The real numerator is (144-25)(9-4)+4(12)(3)(5)(2)=595+1440=2035.
\operatorname{Re}=\dfrac{2035}{169}
The imaginary numerator is 2(-414)=-828.
\operatorname{Im}=\dfrac{2\left[60(9-4)-6(144-25)\right]}{169}=\dfrac{2(300-714)}{169}=\dfrac{-828}{169}
(iv) Use the n=-2 formulas with denominator (x_1^2+y_1^2)^2=169^2=28561. The real numerator is the same 2035.
\operatorname{Re}=\dfrac{2035}{28561}
The imaginary numerator changes sign: -2(-414)=828.
\operatorname{Im}=\dfrac{-2(300-714)}{169^2}=\dfrac{828}{28561}

Check by direct division: \dfrac{12+5i}{3+2i}=\dfrac{46-9i}{13}, so (iii) =\left(\dfrac{46-9i}{13}\right)^2=\dfrac{2116-81-828i}{169}=\dfrac{2035-828i}{169}, as found.

\dfrac{46}{13},-\dfrac{9}{13};\ \dfrac{46}{169},\dfrac{9}{169};\ \dfrac{2035}{169},-\dfrac{828}{169};\ \dfrac{2035}{28561},\dfrac{828}{28561}

Equations and inequations in x and y

A condition on z is turned into a condition on the real numbers x and y by putting z=x+iy, so that \bar z=x-iy, and by using |x+iy|=\sqrt{x^2+y^2}.

Example 13

Write the equation |z-2i|=|\bar z+3| in terms of x and y, by taking z=x+iy.

Put z=x+iy and \bar z=x-iy in the equation.
|x+iy-2i|=|x-iy+3|
Group the real and imaginary parts inside each modulus.
|x+i(y-2)|=|(x+3)-iy|
Write each modulus as a square root of (real part)^2 + (imaginary part)^2.
\sqrt{x^2+(y-2)^2}=\sqrt{(x+3)^2+(-y)^2}
Square both sides.
x^2+(y-2)^2=(x+3)^2+y^2
Expand both brackets.
x^2+y^2-4y+4=x^2+6x+9+y^2
Cancel x^2 and y^2 on both sides and collect everything on one side.
-4y+4=6x+9\ \Rightarrow\ 6x+4y+5=0

The equation 6x+4y+5=0 is a straight line. It is the set of points z equally distant from 2i=(0,2) and -3=(-3,0), since |\bar z+3|=|z+3|.

6x+4y+5=0

Example 14

Write the inequation \operatorname{Re}(z-3)\le2 in terms of x and y, by taking z=x+iy.

Put z=x+iy.
\operatorname{Re}(x+iy-3)\le2
Collect the real part x-3 and the imaginary part y.
\operatorname{Re}\{(x-3)+iy\}\le2
The real part of (x-3)+iy is x-3, and y does not appear.
x-3\le2\ \Rightarrow\ x\le5

The inequation x\le5 describes the half-plane to the left of the line x=5, boundary included.

x\le5

Worked example: an equation with moduli in x and y

Write |z+1|=|z-i| in terms of x and y, taking z=x+iy, and say what kind of curve it is.

The line y=-x is the set of all z=x+iy with |z+1|=|z-i|. The numbers -1 and i are marked: each point of the line is equally far from both.

Put z=x+iy and collect the real and imaginary parts inside each modulus.
z+1=(x+1)+iy,\qquad z-i=x+i(y-1)
Use |a+ib|=\sqrt{a^2+b^2} on both sides.
\sqrt{(x+1)^2+y^2}=\sqrt{x^2+(y-1)^2}
Both sides are non-negative, so squaring is safe.
(x+1)^2+y^2=x^2+(y-1)^2
Expand the brackets.
x^2+2x+1+y^2=x^2+y^2-2y+1
Cancel x^2, y^2 and 1 from both sides.
2x=-2y\;\Longrightarrow\;x+y=0
Check with a point of the line, z=1-i: z+1=2-i and z-i=1-2i have the same modulus.
|2-i|=\sqrt5=|1-2i|
So the condition is a straight line through the origin (see the graph).

x+y=0\quad(\text{the line }y=-x)

1.6Solution of equationsOften examined

More than one equation together form a system of equations. The values of the variables that satisfy all the equations at the same time are the simultaneous solutions. If z and \omega are complex variables, an equation of the form az+b\omega=p is an equation with complex variables z and \omega; here a and b cannot both be zero. If a,b are themselves complex numbers, the equation is a linear equation in two variables with complex coefficients. This section solves a system of two such equations in two variables.

Consistent system: A system of equations that has at least one solution.
Inconsistent system: A system of equations that has no solution.

Working rule: elimination method (1.6.1)

Consider the two linear equations a_1z+b_1\omega=p_1 and a_2z+b_2\omega=p_2.

Step 1. Multiply one equation, or both, by suitable numbers so that the coefficients of one of the variables become the same.
Step 2. Add or subtract the equations of Step 1 to eliminate the term of the variable with the same coefficients.
Step 3. The equation of Step 2 has only one variable. Find the value of this variable.
Step 4. Substitute the value found in Step 3 in either given equation and get the value of the other variable.
Step 5. Write the values of z and \omega as the ordered pair (z,\omega), the solution of the system.

Working rule: substitution method (1.6.2)

Step 1. From either equation, write one variable in terms of the other.
Step 2. Substitute it in the equation not yet used.
Step 3. That equation now has one variable only. Find its value.
Step 4. Substitute this value in either given equation and get the value of the other variable.
Step 5. Write the values of z and \omega as the ordered pair (z,\omega), the solution of the system.

Example 15

Solve the following system of simultaneous equations: 2z-(1-3i)\omega=1+2i,\quad (1+i)z+(2-i)\omega=2+i.

The equations are

2z-(1-3i)\omega=1+2i\qquad(1)\qquad\qquad (1+i)z+(2-i)\omega=2+i\qquad(2)

**Step 1: make the coefficients of z equal.** Multiply (1) by (1+i) and (2) by 2. Both now have 2(1+i)z.
\begin{aligned}2(1+i)z-(1+i)(1-3i)\omega&=(1+i)(1+2i)\\ 2(1+i)z+2(2-i)\omega&=2(2+i)\end{aligned}
Step 2: subtract. Subtract the second equation from the first. The z terms cancel.
-\left[(1+i)(1-3i)+2(2-i)\right]\omega=(1+i)(1+2i)-2(2+i)
Evaluate the brackets. On the left (1+i)(1-3i)=1-3i+i-3i^2=4-2i and 2(2-i)=4-2i. On the right (1+i)(1+2i)=1+2i+i+2i^2=-1+3i and 2(2+i)=4+2i.
-(8-4i)\,\omega=(-1+3i)-(4+2i)=-5+i
**Step 3: solve for \omega.** Divide by -8+4i and rationalise with -8-4i.
\omega=\dfrac{-5+i}{-8+4i}\times\dfrac{-8-4i}{-8-4i}=\dfrac{40+20i-8i-4i^2}{(-8)^2-(4i)^2}=\dfrac{40+12i+4}{64+16}
Simplify.
\omega=\dfrac{44+12i}{80}=\dfrac{11}{20}+\dfrac{3}{20}i
Step 4: substitute in (1) to find z.
2z-(1-3i)\left(\dfrac{11+3i}{20}\right)=1+2i
Multiply both sides by 20.
40z-(1-3i)(11+3i)=20(1+2i)
Expand: (1-3i)(11+3i)=11+3i-33i-9i^2=20-30i.
40z-(20-30i)=20+40i
Move (20-30i) to the right and divide by 40.
40z=40+10i\ \Rightarrow\ z=1+\dfrac14 i
Step 5: write the solution as an ordered pair (z,\omega). Check in (2): (1+i)\left(1+\tfrac i4\right)=\tfrac34+\tfrac54 i and (2-i)\left(\tfrac{11}{20}+\tfrac{3}{20}i\right)=\tfrac54-\tfrac14 i, which add to 2+i.

(z,\omega)=\left(1+\tfrac14 i,\ \tfrac{11}{20}+\tfrac{3}{20}i\right)

Challenge: Cramer's rule

The book asks: solve the system of Example 15 by Cramer's rule.

For a_1z+b_1\omega=p_1 and a_2z+b_2\omega=p_2, Cramer's rule gives z=\dfrac{\Delta_z}{\Delta} and \omega=\dfrac{\Delta_\omega}{\Delta}, where \Delta=a_1b_2-a_2b_1, \Delta_z=p_1b_2-p_2b_1 and \Delta_\omega=a_1p_2-a_2p_1. Here a_1=2, b_1=-(1-3i), p_1=1+2i, a_2=1+i, b_2=2-i, p_2=2+i.

Find \Delta.
\Delta=2(2-i)+(1+i)(1-3i)=(4-2i)+(4-2i)=8-4i
Find \Delta_z.
\Delta_z=(1+2i)(2-i)+(2+i)(1-3i)=(4+3i)+(5-5i)=9-2i
Find \Delta_\omega.
\Delta_\omega=2(2+i)-(1+i)(1+2i)=(4+2i)-(-1+3i)=5-i
Divide and rationalise with 8+4i.
z=\dfrac{9-2i}{8-4i}=\dfrac{(9-2i)(8+4i)}{80}=\dfrac{80+20i}{80}=1+\dfrac14 i
Similarly for \omega.
\omega=\dfrac{5-i}{8-4i}=\dfrac{(5-i)(8+4i)}{80}=\dfrac{44+12i}{80}=\dfrac{11}{20}+\dfrac{3}{20}i

(z,\omega)=\left(1+\tfrac14 i,\ \tfrac{11}{20}+\tfrac{3}{20}i\right)

1.7Complex polynomialOften examined

If z is a complex variable, the expression a_0+a_1z+a_2z^2+\cdots+a_nz^n is called a complex polynomial of degree n if a_n\neq0 and n is a non-negative integer. The a_0,a_1,\ldots,a_n are constants and may be real or complex. The polynomial is denoted by

P(z)=a_0+a_1z+a_2z^2+\cdots+a_nz^n

When n=1 the polynomial a_0+a_1z is a linear polynomial. The aim here is to write polynomials of two types as a product of linear factors:

P(z)=z^2+a^2, where a is a real number;
P(z)=az^3+bz^2+cz+d, where a,b,c,d are real numbers.

Factorisation of z^2+a^2 (1.7.1)

This type is simple. Write a^2 as -i^2a^2, because i^2=-1, and use the difference of two squares A^2-B^2=(A+B)(A-B) with A=z and B=ia.

z^2+a^2=z^2-i^2a^2=z^2-(ia)^2=(z+ia)(z-ia)

So (z+ia)(z-ia) are the required linear factors of z^2+a^2.

Factorisation of az^3+bz^2+cz+d, a\neq0 (1.7.2)

First find one factor with the help of the factor theorem, then divide by it using synthetic division to get the depressed (quadratic) equation, and factorise that. Recall that z-\alpha is a factor of P(z) if and only if P(\alpha)=0; we then say \alpha is a root or zero of the polynomial.

For example, z-2 is a factor of P(z)=2z^3+3z^2+6z-40 since P(2)=2(2)^3+3(2)^2+6(2)-40=16+12+12-40=0.

Example 16

Factorise the polynomial P(z)=z^3+2z^2-5z-6.

The product of the coefficient of z^3 and the last term is (1)(6)=6. The possible roots are the factors of 6.
\pm1,\ \pm2,\ \pm3,\ \pm6
Try z=-1.
P(-1)=(-1)^3+2(-1)^2-5(-1)-6=-1+2+5-6=0
So z-(-1)=z+1 is a factor. Divide P(z) by z+1 using synthetic division with the root -1: bring down the first coefficient, multiply by -1, add to the next coefficient, and repeat. The table is below. The last number 0 is the remainder.

Synthetic division of z^3+2z^2-5z-6 by z+1. The bottom row 1,\,1,\,-6 gives the quotient z^2+z-6; the final 0 is the remainder.
root -1	1	2	-5	-6
multiply by -1		-1	-1	6
sum (bring down and add)	1	1	-6	0

Synthetic division of z^3+2z^2-5z-6 by z+1. The bottom row 1,\,1,\,-6 gives the quotient z^2+z-6; the final 0 is the remainder.
root -1	multiply by -1	sum (bring down and add)
1		1
2	-1	1
-5	-1	-6
-6	6	0

Read the quotient (the depressed polynomial) from the bottom row.
z^3+2z^2-5z-6=(z+1)(z^2+z-6)
Factorise z^2+z-6: split the middle term z=3z-2z and take common factors.
z^2+z-6=z^2+3z-2z-6=z(z+3)-2(z+3)=(z+3)(z-2)
Combine.
P(z)=(z+1)(z+3)(z-2)

z^3+2z^2-5z-6=(z+1)(z+3)(z-2)

Solution by the completing the square method (1.7.3)

Example 17

Solve the equation 2z^2-6z-9=0 by the completing square method.

Divide both sides by 2, the coefficient of z^2, so that the coefficient of z^2 becomes 1.
z^2-3z-\dfrac92=0
Move the constant to the right side.
z^2-3z=\dfrac92
Add the square of half the coefficient of z, namely \left(-\dfrac32\right)^2, to both sides. The left side becomes a perfect square.
z^2-3z+\left(-\dfrac32\right)^2=\dfrac92+\left(-\dfrac32\right)^2\ \Rightarrow\ \left(z-\dfrac32\right)^2=\dfrac92+\dfrac94=\dfrac{27}{4}
Take the square root of both sides. \sqrt{27}=3\sqrt3.
z-\dfrac32=\pm\dfrac{3\sqrt3}{2}
Add \dfrac32.
z=\dfrac32\pm\dfrac{3\sqrt3}{2}=\dfrac{3\pm3\sqrt3}{2}

\text{S.S.}=\left\{\dfrac{3\pm3\sqrt3}{2}\right\}

The opening box of the unit gives two more cases as the aim of this section, where the roots are complex. They use the same two ideas, completing the square and the factor theorem:

From the opening box: z^2-2z+5=0

Solve z^2-2z+5=0 by completing the square and write it as a product of linear factors.

Move the constant and add (-1)^2=1 to both sides.
z^2-2z+1=-5+1\ \Rightarrow\ (z-1)^2=-4
The right side is negative, so use -4=(2i)^2. Then (z-1)^2-(2i)^2=0, which factorises by the difference of two squares.
(z-1-2i)(z-1+2i)=0
Each factor gives a root.
z=1+2i\quad\text{or}\quad z=1-2i

z^2-2z+5=(z-1-2i)(z-1+2i),\quad z=1\pm2i

From the opening box: z^3-3z^2+z+5

Write P(z)=z^3-3z^2+z+5 as a product of linear factors.

Try the factors of 5. For z=-1:
P(-1)=-1-3-1+5=0
So z+1 is a factor. Synthetic division with root -1 has bottom row 1,\,-4,\,5,\,0.
z^3-3z^2+z+5=(z+1)(z^2-4z+5)
Factorise z^2-4z+5 by completing the square: (z-2)^2=-1=i^2.
z^2-4z+5=(z-2-i)(z-2+i)
Combine.
P(z)=(z+1)(z-2-i)(z-2+i)

z^3-3z^2+z+5=(z+1)(z-2-i)(z-2+i)

Worked example: completing the square with a leading coefficient

Solve 4z^2-8z+13=0 by completing the square and write the left side as a product of linear factors.

Divide by the leading coefficient 4 so that the coefficient of z^2 is 1.
z^2-2z+\dfrac{13}4=0
Move the constant to the right.
z^2-2z=-\dfrac{13}4
Add the square of half the coefficient of z, that is (-1)^2=1, to both sides.
z^2-2z+1=1-\dfrac{13}4=-\dfrac94
The left side is a perfect square. Write -\tfrac94=\left(\tfrac{3i}2\right)^2, since i^2=-1.
(z-1)^2=\left(\dfrac{3i}2\right)^2
Take square roots, keeping both signs.
z-1=\pm\dfrac{3i}2\;\Longrightarrow\;z=1\pm\dfrac32 i
Check using the roots: their sum should be -b/a=2 and their product c/a=\tfrac{13}4.
\left(1+\tfrac32i\right)+\left(1-\tfrac32i\right)=2,\quad \left(1+\tfrac32i\right)\left(1-\tfrac32i\right)=1+\tfrac94=\tfrac{13}4
Each root gives a linear factor, and the leading coefficient 4 stays in front.
4z^2-8z+13=4\left(z-1-\tfrac32i\right)\left(z-1+\tfrac32i\right)

z=1\pm\dfrac32 i

1.8Polar coordinate system

Another way to locate a point in the plane is the polar coordinate system. It consists of a fixed point O called the pole and the horizontal line emerging from the pole, called the initial line (or polar axis). For a point P in the plane, let r be its distance from the pole and let \theta be the angle measured anticlockwise from the initial line to the line \overrightarrow{OP}. Then the ordered pair (r,\theta) gives the polar coordinates of P.

Polar coordinates (r,\theta) of a point P: r is the distance OP and \theta is the angle from the initial line to OP, measured anticlockwise.

The angle \theta is also called the argument of z, written \arg(z).

For z=0, \arg(z) is undefined, so it is understood that z\neq0 whenever the polar system is used.
If a complex number z=x+iy has polar coordinates (r,\theta), then its conjugate \bar z=x-iy has polar coordinates (r,-\theta).

1.9.1Polar representation of a complex numberOften examined

Consider a complex number z=x+iy in Cartesian form. Draw it as the point P(x,y) on the complex plane. Let r=|z| and let \theta be the angle, in the positive direction, which \overrightarrow{OP} makes with the initial line (the x-axis). Draw PL perpendicular to the initial line.

Derivation Polar form of a complex number

Aim. Write z=x+iy in terms of r=|z| and \theta.

The point P(x,y) for z=x+iy with r=|z|=OP and \theta the angle OP makes with the initial line (the x-axis). PL is perpendicular to the x-axis.

Apply Pythagoras' theorem to the right-angled triangle OLP, with OL=x, LP=y and OP=r=|z|.
|OL|^2+|LP|^2=|OP|^2\ \Rightarrow\ x^2+y^2=|z|^2=r^2\ \Rightarrow\ r=\sqrt{x^2+y^2}
In the triangle, \cos\theta=\dfrac{OL}{OP}=\dfrac{x}{r}. Multiply by r.
\dfrac{x}{|z|}=\cos\theta\ \Rightarrow\ x=r\cos\theta
Similarly \sin\theta=\dfrac{LP}{OP}=\dfrac{y}{r}.
\dfrac{y}{r}=\sin\theta\ \Rightarrow\ y=r\sin\theta
Substitute x and y in z=x+iy and take r out as a common factor.
z=r\cos\theta+ir\sin\theta=r(\cos\theta+i\sin\theta)

Resultz=r(\cos\theta+i\sin\theta),\qquad r=|z|=\sqrt{x^2+y^2},\quad \theta=\arg(z)

This form of the complex number is called its polar form. It tells the distance from the origin (r) and the direction (\theta) instead of the horizontal and vertical steps (x and y).

1.9.2Principal argumentOften examined

The symbol \arg(z) really stands for a set of values. The argument \theta that lies in the interval -\pi<\theta\le\pi is called the principal value of \arg(z), or the principal argument of z. It is unique and is written \operatorname{Arg}(z):

-\pi<\operatorname{Arg}(z)\le\pi

Let \alpha be the acute angle between \overrightarrow{OP} and the x-axis, found from \alpha=\tan^{-1}\left|\dfrac{y}{x}\right| (it uses the absolute value, so \alpha is positive). The principal argument \theta depends on the quadrant of z:

In the first quadrant the value is \theta, in the second \pi-\theta, in the fourth -\theta and in the third \theta-\pi (the book's figure, with \theta the acute angle).
Quadrant of z	Principal argument
First	\theta=\alpha
Second	\theta=\pi-\alpha
Third	\theta=\alpha-\pi=-(\pi-\alpha)
Fourth	\theta=-\alpha

Principal argument in each quadrant, in terms of the acute reference angle \theta (called \alpha in the table).

Example 18

Find the modulus and principal argument of the following complex numbers: (i) \sqrt3+i, (ii) -\sqrt3+i, (iii) -\sqrt3-i, (iv) \sqrt3-i.

All four have |x|=\sqrt3 and |y|=1, so they have the same modulus and the same reference angle.

Modulus (the same for all four).
r=\sqrt{x^2+y^2}=\sqrt{(\sqrt3)^2+1^2}=\sqrt{3+1}=2
Reference angle (the same for all four).
\alpha=\tan^{-1}\left|\dfrac yx\right|=\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}
(i) \sqrt3+i is in the first quadrant (x positive, y positive), so \theta=\alpha.
\operatorname{Arg}(\sqrt3+i)=\dfrac\pi6
(ii) -\sqrt3+i is in the second quadrant, so \theta=\pi-\alpha.
\operatorname{Arg}(-\sqrt3+i)=\pi-\dfrac\pi6=\dfrac{5\pi}{6}
(iii) -\sqrt3-i is in the third quadrant, so \theta=\alpha-\pi.
\operatorname{Arg}(-\sqrt3-i)=-\pi+\dfrac\pi6=-\dfrac{5\pi}{6}
(iv) \sqrt3-i is in the fourth quadrant, so \theta=-\alpha.
\operatorname{Arg}(\sqrt3-i)=-\dfrac\pi6

The four numbers of Example 18 all lie at distance 2 from the origin and make the same acute angle \pi/6 with the x-axis, one in each quadrant.

In all four cases the moduli are equal, but the arguments depend on the quadrant in which the complex number lies.

2\ \text{and}\ \dfrac\pi6;\quad 2\ \text{and}\ \dfrac{5\pi}6;\quad 2\ \text{and}\ -\dfrac{5\pi}6;\quad 2\ \text{and}\ -\dfrac\pi6

Example 19

Represent the complex number (i) -1-i, (ii) 1+i\sqrt3 in polar form.

(i) Let -1-i=r(\cos\theta+i\sin\theta).

Modulus.
r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt2
Reference angle.
\alpha=\tan^{-1}\left|\dfrac yx\right|=\tan^{-1}1=\dfrac\pi4
-1-i lies in the third quadrant, so the principal argument is \alpha-\pi.
\theta=\alpha-\pi=\dfrac\pi4-\pi=-\dfrac{3\pi}{4}
Write the polar form. Using \cos(-\phi)=\cos\phi and \sin(-\phi)=-\sin\phi it can also be written with positive angles.
-1-i=\sqrt2\left[\cos\left(-\dfrac{3\pi}{4}\right)+i\sin\left(-\dfrac{3\pi}{4}\right)\right]=\sqrt2\left(\cos\dfrac{3\pi}{4}-i\sin\dfrac{3\pi}{4}\right)
Since \sin(\theta+2k\pi)=\sin\theta and \cos(\theta+2k\pi)=\cos\theta (key fact), the general polar form adds 2k\pi to the angle, k an integer.
-1-i=\sqrt2\left[\cos\left(\dfrac{3\pi}{4}+2k\pi\right)-i\sin\left(\dfrac{3\pi}{4}+2k\pi\right)\right]

(ii) 1+i\sqrt3.

Modulus.
r=|z|=\sqrt{1^2+(\sqrt3)^2}=2
Both parts are positive (first quadrant), so \theta=\alpha.
\theta=\tan^{-1}(\sqrt3)=\dfrac\pi3,\qquad \arg(z)=\dfrac\pi3
Polar form, and the general form with 2k\pi.
1+i\sqrt3=2\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)=2\left[\cos\left(\dfrac\pi3+2k\pi\right)+i\sin\left(\dfrac\pi3+2k\pi\right)\right]

-1-i=\sqrt2\left(\cos\tfrac{3\pi}{4}-i\sin\tfrac{3\pi}{4}\right);\quad 1+i\sqrt3=2\left(\cos\tfrac\pi3+i\sin\tfrac\pi3\right)

Example 20

Find the principal argument of z=\dfrac{-2}{1+i\sqrt3}.

Use \arg\left(\dfrac{z_1}{z_2}\right)=\arg z_1-\arg z_2.
\arg z=\arg(-2)-\arg(1+i\sqrt3)
-2 lies on the negative real axis, so its argument is \pi (the book writes \pi-\tan^{-1}(0/2)=\pi). For 1+i\sqrt3 the argument is \tan^{-1}\dfrac{\sqrt3}{1}=\dfrac\pi3.
\arg z=\left[\pi-\tan^{-1}\left(\dfrac02\right)\right]-\tan^{-1}\left(\dfrac{\sqrt3}{1}\right)=\pi-\dfrac\pi3=\dfrac{2\pi}{3}
So one value of \arg z is \dfrac{2\pi}3. It lies between -\pi and \pi, so it is the principal argument.
\operatorname{Arg}z=\dfrac{2\pi}{3}
Check by direct division: z=\dfrac{-2(1-i\sqrt3)}{4}=\dfrac{-1+i\sqrt3}{2}, in the second quadrant with reference angle \tan^{-1}\sqrt3=\dfrac\pi3, so \theta=\pi-\dfrac\pi3=\dfrac{2\pi}3.

\operatorname{Arg}z=\dfrac{2\pi}{3}

Worked example: principal argument in the third quadrant

Find the modulus and the principal argument of z=-2-2\sqrt3\,i, and write z in polar form.

z=-2-2\sqrt3\,i lies in the third quadrant. The acute angle \alpha=\pi/3 is measured from the negative x-axis, but the principal argument \theta=-2\pi/3 is measured clockwise from the positive x-axis.

Modulus from r=\sqrt{x^2+y^2} with x=-2, y=-2\sqrt3.
r=\sqrt{4+12}=\sqrt{16}=4
Acute reference angle, using the absolute value of y/x so that \alpha is positive.
\alpha=\tan^{-1}\left|\dfrac{-2\sqrt3}{-2}\right|=\tan^{-1}\sqrt3=\dfrac\pi3
Both x and y are negative, so z is in the third quadrant. There the table gives \alpha-\pi.
\operatorname{Arg}(z)=\dfrac\pi3-\pi=-\dfrac{2\pi}3
Check that it lies in (-\pi,\pi]: yes. Note that \tan^{-1} of y/x=\sqrt3 alone gives \pi/3, which is the first quadrant and wrong.
Check by returning to x+iy: 4\cos\left(-\tfrac{2\pi}3\right)=4\left(-\tfrac12\right)=-2 and 4\sin\left(-\tfrac{2\pi}3\right)=4\left(-\tfrac{\sqrt3}2\right)=-2\sqrt3.
Write the polar form z=r(\cos\theta+i\sin\theta).
z=4\left[\cos\left(-\dfrac{2\pi}3\right)+i\sin\left(-\dfrac{2\pi}3\right)\right]

r=4,\quad \operatorname{Arg}(z)=-\dfrac{2\pi}3

1.9.3Properties of complex numbers in polar formOften examined

Let z=r(\cos\theta+i\sin\theta), z_1=r_1(\cos\theta_1+i\sin\theta_1) and z_2=r_2(\cos\theta_2+i\sin\theta_2). The five properties below show how the operations look in polar form.

Derivation Property 1: reciprocal

Aim. If z=r(\cos\theta+i\sin\theta), show that z^{-1}=\dfrac1r(\cos\theta-i\sin\theta).

Write z^{-1}=\dfrac1z.
z^{-1}=\dfrac1z=\dfrac{1}{r(\cos\theta+i\sin\theta)}
Multiply the top and bottom by the conjugate \cos\theta-i\sin\theta.
=\dfrac{1}{r(\cos\theta+i\sin\theta)}\times\dfrac{\cos\theta-i\sin\theta}{\cos\theta-i\sin\theta}=\dfrac{\cos\theta-i\sin\theta}{r(\cos^2\theta-i^2\sin^2\theta)}
Since i^2=-1, the denominator is r(\cos^2\theta+\sin^2\theta)=r.
z^{-1}=\dfrac1r(\cos\theta-i\sin\theta)

Resultz^{-1}=\dfrac1r(\cos\theta-i\sin\theta)

The reciprocal has modulus \dfrac1r and argument -\theta.

Derivation Property 2: sum

Aim. Show that z_1+z_2=(r_1\cos\theta_1+r_2\cos\theta_2)+i(r_1\sin\theta_1+r_2\sin\theta_2).

Write the sum and open the brackets.
z_1+z_2=r_1(\cos\theta_1+i\sin\theta_1)+r_2(\cos\theta_2+i\sin\theta_2)=r_1\cos\theta_1+ir_1\sin\theta_1+r_2\cos\theta_2+ir_2\sin\theta_2
Group the real parts and the imaginary parts.
z_1+z_2=(r_1\cos\theta_1+r_2\cos\theta_2)+i(r_1\sin\theta_1+r_2\sin\theta_2)

Resultz_1+z_2=(r_1\cos\theta_1+r_2\cos\theta_2)+i(r_1\sin\theta_1+r_2\sin\theta_2)

Derivation Property 3: difference

Aim. Show that z_1-z_2=(r_1\cos\theta_1-r_2\cos\theta_2)+i(r_1\sin\theta_1-r_2\sin\theta_2).

Write the difference and open the brackets, changing the sign of both terms of z_2.
z_1-z_2=r_1\cos\theta_1+ir_1\sin\theta_1-r_2\cos\theta_2-ir_2\sin\theta_2
Group the real parts and the imaginary parts.
z_1-z_2=(r_1\cos\theta_1-r_2\cos\theta_2)+i(r_1\sin\theta_1-r_2\sin\theta_2)

Resultz_1-z_2=(r_1\cos\theta_1-r_2\cos\theta_2)+i(r_1\sin\theta_1-r_2\sin\theta_2)

Derivation Property 4: product

Aim. Show that z_1z_2=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right].

Multiply the two polar forms and take r_1r_2 out.
z_1z_2=r_1(\cos\theta_1+i\sin\theta_1)\,r_2(\cos\theta_2+i\sin\theta_2)=r_1r_2\left[(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\right]
Expand the brackets.
=r_1r_2\left[\cos\theta_1\cos\theta_2+i\cos\theta_1\sin\theta_2+i\sin\theta_1\cos\theta_2+i^2\sin\theta_1\sin\theta_2\right]
Replace i^2 by -1 and group the real and imaginary parts.
=r_1r_2\left[(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2)\right]
Use the addition formulas \cos(A+B)=\cos A\cos B-\sin A\sin B and \sin(A+B)=\sin A\cos B+\cos A\sin B.
z_1z_2=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]

Resultz_1z_2=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]=r_1r_2\,e^{i(\theta_1+\theta_2)}

To multiply, multiply the moduli and add the arguments.

Derivation Property 5: division

Aim. Show that \dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\left[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right].

Write the quotient and multiply the top and bottom by the conjugate \cos\theta_2-i\sin\theta_2 of the denominator's bracket.
\dfrac{z_1}{z_2}=\dfrac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2(\cos\theta_2+i\sin\theta_2)}=\dfrac{r_1}{r_2}\left[\dfrac{(\cos\theta_1+i\sin\theta_1)(\cos\theta_2-i\sin\theta_2)}{(\cos\theta_2+i\sin\theta_2)(\cos\theta_2-i\sin\theta_2)}\right]
Expand the numerator, and use (\cos\theta_2)^2-(i\sin\theta_2)^2 for the denominator.
=\dfrac{r_1}{r_2}\left[\dfrac{\cos\theta_1\cos\theta_2-i\cos\theta_1\sin\theta_2+i\sin\theta_1\cos\theta_2-i^2\sin\theta_1\sin\theta_2}{\cos^2\theta_2-i^2\sin^2\theta_2}\right]
Replace i^2 by -1. The denominator becomes \cos^2\theta_2+\sin^2\theta_2=1. Group the numerator.
=\dfrac{r_1}{r_2}\left[(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2)\right]
Use \cos(A-B)=\cos A\cos B+\sin A\sin B and \sin(A-B)=\sin A\cos B-\cos A\sin B.
\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\left[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right]

Result\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\left[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right]=\dfrac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}

To divide, divide the moduli and subtract the arguments.

Example 21

Find the product \dfrac23\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)\times6\left(\cos\dfrac{5\pi}6+i\sin\dfrac{5\pi}6\right) in rectangular form.

Use Property 4: multiply the moduli, add the arguments.
=\dfrac23\times6\left[\cos\left(\dfrac\pi3+\dfrac{5\pi}6\right)+i\sin\left(\dfrac\pi3+\dfrac{5\pi}6\right)\right]=4\left[\cos\dfrac{7\pi}6+i\sin\dfrac{7\pi}6\right]
Write \dfrac{7\pi}6=\pi+\dfrac\pi6. Rotating by \pi changes the sign of both cosine and sine.
=4\left[\cos\left(\pi+\dfrac\pi6\right)+i\sin\left(\pi+\dfrac\pi6\right)\right]=-4\cos\dfrac\pi6-4i\sin\dfrac\pi6
Use \cos\dfrac\pi6=\dfrac{\sqrt3}2 and \sin\dfrac\pi6=\dfrac12.
=-4\left(\dfrac{\sqrt3}{2}\right)-4i\left(\dfrac12\right)=-2\sqrt3-2i

-2\sqrt3-2i

Example 22

Find the quotient \dfrac{2\left(\cos\frac{9\pi}4+i\sin\frac{9\pi}4\right)}{4\left[\cos\left(-\frac{3\pi}4\right)+i\sin\left(-\frac{3\pi}4\right)\right]} in rectangular form.

Use Property 5: divide the moduli, subtract the arguments.
=\dfrac24\left[\cos\left(\dfrac{9\pi}4-\left(-\dfrac{3\pi}4\right)\right)+i\sin\left(\dfrac{9\pi}4-\left(-\dfrac{3\pi}4\right)\right)\right]
Add the angles: \dfrac{9\pi}4+\dfrac{3\pi}4=\dfrac{12\pi}4=3\pi.
=\dfrac12\left(\cos3\pi+i\sin3\pi\right)
\cos3\pi=-1 and \sin3\pi=0.
=\dfrac12(-1+0)=-\dfrac12

-\dfrac12

Example 23

If z=x+iy and \arg\left(\dfrac{z-1}{z+1}\right)=\dfrac\pi2, show that x^2+y^2=1.

Put z=x+iy and rationalise with the conjugate of (x+1)+iy.
\dfrac{z-1}{z+1}=\dfrac{(x-1)+iy}{(x+1)+iy}\times\dfrac{(x+1)-iy}{(x+1)-iy}
Expand the numerator: (x-1)(x+1)-iy(x-1)+iy(x+1)+y^2=(x^2-1+y^2)+2iy. The denominator is (x+1)^2+y^2.
\dfrac{z-1}{z+1}=\dfrac{(x^2+y^2-1)+i(2y)}{(x+1)^2+y^2}
An argument of \dfrac\pi2 means the number lies on the positive imaginary axis: its real part is zero (and the imaginary part is positive). Since \tan\dfrac\pi2 is undefined, the real part divided by the denominator must be zero.
\dfrac{x^2+y^2-1}{(x+1)^2+y^2}=0
A fraction is zero when its numerator is zero.
x^2+y^2-1=0\ \Rightarrow\ x^2+y^2=1

The condition describes the unit circle. The imaginary part 2y must be positive for the argument to be +\pi/2, so the points of the number lie on the upper half (y>0), drawn solid.

x^2+y^2=1

Example 24

Find the equation in Cartesian form if z=x+iy and \arg(z-2)-\arg(z+2)=\dfrac\pi4.

Put z=x+iy and group the real and imaginary parts.
\arg\left((x-2)+iy\right)-\arg\left((x+2)+iy\right)=\dfrac\pi4
Write each argument as an inverse tangent (the reference angle, taking the argument in the first quadrant: \tan\arg=\dfrac{\text{imaginary}}{\text{real}}).
\tan^{-1}\dfrac{y}{x-2}-\tan^{-1}\dfrac{y}{x+2}=\dfrac\pi4
Take the tangent of both sides.
\tan\left(\tan^{-1}\dfrac{y}{x-2}-\tan^{-1}\dfrac{y}{x+2}\right)=\tan\dfrac\pi4=1
Use \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}.
\dfrac{\dfrac{y}{x-2}-\dfrac{y}{x+2}}{1+\left(\dfrac{y}{x-2}\right)\left(\dfrac{y}{x+2}\right)}=1
Multiply the top and bottom by (x-2)(x+2) to clear the small fractions.
\dfrac{y(x+2)-y(x-2)}{(x-2)(x+2)+y^2}=1
Expand: xy+2y-xy+2y=4y and (x-2)(x+2)=x^2-4.
\dfrac{4y}{x^2-4+y^2}=1\ \Rightarrow\ 4y=x^2+y^2-4
Bring everything to one side.
x^2+y^2-4y-4=0

The locus is a circle with centre (0,2) and radius 2\sqrt2, passing through 2 and -2. Only the arc above the x-axis (solid) gives the angle +\pi/4; the dashed arc would give a different angle.

x^2+y^2-4y-4=0

Worked example: product and quotient in polar form, bringing the argument back

Let z_1=2\left(\cos\dfrac{3\pi}4+i\sin\dfrac{3\pi}4\right) and z_2=5\left(\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3\right). Find z_1z_2 and \dfrac{z_1}{z_2} in polar form with the principal argument.

Product: multiply the moduli and add the arguments (Property 4).
|z_1z_2|=2\times5=10,\qquad \theta_1+\theta_2=\dfrac{3\pi}4+\dfrac{2\pi}3=\dfrac{9\pi+8\pi}{12}=\dfrac{17\pi}{12}
\dfrac{17\pi}{12}>\pi, so it is not the principal argument. Subtract 2\pi; sine and cosine do not change.
\dfrac{17\pi}{12}-2\pi=\dfrac{17\pi-24\pi}{12}=-\dfrac{7\pi}{12}
Write the product.
z_1z_2=10\left[\cos\left(-\dfrac{7\pi}{12}\right)+i\sin\left(-\dfrac{7\pi}{12}\right)\right]
Quotient: divide the moduli and subtract the arguments (Property 5).
\left|\dfrac{z_1}{z_2}\right|=\dfrac25,\qquad \theta_1-\theta_2=\dfrac{9\pi-8\pi}{12}=\dfrac\pi{12}
\dfrac\pi{12} is already in (-\pi,\pi], so no adjustment is needed.
\dfrac{z_1}{z_2}=\dfrac25\left(\cos\dfrac\pi{12}+i\sin\dfrac\pi{12}\right)

z_1z_2=10\left[\cos\left(-\tfrac{7\pi}{12}\right)+i\sin\left(-\tfrac{7\pi}{12}\right)\right],\quad \dfrac{z_1}{z_2}=\dfrac25\left(\cos\tfrac\pi{12}+i\sin\tfrac\pi{12}\right)

1.10Application of complex numbers in real world

Complex numbers are used in many real-life situations such as cryptography, wave phenomena, pressure and velocity of a fluid, and the calculation of voltage and current in circuits. These applications are of higher level and are discussed in higher classes. Here complex numbers are used through simple examples from simple harmonic motion, electrical engineering and waves.

Simple harmonic motion

In simple harmonic motion we determine the position of a microscopic particle from its mean position. The equation which gives the position of the particle from the mean position is

x=x_{\max}\,e^{i\theta}

(1)

where x is the displacement from the mean position, x_{\max} is the amplitude and e^{i\theta}=\cos\theta+i\sin\theta is a complex number (Euler's equation).

Example 25

A micro particle is performing to and fro motion. Find its position at an angle of \dfrac\pi2, when its amplitude is 0.05\,\mathrm{mm}.

Write down the data.
x_{\max}=0.05,\qquad \theta=\dfrac\pi2
Use x=x_{\max}e^{i\theta} and write e^{i\theta} as \cos\theta+i\sin\theta.
x=0.05\,e^{i\pi/2}=0.05\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)
\cos\dfrac\pi2=0 and \sin\dfrac\pi2=1.
x=0.05(0+i)=0.05\,i\ \text{mm}

The position 0.05i is purely imaginary. As the book remarks, the particle is at a position where its real displacement is zero, a position we cannot read on the real line.

x=0.05\,i\ \mathrm{mm}

The formula can also be written x=x_{\max}e^{i\omega t}, where \omega is the angular velocity and t is the time. Also \omega=2\pi f, where f is the frequency of the particle.

Electrical engineering

The flow of electricity I in a circuit, the resistance to flow Z (called impedance) and the electromotive force E (called voltage) are related by

E=I\,Z

Electrical engineers use j for the imaginary unit; here it is written i for understanding.

Example 26

An electrical engineer is designing a circuit that is to have a current of (6-8i) amps. If the impedance is (14+8i), find the voltage.

Write down the data.
I=6-8i,\qquad Z=14+8i
Use E=IZ and multiply the brackets term by term.
E=(6-8i)(14+8i)=84+48i-112i-64i^2
Replace i^2 by -1, so -64i^2=+64, and collect the parts.
E=84+64+(48-112)i=148-64i

E=148-64i\ \text{volts}

Complex numbers in the wave phenomenon

Example 27

Compute the potential difference across two AC power supplies with respect to time, when V_A=3\cos(t+30^\circ) and V_B=4\sin(t+120^\circ).

To find the total potential difference, simply adding V_A and V_B is not correct, so the following method is used. Each voltage is written as the real part of a complex number, using e^{i\phi}=\cos\phi+i\sin\phi.

Let Z_A=3\{\cos(t+30^\circ)+i\sin(t+30^\circ)\} and Z_B=4\{\cos(t+120^\circ)+i\sin(t+120^\circ)\}. Then V_A=\operatorname{Re}(Z_A) and, in the book's method, V_B=\operatorname{Re}(Z_B). In exponential form:
Z_A=3e^{i(t+30^\circ)},\qquad Z_B=4e^{i(t+120^\circ)}
Add the complex numbers and take out the common factor e^{it}.
Z_{\text{total}}=Z_A+Z_B=e^{it}\left(3e^{30^\circ i}+4e^{120^\circ i}\right)=e^{it}\left\{3(\cos30^\circ+i\sin30^\circ)+4(\cos120^\circ+i\sin120^\circ)\right\}
Put in the values \cos30^\circ=\dfrac{\sqrt3}2, \sin30^\circ=\dfrac12, \cos120^\circ=-\dfrac12, \sin120^\circ=\dfrac{\sqrt3}2.
Z_{\text{total}}=e^{it}\left\{3\left(\dfrac{\sqrt3}2+\dfrac i2\right)+4\left(-\dfrac12+i\dfrac{\sqrt3}2\right)\right\}=e^{it}\left\{\left(\dfrac{3\sqrt3}2-2\right)+i\left(\dfrac32+2\sqrt3\right)\right\}
Evaluate numerically.
Z_{\text{total}}=e^{it}\{0.5980762+4.964102\,i\}
The modulus is r and the argument is \phi with \tan\phi=\dfrac{4.964102}{0.5980762}.
\begin{aligned}r&=\sqrt{(0.5980762)^2+(4.964102)^2}=\sqrt{25}=5\\ \phi&=\tan^{-1}\left(\dfrac{4.964102}{0.5980762}\right)=83.13^\circ\end{aligned}
So Z_{\text{total}}=5e^{i(t+\phi)} and the total potential difference is the real part.
V_{\text{total}}=\operatorname{Re}(Z_{\text{total}})=5\cos(t+83.13^\circ)

V_{\text{total}}=5\cos(t+83.13^\circ)\ \text{(book's method; see note)}

Complex numbers in a resistor in an AC circuit

Consider a simple AC circuit with a resistor, as shown in the figure. Let v(t) be an AC voltage source given by v(t)=V_0\cos(\omega t)=\operatorname{Re}(V_0e^{i\omega t}), where V_0 and \omega are real quantities.

A simple AC circuit: a voltage source v(t) drives a current i through a resistor R, across which the voltage is v(t)_R.

Derivation Ohm's law with complex quantities

Aim. Relate the current i, the complex voltage V_R and the resistance R of the AC circuit.

The relation between the current i through the resistor and the voltage v(t)_R across it is Ohm's law.
v(t)_R=Ri
In the single loop the source voltage equals the voltage across the resistor, and the source is V_0\cos\omega t.
v(t)=v(t)_R=V_0\cos(\omega t)=\operatorname{Re}(V_0e^{i\omega t})
Combine the two relations.
Ri=\operatorname{Re}(V_0e^{i\omega t})\qquad(i)
Let V_R=V_0e^{i\omega t} and rewrite (i).
Ri=\operatorname{Re}(V_R)
R is a real quantity, so it can be taken inside the real part and the current is written as
i=\operatorname{Re}\left(\dfrac{V_R}{R}\right)
Let Z_R be the impedance of the resistor, defined by Z_R=R. Since R is real, the impedance of a resistor is a real number.
i=\operatorname{Re}\left(\dfrac{V_R}{Z_R}\right)
Let the complex current be I=\dfrac{V_R}{Z_R}.
I=\dfrac{V_R}{Z_R}

ResultI=\dfrac{V_R}{Z_R},\qquad i=\operatorname{Re}(I)

This is similar to Ohm's law in DC circuits. The relation between the complex quantities I, V_R and R makes calculations much easier: do all the calculations with the complex impedance, voltage and current, and then take the real part as the final answer.

Formula sheet

Imaginary unit

i^2=-1,\qquad i=\sqrt{-1}

i: the imaginary unit

Standard form

z=x+iy=(x,y)

x=\operatorname{Re}(z) (real part), y=\operatorname{Im}(z) (imaginary part)

Equality of complex numbers

x_1+iy_1=x_2+iy_2\iff x_1=x_2\ \text{and}\ y_1=y_2

Real parts equal and imaginary parts equal

Sum and difference

z_1\pm z_2=(x_1\pm x_2)+i(y_1\pm y_2)

z_1=x_1+iy_1, z_2=x_2+iy_2

Product

z_1z_2=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)

Expand and use i^2=-1

Quotient

\dfrac{z_1}{z_2}=\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2}+i\,\dfrac{x_2y_1-x_1y_2}{x_2^2+y_2^2}

Multiply top and bottom by the conjugate x_2-iy_2 of the denominator

Conjugate

\bar z=x-iy

The mirror image of z in the x-axis

Conjugate of product and quotient

\overline{z_1z_2}=\bar z_1\bar z_2,\qquad \overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar z_1}{\bar z_2}

Key facts

Modulus

|z|=\sqrt{x^2+y^2}

Distance of z from the origin; |z|\ge0 and |z|=0\iff z=0

Modulus of product and quotient

|z_1z_2|=|z_1||z_2|,\qquad \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}

Key facts

Parts of z^{-1}

\operatorname{Re}(z^{-1})=\dfrac{\operatorname{Re}z}{|z|^2},\quad \operatorname{Im}(z^{-1})=-\dfrac{\operatorname{Im}z}{|z|^2}

Parts of z^2

\operatorname{Re}(z^2)=(\operatorname{Re}z)^2-(\operatorname{Im}z)^2,\quad \operatorname{Im}(z^2)=2\operatorname{Re}z\operatorname{Im}z

Parts of z_1/z_2

\operatorname{Re}\dfrac{z_1}{z_2}=\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2},\quad \operatorname{Im}\dfrac{z_1}{z_2}=\dfrac{x_2y_1-x_1y_2}{x_2^2+y_2^2}

Factorisation of z^2+a^2

z^2+a^2=(z+ia)(z-ia)

a real

Factor theorem

z-\alpha\ \text{is a factor of}\ P(z)\iff P(\alpha)=0

\alpha is a root (zero) of P(z)

Polar form

z=r(\cos\theta+i\sin\theta)=re^{i\theta},\quad r=\sqrt{x^2+y^2}

x=r\cos\theta, y=r\sin\theta, \theta=\arg z

Principal argument

-\pi<\operatorname{Arg}z\le\pi;\quad \theta=\alpha,\ \pi-\alpha,\ \alpha-\pi,\ -\alpha

\alpha=\tan^{-1}|y/x|; quadrants I, II, III, IV

Unit: radians

Reciprocal in polar form

z^{-1}=\dfrac1r(\cos\theta-i\sin\theta)

Product in polar form

z_1z_2=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]

Multiply moduli, add arguments

Quotient in polar form

\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\left[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right]

Divide moduli, subtract arguments

Euler's formula

e^{i\theta}=\cos\theta+i\sin\theta

Simple harmonic motion

x=x_{\max}e^{i\theta}=x_{\max}e^{i\omega t},\quad \omega=2\pi f

x_{\max}: amplitude, \omega: angular velocity, f: frequency

Voltage, current, impedance

E=IZ,\qquad I=\dfrac{V_R}{Z_R}

E: voltage, I: current, Z: impedance

Key terms

Complex number: An expression x+iy with x,y\in\mathbb{R} and i^2=-1; also written as the ordered pair (x,y).
Imaginary unit: The number i with i^2=-1.
Real part and imaginary part: For z=x+iy, \operatorname{Re}(z)=x and \operatorname{Im}(z)=y.
Pure imaginary number: A complex number whose real part is zero.
Conjugate: \bar z=x-iy for z=x+iy; the mirror image of z in the x-axis.
Modulus (magnitude): |z|=\sqrt{x^2+y^2}, the distance of z from the origin.
Rationalisation: Multiplying the numerator and denominator by the conjugate of the denominator so that the denominator becomes a real number.
Consistent system: A system of equations that has at least one solution.
Inconsistent system: A system of equations that has no solution.
Complex polynomial: P(z)=a_0+a_1z+\cdots+a_nz^n with constant (real or complex) coefficients and a_n\neq0.
Synthetic division: A short method of dividing a polynomial by z-\alpha using only its coefficients.
Polar coordinates: The pair (r,\theta) giving the distance from the pole and the angle from the initial line, measured anticlockwise.
Argument: The angle \theta of z, written \arg(z); it is defined up to adding 2k\pi.
Principal argument: The value of \arg(z) in the interval -\pi<\theta\le\pi, written \operatorname{Arg}(z).
Polar form: z=r(\cos\theta+i\sin\theta) with r=|z| and \theta=\arg z.
Euler's formula: e^{i\theta}=\cos\theta+i\sin\theta (also written \operatorname{cis}\theta).
Impedance: The resistance to the flow of current in a circuit, Z, with E=IZ.

Exercise 1.1p. 10–11

Powers of i, the four operations on complex numbers, equality of complex numbers, conjugates and moduli. Divide by multiplying the top and bottom by the conjugate of the denominator, and use i^2=-1 everywhere.

Evaluate the following:

Solution

(i) i^{31}

Powers of i repeat in a cycle of 4: i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1. Divide the index by 4 and keep the remainder.
31=4(7)+3
Write the power as (i^4)^7\cdot i^3 and use i^4=1.
i^{31}=(i^4)^7\,i^3=(1)^7(-i)=-i

i^{31}=-i

(ii) (-i)^6

The index is even, so the minus sign disappears: (-i)^6=(-1)^6i^6=i^6.
(-i)^6=i^6
Write 6=4+2 and use i^4=1, i^2=-1.
i^6=i^4\cdot i^2=(1)(-1)=-1

(-i)^6=-1

(iii) (-1)^{-13/2}

Write -1=i^2 so the fractional power can be handled with powers of i.
(-1)^{-\frac{13}{2}}=(i^2)^{-\frac{13}{2}}=i^{-13}
A negative index means a reciprocal. Reduce i^{13} using i^4=1: 13=4(3)+1.
i^{-13}=\dfrac{1}{i^{13}}=\dfrac{1}{(i^4)^3\,i}=\dfrac1i
Rationalise by multiplying the top and bottom by i (since i\cdot i=-1).
\dfrac1i=\dfrac{i}{i^2}=\dfrac{i}{-1}=-i

(-1)^{-13/2}=-i

(iv) \dfrac{2}{(-1)^{3/2}}

Write -1=i^2, so (-1)^{3/2}=(i^2)^{3/2}=i^3.
(-1)^{\frac32}=i^3=-i
Substitute and rationalise: multiply the top and bottom by i.
\dfrac{2}{-i}=\dfrac{2\cdot i}{-i\cdot i}=\dfrac{2i}{-i^2}=\dfrac{2i}{1}=2i

\dfrac{2}{(-1)^{3/2}}=2i

(v) i^{23}+i^{58}+i^{21}

Reduce each power with i^4=1.
i^{23}=i^{20}\,i^3=-i,\qquad i^{58}=i^{56}\,i^2=-1,\qquad i^{21}=i^{20}\,i=i
Add the three values; -i and +i cancel.
i^{23}+i^{58}+i^{21}=-i-1+i=-1

i^{23}+i^{58}+i^{21}=-1

Write the following complex numbers in the form x+iy.

Solution

(i) (3+2i)+(2+4i)

Add the real parts and the imaginary parts separately (section 1.2.1).
(3+2)+(2+4)i=5+6i

5+6i

(ii) (4+3i)-(2+5i)

Subtract the real parts and the imaginary parts separately (section 1.2.2).
(4-2)+(3-5)i=2-2i

2-2i

(iii) (4+7i)+(4-7i)

Add real and imaginary parts. The imaginary parts cancel.
(4+4)+(7-7)i=8+0i

8+0i

(iv) (2+5i)-(2-5i)

Subtract real and imaginary parts. The real parts cancel.
(2-2)+(5-(-5))i=0+10i

0+10i

(v) (3+2i)(4-3i)

Multiply term by term (section 1.2.3).
(3+2i)(4-3i)=12-9i+8i-6i^2
Use i^2=-1 and collect like terms.
=12-i+6=18-i

18-i

(vi) (3,2)\div(3,-1)

The ordered pairs stand for 3+2i and 3-i.
\dfrac{3+2i}{3-i}
Multiply the top and bottom by the conjugate of the denominator, 3+i (section 1.2.4).
\dfrac{(3+2i)(3+i)}{(3-i)(3+i)}=\dfrac{9+3i+6i+2i^2}{9-i^2}
Simplify using i^2=-1.
=\dfrac{9+9i-2}{9+1}=\dfrac{7+9i}{10}=\dfrac{7}{10}+\dfrac{9}{10}i

\dfrac{7}{10}+\dfrac{9}{10}i

(vii) (1+i)(1-i)(2+i)

First multiply the conjugate pair; (a+b)(a-b)=a^2-b^2.
(1+i)(1-i)=1-i^2=1+1=2
Now multiply by (2+i).
2(2+i)=4+2i

4+2i

(viii) \dfrac{1}{2+3i}

Multiply the top and bottom by the conjugate of the denominator, 2-3i.
\dfrac{1}{2+3i}\cdot\dfrac{2-3i}{2-3i}=\dfrac{2-3i}{4-9i^2}
Use i^2=-1, so the denominator is 4+9=13.
=\dfrac{2-3i}{13}=\dfrac{2}{13}-\dfrac{3}{13}i

\dfrac{2}{13}-\dfrac{3}{13}i

Simplify the following:

Solution

(i) \dfrac{(2+i)(3-2i)}{1+i}

Expand the numerator.
(2+i)(3-2i)=6-4i+3i-2i^2=6-i+2=8-i
Multiply the top and bottom by the conjugate 1-i.
\dfrac{8-i}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{8-8i-i+i^2}{1-i^2}=\dfrac{7-9i}{2}
Write in the form x+iy.
=\dfrac72-\dfrac92 i

\dfrac72-\dfrac92i

(ii) \dfrac{1+i}{(2+i)^2}

Square the denominator.
(2+i)^2=4+4i+i^2=3+4i
Multiply the top and bottom by the conjugate 3-4i.
\dfrac{1+i}{3+4i}\cdot\dfrac{3-4i}{3-4i}=\dfrac{3-4i+3i-4i^2}{9+16}
Simplify.
=\dfrac{7-i}{25}=\dfrac{7}{25}-\dfrac{1}{25}i

\dfrac{7}{25}-\dfrac{1}{25}i

(iii) \dfrac{1}{3+i}-\dfrac{1}{3-i}

Take the common denominator (3+i)(3-i)=9-i^2=10.
\dfrac{1}{3+i}-\dfrac{1}{3-i}=\dfrac{(3-i)-(3+i)}{(3+i)(3-i)}
Simplify the numerator and the denominator.
=\dfrac{-2i}{10}=-\dfrac{1}{5}i

-\dfrac15 i

(iv) (1+i)^{-2}+(1-i)^{-2}

Square the brackets first: (1+i)^2=1+2i+i^2=2i and (1-i)^2=1-2i+i^2=-2i.
(1+i)^{-2}=\dfrac{1}{2i},\qquad(1-i)^{-2}=\dfrac{1}{-2i}
The two terms are negatives of each other, so they cancel.
\dfrac{1}{2i}-\dfrac{1}{2i}=0

(v) (2+i)^2+\dfrac{7-4i}{2+i}

Expand the square.
(2+i)^2=4+4i+i^2=3+4i
Rationalise the fraction with the conjugate 2-i.
\dfrac{7-4i}{2+i}\cdot\dfrac{2-i}{2-i}=\dfrac{14-7i-8i+4i^2}{4+1}=\dfrac{10-15i}{5}=2-3i
Add.
(3+4i)+(2-3i)=5+i

5+i

Find the values of the real numbers x and y in each of the following:

Solution

Method. Reduce each equation to the form A+iB=0 (or A+iB=C+iD) with A,B,C,D real in terms of x and y, then equate the real parts and the imaginary parts (section 1.1.2: two complex numbers are equal if and only if their real parts and imaginary parts are equal). This gives two linear equations in x and y.

(i) (2+3i)x+(1+3i)y+2=0

Separate the real and the imaginary parts (note x,y are real).
(2x+y+2)+i(3x+3y)=0+0i
Equate the real parts and the imaginary parts.
2x+y+2=0\quad(1),\qquad 3x+3y=0\quad(2)
From (2), y=-x. Substitute into (1).
2x-x+2=0\;\Rightarrow\;x=-2,\qquad y=2

x=-2,\ y=2

(ii) \dfrac{x}{1+i}+\dfrac{y}{1-2i}=1

Rationalise each fraction with the conjugate of its denominator.
\dfrac{x(1-i)}{2}+\dfrac{y(1+2i)}{5}=1
Separate the real and imaginary parts.
\left(\dfrac x2+\dfrac y5\right)+i\left(-\dfrac x2+\dfrac{2y}{5}\right)=1+0i
Equate the parts.
\dfrac x2+\dfrac y5=1\quad(1),\qquad -\dfrac x2+\dfrac{2y}{5}=0\quad(2)
Add (1) and (2) to eliminate x.
\dfrac{3y}{5}=1\;\Rightarrow\;y=\dfrac53
Put y=\frac53 in (2): \frac x2=\frac25\cdot\frac53=\frac23.
x=\dfrac43

x=\dfrac43,\ y=\dfrac53

(iii) \dfrac{x}{2+i}=\dfrac{1-5i}{3-2i}+\dfrac{y}{2-i}

Rationalise each fraction.
\dfrac{x(2-i)}{5}=\dfrac{(1-5i)(3+2i)}{13}+\dfrac{y(2+i)}{5}
Expand (1-5i)(3+2i)=3+2i-15i-10i^2=13-13i, so that fraction is 1-i.
\dfrac{2x}{5}-\dfrac{x}{5}i=1-i+\dfrac{2y}{5}+\dfrac{y}{5}i
Equate the real parts and the imaginary parts.
\dfrac{2x}{5}=1+\dfrac{2y}{5}\;\Rightarrow\;2x-2y=5\quad(1)
Similarly for the imaginary parts.
-\dfrac{x}{5}=-1+\dfrac{y}{5}\;\Rightarrow\;x+y=5\quad(2)
From (1), x-y=\frac52. Add this to (2): 2x=\frac{15}{2}.
x=\dfrac{15}{4},\qquad y=5-\dfrac{15}{4}=\dfrac54

x=\dfrac{15}{4},\ y=\dfrac54

(iv) x(1+i)^2+y(2-i)^2=3+10i

Expand the squares.
(1+i)^2=1+2i+i^2=2i,\qquad(2-i)^2=4-4i+i^2=3-4i
Substitute and separate the parts.
2ix+y(3-4i)=3y+i(2x-4y)=3+10i
Equate the real parts and the imaginary parts.
3y=3\;\Rightarrow\;y=1,\qquad 2x-4y=10\;\Rightarrow\;2x=14\;\Rightarrow\;x=7

x=7,\ y=1

Find the complex number z if 4z-3\bar z=\dfrac{1-18i}{2-i}.

Solution

Simplify the right side by multiplying the top and bottom by the conjugate 2+i.
\dfrac{1-18i}{2-i}=\dfrac{(1-18i)(2+i)}{4+1}=\dfrac{2+i-36i-18i^2}{5}=\dfrac{20-35i}{5}=4-7i
Let z=x+iy, so \bar z=x-iy, and work out the left side.
4z-3\bar z=4x+4iy-3x+3iy=x+7iy
Equate the real parts and the imaginary parts.
x=4,\qquad 7y=-7\;\Rightarrow\;y=-1
Check by substituting z=4-i: 4z=16-4i and 3\bar z=12+3i, so 4z-3\bar z=4-7i, which equals the right side.
z=4-i

z=4-i

Book answer. 4-\dfrac57 i

Note. Our answer z=4-i differs from the key's 4-\frac57i. Substituting z=4-i gives 4z-3\bar z=4-7i, exactly the value of \frac{1-18i}{2-i}, so the answer 4-i is correct. The printed key value is a misprint.

Find the conjugate of the following complex numbers:

Solution

Method. The conjugate of x+iy is \bar z=x-iy (section 1.3): change the sign of the imaginary part and leave the real part unchanged.

(i) 4-3i

Change the sign of the imaginary part.
\overline{4-3i}=4+3i

4+3i

(ii) 3i+8

Write in the form x+iy first: 8+3i. Then change the sign of the imaginary part.
\overline{8+3i}=8-3i

8-3i

(iii) 2+\sqrt{\dfrac{-1}{5}}

Write the square root of a negative number with i.
\sqrt{\dfrac{-1}{5}}=\dfrac{\sqrt{-1}}{\sqrt5}=\dfrac{i}{\sqrt5},\quad\text{so } z=2+\dfrac{i}{\sqrt5}
Change the sign of the imaginary part.
\bar z=2-\dfrac{i}{\sqrt5}

2-\dfrac{i}{\sqrt5}

(iv) \dfrac{5i}{2}-\dfrac78

Write in the form x+iy.
z=-\dfrac78+\dfrac52 i
Change the sign of the imaginary part.
\bar z=-\dfrac78-\dfrac52 i

-\dfrac78-\dfrac52 i

Find the magnitude of the following complex numbers:

Solution

Method. The magnitude (modulus) of z=x+iy is |z|=\sqrt{x^2+y^2} (section 1.4). For a product or quotient we can use |z_1z_2|=|z_1||z_2| and \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}.

(i) 11+12i

Use |z|=\sqrt{x^2+y^2}.
|z|=\sqrt{11^2+12^2}=\sqrt{121+144}=\sqrt{265}

\sqrt{265}

(ii) (2+3i)-(2+6i)

Simplify first.
(2+3i)-(2+6i)=-3i
Its real part is 0 and its imaginary part is -3.
|z|=\sqrt{0^2+(-3)^2}=3

(iii) (2-i)(6+3i)

Use |z_1z_2|=|z_1||z_2|.
|2-i|=\sqrt{4+1}=\sqrt5,\qquad|6+3i|=\sqrt{36+9}=\sqrt{45}=3\sqrt5
Multiply the magnitudes.
|z|=\sqrt5\cdot3\sqrt5=15
Check by multiplying out: (2-i)(6+3i)=12+6i-6i-3i^2=15, and |15|=15.

(iv) \dfrac{3-2i}{2+i}

Use \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}.
|3-2i|=\sqrt{13},\qquad|2+i|=\sqrt5
Divide.
|z|=\dfrac{\sqrt{13}}{\sqrt5}=\sqrt{\dfrac{13}{5}}

\sqrt{\dfrac{13}{5}}

(v) (\sqrt3-\sqrt{-8})(\sqrt3+\sqrt{-8})

Write \sqrt{-8}=\sqrt8\,i=2\sqrt2\,i.
(\sqrt3-2\sqrt2\,i)(\sqrt3+2\sqrt2\,i)
This is (a-b)(a+b)=a^2-b^2 with b=2\sqrt2\,i and b^2=-8.
=3-(2\sqrt2\,i)^2=3+8=11
The result is a real number, so its magnitude is its absolute value.
|z|=|11|=11

Exercise 1.2p. 15–16

Properties of conjugates and moduli, proofs with z=x+iy, loci from equations and inequations, and real and imaginary parts of (x+iy)^n and \left(\dfrac{x_1+iy_1}{x_2+iy_2}\right)^n. Always put z=x+iy and compare real and imaginary parts.

Show that for any complex number.

Solution

(i) \mathrm{Re}(iz)=-\mathrm{Im}(z)

Let z=x+iy with x,y real, so \mathrm{Im}(z)=y.
Multiply by i and use i^2=-1.
iz=i(x+iy)=ix+i^2y=-y+ix
Read off the real part of iz, which is the term without i.
\mathrm{Re}(iz)=-y=-\mathrm{Im}(z)
Hence the result is proved.

(ii) \mathrm{Im}(iz)=\mathrm{Re}(z)

As in (i), iz=-y+ix for z=x+iy.
iz=-y+ix
The imaginary part of iz is the coefficient of i, which is x=\mathrm{Re}(z).
\mathrm{Im}(iz)=x=\mathrm{Re}(z)
Hence the result is proved.

Use the algebraic properties of complex numbers to prove that: (z_1z_2)(z_3z_4)=(z_1z_3)(z_2z_4)=z_3(z_1z_2)z_4.

Solution

Properties used. Multiplication of complex numbers is associative, (ab)c=a(bc), and commutative, ab=ba. Each step below names the property used.

Start with the left side and apply the associative law to regroup, taking z_1 out.
(z_1z_2)(z_3z_4)=z_1\big[z_2(z_3z_4)\big]
Regroup the inside with the associative law, z_2(z_3z_4)=(z_2z_3)z_4.
=z_1\big[(z_2z_3)z_4\big]
Commutative law: z_2z_3=z_3z_2.
=z_1\big[(z_3z_2)z_4\big]
Associative law again: (z_3z_2)z_4=z_3(z_2z_4).
=z_1\big[z_3(z_2z_4)\big]
Associative law once more: z_1[z_3(z_2z_4)]=(z_1z_3)(z_2z_4). This proves the first equality.
=(z_1z_3)(z_2z_4)
For the second equality start again from (z_1z_2)(z_3z_4). Associative law: (z_1z_2)(z_3z_4)=\big[(z_1z_2)z_3\big]z_4.
(z_1z_2)(z_3z_4)=\big[(z_1z_2)z_3\big]z_4
Commutative law applied to the product of the brackets (z_1z_2) and z_3: (z_1z_2)z_3=z_3(z_1z_2).
=\big[z_3(z_1z_2)\big]z_4=z_3(z_1z_2)z_4
Hence all three expressions are equal.
(z_1z_2)(z_3z_4)=(z_1z_3)(z_2z_4)=z_3(z_1z_2)z_4

Prove that for z\in\mathbb{C}.

Solution

(i) z is real iff z=\bar z

Let z=x+iy, so \bar z=x-iy. We prove both directions of iff.
**If z is real**, then y=0, so z=x and \bar z=x.
z=x=\bar z
Conversely, suppose z=\bar z. Then x+iy=x-iy.
x+iy=x-iy\;\Rightarrow\;2iy=0
Since 2i\ne0, we get y=0, so z=x is real.
y=0
Hence z is real if and only if z=\bar z.

(ii) \dfrac{z-\bar z}{z+\bar z}=i\left(\dfrac{\mathrm{Im}\,z}{\mathrm{Re}\,z}\right)

Let z=x+iy, so \bar z=x-iy. Find the numerator and the denominator.
z-\bar z=(x+iy)-(x-iy)=2iy,\qquad z+\bar z=(x+iy)+(x-iy)=2x
Divide (we need x=\mathrm{Re}\,z\ne0 so that the denominator is not zero) and cancel 2.
\dfrac{z-\bar z}{z+\bar z}=\dfrac{2iy}{2x}=i\,\dfrac{y}{x}
Since y=\mathrm{Im}\,z and x=\mathrm{Re}\,z, this is the required result.
=i\left(\dfrac{\mathrm{Im}\,z}{\mathrm{Re}\,z}\right)

(iii) z is either real or pure imaginary iff (\bar z)^2=z^2

Let z=x+iy. Real means y=0; pure imaginary means x=0. So real or pure imaginary means xy=0.
Compute the two squares.
z^2=x^2-y^2+2ixy,\qquad(\bar z)^2=(x-iy)^2=x^2-y^2-2ixy
Subtract.
z^2-(\bar z)^2=4ixy
If z is real or pure imaginary, then xy=0, so z^2-(\bar z)^2=0, i.e. (\bar z)^2=z^2.
Conversely, if (\bar z)^2=z^2 then 4ixy=0, so xy=0: either y=0 (z real) or x=0 (z pure imaginary).
4ixy=0\;\Rightarrow\;x=0\ \text{or}\ y=0
Hence the statement is proved in both directions.

If z_1=2-3i and |z_1z_2|=16 find |z_2|.

Solution

Find |z_1|.
|z_1|=\sqrt{2^2+(-3)^2}=\sqrt{13}
Use the product rule |z_1z_2|=|z_1||z_2| (Key Facts, section 1.4).
16=\sqrt{13}\,|z_2|
Divide by \sqrt{13}.
|z_2|=\dfrac{16}{\sqrt{13}}

|z_2|=\dfrac{16}{\sqrt{13}}

Book answer. |z_2|=\dfrac{16}{\sqrt{13}}

If z_1 and z_2 are any two complex numbers then prove that |z_1+z_2|^2-|z_1-z_2|^2=4\,\mathrm{Re}(z_1)\mathrm{Re}(z_2)+4\,\mathrm{Im}(z_1)\mathrm{Im}(z_2)

Solution

Let z_1=x_1+iy_1 and z_2=x_2+iy_2.
Find the sum and the difference and apply |z|^2=x^2+y^2.
|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2,\qquad|z_1-z_2|^2=(x_1-x_2)^2+(y_1-y_2)^2
Expand each square.
|z_1+z_2|^2=x_1^2+2x_1x_2+x_2^2+y_1^2+2y_1y_2+y_2^2
and for the difference,
|z_1-z_2|^2=x_1^2-2x_1x_2+x_2^2+y_1^2-2y_1y_2+y_2^2
Subtract. The squared terms cancel and the cross terms double.
|z_1+z_2|^2-|z_1-z_2|^2=4x_1x_2+4y_1y_2
Since x_1=\mathrm{Re}(z_1), x_2=\mathrm{Re}(z_2), y_1=\mathrm{Im}(z_1), y_2=\mathrm{Im}(z_2), this is the required result.
=4\,\mathrm{Re}(z_1)\mathrm{Re}(z_2)+4\,\mathrm{Im}(z_1)\mathrm{Im}(z_2)

Find the value of \lambda; if \left|\dfrac{z_1}{z_2}+\lambda\right|=\sqrt{\lambda+2}; where z_1=3+i and z_2=1+i.

Solution

Find \dfrac{z_1}{z_2} by multiplying the top and bottom by the conjugate 1-i.
\dfrac{z_1}{z_2}=\dfrac{(3+i)(1-i)}{(1+i)(1-i)}=\dfrac{3-3i+i-i^2}{2}=\dfrac{4-2i}{2}=2-i
Substitute and group the real and imaginary parts: \dfrac{z_1}{z_2}+\lambda=(2+\lambda)-i.
|(2+\lambda)-i|=\sqrt{\lambda+2}
Use |x+iy|=\sqrt{x^2+y^2}, then square both sides.
\sqrt{(2+\lambda)^2+1}=\sqrt{\lambda+2}\;\Rightarrow\;(\lambda+2)^2+1=\lambda+2
Expand and collect terms.
\lambda^2+4\lambda+4+1-\lambda-2=0\;\Rightarrow\;\lambda^2+3\lambda+3=0
Solve with the quadratic formula. The discriminant is 9-12=-3.
\lambda=\dfrac{-3\pm\sqrt{-3}}{2}=\dfrac{-3\pm i\sqrt3}{2}
Check: put u=\lambda+2=\dfrac{1\pm i\sqrt3}{2}. Then u^2-u+1=0, which is the equation (\lambda+2)^2+1=\lambda+2 rewritten, so both values satisfy it.

\lambda=\dfrac{-3\pm i\sqrt3}{2}

Book answer. \lambda=\dfrac{-3\pm i\sqrt3}{2}

Note. For a real \lambda the equation \lambda^2+3\lambda+3=0 has no solution (discriminant -3<0), so no real \lambda works. The key's value is the pair of complex roots of this quadratic, which is the answer obtained by squaring and solving as above. We follow the key.

Verify that \sqrt2\,|z|\ge|\mathrm{Re}(z)|+|\mathrm{Im}(z)|. Hint: (Start with (|x|-|y|)^2\ge0)

Solution

Let z=x+iy, so |z|=\sqrt{x^2+y^2}, |\mathrm{Re}(z)|=|x| and |\mathrm{Im}(z)|=|y|. Start with the hint: a square is never negative.
(|x|-|y|)^2\ge0
Expand and use |x|^2=x^2, |y|^2=y^2.
x^2+y^2-2|x||y|\ge0\;\Rightarrow\;x^2+y^2\ge2|x||y|
Add x^2+y^2 to both sides.
2(x^2+y^2)\ge x^2+y^2+2|x||y|=(|x|+|y|)^2
Take the (positive) square root of both sides. Both sides are non-negative, so the inequality is kept.
\sqrt2\sqrt{x^2+y^2}\ge|x|+|y|
Replace the symbols by |z|, \mathrm{Re}(z) and \mathrm{Im}(z).
\sqrt2\,|z|\ge|\mathrm{Re}(z)|+|\mathrm{Im}(z)|
Quick numerical check with z=3+4i: \sqrt2(5)=7.07\ge3+4=7.

Write the following equations and inequations in terms of x and y by taking z=x+iy.

Solution

Method. Replace z by x+iy (and \bar z by x-iy), separate the real and imaginary parts inside each modulus, use |a+ib|=\sqrt{a^2+b^2} and square to remove the root. The graphs show the locus of the points z.

(i) |2z-i|=4

Substitute z=x+iy and group the real and the imaginary parts.
2z-i=2x+i(2y-1)
Use the modulus formula and square both sides.
\sqrt{(2x)^2+(2y-1)^2}=4\;\Rightarrow\;4x^2+(2y-1)^2=16
Expand and collect.
4x^2+4y^2-4y+1=16\;\Rightarrow\;4x^2+4y^2-4y-15=0
Dividing by 4 gives a circle with centre (0,\tfrac12) and radius 2.
x^2+\left(y-\tfrac12\right)^2=4

The points with |2z-i|=4 lie on a circle of radius 2 about \left(0,\tfrac12\right).

4x^2+4y^2-4y-15=0

(ii) |z-1|=|\bar z+i|

Write both sides with z=x+iy, \bar z=x-iy.
z-1=(x-1)+iy,\qquad\bar z+i=x+i(1-y)
Equate the moduli and square both sides.
(x-1)^2+y^2=x^2+(1-y)^2
Expand. The terms x^2, y^2 and 1 cancel.
-2x+1=-2y+1\;\Rightarrow\;x-y=0

The equation |z-1|=|\bar z+i| is the straight line y=x.

x-y=0

(iii) |z-4i|+|z+4i|=10

Write each modulus with z=x+iy.
\sqrt{x^2+(y-4)^2}+\sqrt{x^2+(y+4)^2}=10
Isolate one root and square both sides.
x^2+(y-4)^2=100-20\sqrt{x^2+(y+4)^2}+x^2+(y+4)^2
Expand both brackets. The x^2 and y^2 terms cancel on the two sides.
-8y+16=100-20\sqrt{x^2+(y+4)^2}+8y+16\;\Rightarrow\;20\sqrt{x^2+(y+4)^2}=16y+100
Divide by 4 to get 5\sqrt{x^2+(y+4)^2}=4y+25 and square again.
25\left(x^2+y^2+8y+16\right)=(4y+25)^2=16y^2+200y+625
Collect terms. The 200y terms cancel.
25x^2+9y^2=225

The sum of the distances from (0,4) and (0,-4) is 10: an ellipse with semi-axes 3 and 5.

25x^2+9y^2=225

(iv) \dfrac12\mathrm{Re}(iz)=4

Compute iz.
iz=i(x+iy)=-y+ix
Its real part is -y.
\mathrm{Re}(iz)=-y
Substitute.
\dfrac12(-y)=4\;\Rightarrow\;y=-8

A horizontal line.

y=-8

(v) \mathrm{Im}\left(\dfrac{z-1}{2i}\right)=-5

Write z-1=(x-1)+iy and divide by 2i using \dfrac1i=-i.
\dfrac{(x-1)+iy}{2i}=\dfrac{-i\left[(x-1)+iy\right]}{2}=\dfrac{y}{2}-\dfrac{x-1}{2}\,i
The imaginary part is -\dfrac{x-1}{2}.
-\dfrac{x-1}{2}=-5
Solve.
x-1=10\;\Rightarrow\;x=11

A vertical line.

x=11

(vi) -2\le\mathrm{Im}(z+i)\le3

Compute z+i=x+i(y+1).
\mathrm{Im}(z+i)=y+1
Substitute and subtract 1 from each part of the inequality.
-2\le y+1\le3\;\Rightarrow\;-3\le y\le2

The shaded strip between the lines y=-3 and y=2 (both included).

-3\le y\le2

Book answer. (i) 4x^2+4y^2-4y-15=0; (ii) x-y=0; (iii) 25x^2+9y^2=225; (iv) y=-8; (v) x=11; (vi) -3\le y\le2

Find real and imaginary parts of the followings:

Solution

Method. Reduce each number to the form x+iy by rationalising (multiply the top and bottom by the conjugate of the denominator). Then \mathrm{Re}=x and \mathrm{Im}=y. The formulas of section 1.5 give the same results: \mathrm{Re}(z^{-1})=\dfrac{\mathrm{Re}\,z}{|z|^2} and \mathrm{Im}(z^{-1})=-\dfrac{\mathrm{Im}\,z}{|z|^2}.

(i) (2+4i)^{-1}

Write the inverse as a fraction and rationalise.
(2+4i)^{-1}=\dfrac{1}{2+4i}\cdot\dfrac{2-4i}{2-4i}=\dfrac{2-4i}{4+16}=\dfrac{2-4i}{20}
Separate.
=\dfrac1{10}-\dfrac15 i
Check with the formula: |z|^2=20, so \mathrm{Re}=\dfrac{2}{20}=\dfrac1{10} and \mathrm{Im}=-\dfrac{4}{20}=-\dfrac15.

\mathrm{Re}=\dfrac1{10},\ \mathrm{Im}=-\dfrac15

(ii) (3-\sqrt{-4})^{-2}

Write \sqrt{-4}=2i.
(3-2i)^{-2}=\dfrac{1}{(3-2i)^2}
Square the denominator.
(3-2i)^2=9-12i+4i^2=5-12i
Rationalise.
\dfrac{1}{5-12i}\cdot\dfrac{5+12i}{5+12i}=\dfrac{5+12i}{25+144}=\dfrac{5+12i}{169}
Separate.
=\dfrac{5}{169}+\dfrac{12}{169}i

\mathrm{Re}=\dfrac{5}{169},\ \mathrm{Im}=\dfrac{12}{169}

(iii) \left(\dfrac{7+2i}{3-i}\right)^{-1}

Invert the fraction: the inverse of \dfrac{a}{b} is \dfrac ba.
\left(\dfrac{7+2i}{3-i}\right)^{-1}=\dfrac{3-i}{7+2i}
Rationalise with 7-2i.
\dfrac{(3-i)(7-2i)}{(7+2i)(7-2i)}=\dfrac{21-6i-7i+2i^2}{49+4}
Simplify.
=\dfrac{19-13i}{53}=\dfrac{19}{53}-\dfrac{13}{53}i

\mathrm{Re}=\dfrac{19}{53},\ \mathrm{Im}=-\dfrac{13}{53}

(iv) \left(\dfrac{4+2i}{2+5i}\right)^{-2}

Invert the fraction.
\left(\dfrac{4+2i}{2+5i}\right)^{-2}=\left(\dfrac{2+5i}{4+2i}\right)^{2}
Rationalise the fraction inside with 4-2i.
\dfrac{(2+5i)(4-2i)}{16+4}=\dfrac{8-4i+20i-10i^2}{20}=\dfrac{18+16i}{20}=\dfrac{9+8i}{10}
Square.
\left(\dfrac{9+8i}{10}\right)^2=\dfrac{81+144i+64i^2}{100}=\dfrac{17+144i}{100}
Separate.
=\dfrac{17}{100}+\dfrac{36}{25}i

\mathrm{Re}=\dfrac{17}{100},\ \mathrm{Im}=\dfrac{36}{25}

(v) \left(\dfrac{5-4i}{5+4i}\right)^{2}

Simplify the fraction first: multiply the top and bottom by 5-4i.
\dfrac{5-4i}{5+4i}=\dfrac{(5-4i)^2}{25+16}=\dfrac{25-40i+16i^2}{41}=\dfrac{9-40i}{41}
Square.
\left(\dfrac{9-40i}{41}\right)^2=\dfrac{81-720i+1600i^2}{1681}
Simplify using i^2=-1.
=\dfrac{-1519-720i}{1681}=-\dfrac{1519}{1681}-\dfrac{720}{1681}i

\mathrm{Re}=-\dfrac{1519}{1681},\ \mathrm{Im}=-\dfrac{720}{1681}

(vi) \dfrac{3-7i}{2+5i}

Rationalise with 2-5i.
\dfrac{(3-7i)(2-5i)}{4+25}=\dfrac{6-15i-14i+35i^2}{29}
Simplify.
=\dfrac{-29-29i}{29}=-1-i

\mathrm{Re}=-1,\ \mathrm{Im}=-1

For z_1=-3+2i and z_2=1-3i verify the followings:

Solution

Method. Work out the left side and the right side separately with z_1=-3+2i, z_2=1-3i, and show that they agree. The conjugates are \bar z_1=-3-2i and \bar z_2=1+3i.

(i) |z_1|=|-z_1|=|\bar z_1|=|-\bar z_1|

Compute each modulus with |x+iy|=\sqrt{x^2+y^2}. Here -z_1=3-2i, \bar z_1=-3-2i, -\bar z_1=3+2i.
|z_1|=\sqrt{9+4}=\sqrt{13},\quad|-z_1|=\sqrt{9+4}=\sqrt{13}
and for the conjugates,
|\bar z_1|=\sqrt{9+4}=\sqrt{13},\quad|-\bar z_1|=\sqrt{9+4}=\sqrt{13}
All four are equal to \sqrt{13}. Verified.

(ii) \overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar z_1}{\bar z_2}

Left side. Find \dfrac{z_1}{z_2} by rationalising with 1+3i.
\dfrac{-3+2i}{1-3i}=\dfrac{(-3+2i)(1+3i)}{1+9}=\dfrac{-3-9i+2i+6i^2}{10}=\dfrac{-9-7i}{10}
Take its conjugate.
\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{-9+7i}{10}
Right side.
\dfrac{\bar z_1}{\bar z_2}=\dfrac{-3-2i}{1+3i}=\dfrac{(-3-2i)(1-3i)}{10}=\dfrac{-3+9i-2i+6i^2}{10}=\dfrac{-9+7i}{10}
Both sides are \dfrac{-9+7i}{10}. Verified.

(iii) \overline{z_1z_2}=\bar z_1\bar z_2

Left side. Multiply, then conjugate.
z_1z_2=(-3+2i)(1-3i)=-3+9i+2i-6i^2=3+11i,\quad\overline{z_1z_2}=3-11i
Right side.
\bar z_1\bar z_2=(-3-2i)(1+3i)=-3-9i-2i-6i^2=3-11i
Both sides are 3-11i. Verified.

(iv) \overline{z_1+z_2}=\bar z_1+\bar z_2

Left side.
z_1+z_2=(-3+1)+(2-3)i=-2-i,\quad\overline{z_1+z_2}=-2+i
Right side.
\bar z_1+\bar z_2=(-3-2i)+(1+3i)=-2+i
Both sides are -2+i. Verified.

(v) |z_1z_2|=|z_1||z_2|

Left side. Use z_1z_2=3+11i from (iii).
|z_1z_2|=\sqrt{9+121}=\sqrt{130}
Right side.
|z_1||z_2|=\sqrt{13}\cdot\sqrt{1+9}=\sqrt{13}\sqrt{10}=\sqrt{130}
Both sides are \sqrt{130}. Verified.

(vi) |z_1+z_2|\le|z_1|+|z_2|

Left side. Use z_1+z_2=-2-i from (iv).
|z_1+z_2|=\sqrt{4+1}=\sqrt5\approx2.24
Right side.
|z_1|+|z_2|=\sqrt{13}+\sqrt{10}\approx3.61+3.16=6.77
Since 2.24\le6.77 the inequality holds. Verified.

The vectors z_1, z_2 and z_1+z_2 form a triangle, and one side is shorter than the sum of the other two (triangle inequality).

Exercise 1.3p. 19–20

Factorising z^2+a^2 and cubics into linear factors, completing the square and the quadratic formula for complex roots, and solving two linear equations in z and \omega by elimination.

12026 Annual I, Set D, MCQ 1

Factorize the following polynomials into linear functions:

Solution

Method. For z^2+a^2 use z^2+a^2=z^2-(ia)^2=(z-ia)(z+ia) (section 1.7.1). For a cubic, find one root a by the factor theorem (z-a is a factor iff P(a)=0; try the factors of the constant term), divide by synthetic division, then factorise the quotient (section 1.7.2).

(i) z^2+169

Write 169=13^2=-(13i)^2 since i^2=-1, and use the difference of squares.
z^2+169=z^2-(13i)^2=(z-13i)(z+13i)

(z-13i)(z+13i)

(ii) 2z^2+18

Take out the common factor 2.
2z^2+18=2(z^2+9)
Write 9=3^2 and use z^2+a^2=(z-ia)(z+ia) with a=3.
=2(z-3i)(z+3i)

2(z-3i)(z+3i)

(iii) 3z^2+363

Take out the common factor 3.
3z^2+363=3(z^2+121)
Use z^2+a^2=(z-ia)(z+ia) with a=11.
=3(z-11i)(z+11i)

3(z-11i)(z+11i)

(iv) z^2+\dfrac{3}{25}

Here a^2=\dfrac{3}{25}, so a=\dfrac{\sqrt3}{5}.
z^2+\dfrac3{25}=z^2+\left(\dfrac{\sqrt3}{5}\right)^2
Use z^2+a^2=(z-ia)(z+ia).
=\left(z-\dfrac{\sqrt3}{5}i\right)\left(z+\dfrac{\sqrt3}{5}i\right)

\left(z-\tfrac{\sqrt3}{5}i\right)\left(z+\tfrac{\sqrt3}{5}i\right)

(v) 2z^3+3z^2-10z-15

Group the first two and the last two terms and take out the common factors.
2z^3+3z^2-10z-15=z^2(2z+3)-5(2z+3)
Take out the common bracket (2z+3).
=(2z+3)(z^2-5)
Factorise the difference of squares, z^2-5=z^2-(\sqrt5)^2.
=(2z+3)(z-\sqrt5)(z+\sqrt5)
Check with the factor theorem: P\left(-\tfrac32\right)=2\left(-\tfrac{27}{8}\right)+3\left(\tfrac94\right)+15-15=-\tfrac{27}{4}+\tfrac{27}{4}=0, so 2z+3 is a factor.

(2z+3)(z-\sqrt5)(z+\sqrt5)

(vi) z^3-7z+6

The constant term is 6, so try the factors \pm1,\pm2,\pm3,\pm6. Test z=1 (the coefficient of z^2 is 0).
P(1)=1-7+6=0
So z-1 is a factor. Divide by synthetic division with root 1 (coefficients 1,\ 0,\ -7,\ 6).
\begin{array}{r|rrrr}1&1&0&-7&6\\ &&1&1&-6\\ \hline &1&1&-6&0\end{array}
The quotient is z^2+z-6. Factorise it by splitting the middle term.
z^2+z-6=z^2+3z-2z-6=(z+3)(z-2)
Combine the factors.
z^3-7z+6=(z-1)(z-2)(z+3)

(z-1)(z-2)(z+3)

(vii) z^3+2z^2-23z-60

Try factors of 60. P(-3)=-27+18+69-60=0, so z+3 is a factor.
P(-3)=0
Synthetic division with root -3 (coefficients 1,\ 2,\ -23,\ -60).
\begin{array}{r|rrrr}-3&1&2&-23&-60\\ &&-3&3&60\\ \hline &1&-1&-20&0\end{array}
The quotient is z^2-z-20.
z^2-z-20=z^2-5z+4z-20=(z-5)(z+4)
Combine the factors.
z^3+2z^2-23z-60=(z+3)(z+4)(z-5)

(z+3)(z+4)(z-5)

(viii) 2z^3+9z^2-11z-30

Try z=2.
P(2)=16+36-22-30=0
So z-2 is a factor. Synthetic division with root 2 (coefficients 2,\ 9,\ -11,\ -30).
\begin{array}{r|rrrr}2&2&9&-11&-30\\ &&4&26&30\\ \hline &2&13&15&0\end{array}
The quotient is 2z^2+13z+15. Split the middle term: 13=10+3 and 10\times3=30=2\times15.
2z^2+13z+15=2z^2+10z+3z+15=2z(z+5)+3(z+5)=(2z+3)(z+5)
Combine the factors.
2z^3+9z^2-11z-30=(z-2)(z+5)(2z+3)
Check by expanding: (z+5)(2z+3)=2z^2+13z+15 and (z-2)(2z^2+13z+15)=2z^3+9z^2-11z-30.

(z-2)(z+5)(2z+3)

(ix) z^2-7z-8

Find two numbers with product -8 and sum -7: they are -8 and +1. Split the middle term.
z^2-7z-8=z^2-8z+z-8=z(z-8)+1(z-8)=(z-8)(z+1)

(z-8)(z+1)

(x) 4z^2-7z-11

Product of the end coefficients is 4\times(-11)=-44. Find two numbers with product -44 and sum -7: they are -11 and +4.
Split the middle term and factorise by grouping.
4z^2-7z-11=4z^2-11z+4z-11=z(4z-11)+1(4z-11)=(z+1)(4z-11)

(z+1)(4z-11)

Solve the following equations by completing square method

Solution

Method (section 1.7.3). Divide by the coefficient of z^2 so that it is 1. Move the constant to the right side. Add the square of half the coefficient of z to both sides, write the left side as a perfect square, and take the square root of both sides.

(i) z^2-6z+2=0

Move the constant to the right side.
z^2-6z=-2
Add \left(\tfrac{-6}{2}\right)^2=9 to both sides.
z^2-6z+9=-2+9\;\Rightarrow\;(z-3)^2=7
Take the square root of both sides.
z-3=\pm\sqrt7\;\Rightarrow\;z=3\pm\sqrt7

z=3\pm\sqrt7

(ii) -\dfrac12z^2-5z+2=0

Divide by -\tfrac12 (that is, multiply by -2).
z^2+10z-4=0
Move the constant to the right side.
z^2+10z=4
Add \left(\tfrac{10}{2}\right)^2=25 to both sides.
z^2+10z+25=4+25\;\Rightarrow\;(z+5)^2=29
Take the square root of both sides.
z+5=\pm\sqrt{29}\;\Rightarrow\;z=-5\pm\sqrt{29}

z=-5\pm\sqrt{29}

(iii) 4z^2+5z=14

Divide by 4.
z^2+\dfrac54z=\dfrac{14}{4}=\dfrac72
Add \left(\tfrac{5}{8}\right)^2=\tfrac{25}{64} to both sides (half of \tfrac54 is \tfrac58).
z^2+\dfrac54z+\dfrac{25}{64}=\dfrac72+\dfrac{25}{64}\;\Rightarrow\;\left(z+\dfrac58\right)^2=\dfrac{224+25}{64}=\dfrac{249}{64}
Take the square root of both sides.
z+\dfrac58=\pm\dfrac{\sqrt{249}}{8}\;\Rightarrow\;z=\dfrac{-5\pm\sqrt{249}}{8}

z=\dfrac{-5\pm\sqrt{249}}{8}

(iv) z^2=5z-3

Bring the z term to the left side.
z^2-5z=-3
Add \left(\tfrac52\right)^2=\tfrac{25}{4} to both sides.
z^2-5z+\dfrac{25}4=-3+\dfrac{25}{4}\;\Rightarrow\;\left(z-\dfrac52\right)^2=\dfrac{13}{4}
Take the square root of both sides.
z-\dfrac52=\pm\dfrac{\sqrt{13}}{2}\;\Rightarrow\;z=\dfrac{5\pm\sqrt{13}}{2}

z=\dfrac{5\pm\sqrt{13}}{2}

Solve the following quadratic equations:

Solution

Method. Clear any fraction, then use the quadratic formula z=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} (it comes from completing the square). When b^2-4ac<0 the roots are complex: \sqrt{-k}=i\sqrt k.

(i) \dfrac13z^2+2z-16=0

Multiply by 3 to clear the fraction.
z^2+6z-48=0\quad(a=1,\ b=6,\ c=-48)
Discriminant.
b^2-4ac=36+192=228=4\times57
Apply the formula and cancel 2.
z=\dfrac{-6\pm2\sqrt{57}}{2}=-3\pm\sqrt{57}

z=-3\pm\sqrt{57}

(ii) z^2-\dfrac12z+17=0

Multiply by 2 to clear the fraction.
2z^2-z+34=0\quad(a=2,\ b=-1,\ c=34)
Discriminant. It is negative, so the roots are complex.
b^2-4ac=1-272=-271
Apply the formula with \sqrt{-271}=i\sqrt{271}.
z=\dfrac{1\pm i\sqrt{271}}{4}

z=\dfrac{1\pm i\sqrt{271}}{4}

(iii) z^2-6z+25=0

Here a=1,\ b=-6,\ c=25. Discriminant.
b^2-4ac=36-100=-64
Apply the formula with \sqrt{-64}=8i.
z=\dfrac{6\pm8i}{2}=3\pm4i
Check: (3+4i)^2-6(3+4i)+25=(-7+24i)-18-24i+25=0.

z=3\pm4i

(iv) z^2-9z+11=0

Here a=1,\ b=-9,\ c=11. Discriminant.
b^2-4ac=81-44=37
Apply the formula.
z=\dfrac{9\pm\sqrt{37}}{2}

z=\dfrac{9\pm\sqrt{37}}{2}

Solve the simultaneous system of linear equations with complex coefficients:

Solution

Method (elimination, section 1.6.1). Multiply the equations by suitable numbers so that the coefficients of one unknown become the same, subtract to eliminate it, find the other unknown, and repeat for the second unknown (or substitute). Rationalise each quotient with the conjugate of its denominator. The solution is the ordered pair (z,\omega). Here the letter w stands for \omega.

(i) (1-i)z+(1+i)\omega=3;\ \ 2z-(2+5i)\omega=2+3i

The coefficients are already simple. Label the equations.
(1-i)z+(1+i)w=3\quad(1)\qquad\qquad 2z-(2+5i)w=2+3i\quad(2)
**Eliminate w.** Multiply (1) by the coefficient of w in (2) and (2) by the coefficient of w in (1), so that both equations have the same w-term.
\begin{aligned}(1)\times\left(-2 - 5i\right):&\ \left(-7 - 3i\right)\,z+\left(3 - 7i\right)\,w=\left(-6 - 15i\right)\\ (2)\times\left(1 + i\right):&\ \left(2 + 2i\right)\,z+\left(3 - 7i\right)\,w=\left(-1 + 5i\right)\end{aligned}
Subtract the second equation from the first. The w terms cancel.
\big[\left(-7 - 3i\right)-\left(2 + 2i\right)\big]z=\left(-6 - 15i\right)-\left(-1 + 5i\right)\;\Rightarrow\;\left(-9 - 5i\right)\,z=\left(-5 - 20i\right)
Divide by the coefficient of z and rationalise with its conjugate -9 + 5i.
z=\dfrac{-5 - 20i}{-9 - 5i}=\dfrac{\left(-5 - 20i\right)\left(-9 + 5i\right)}{106}=\dfrac{145 + 155i}{106}=\dfrac{145}{106} + \dfrac{155}{106}i
**Eliminate z.** Multiply (1) by the coefficient of z in (2) and (2) by the coefficient of z in (1).
\begin{aligned}(1)\times2:&\ \left(2 - 2i\right)\,z+\left(2 + 2i\right)\,w=6\\ (2)\times\left(1 - i\right):&\ \left(2 - 2i\right)\,z+\left(-7 - 3i\right)\,w=\left(5 + i\right)\end{aligned}
Subtract the first from the second. The z terms cancel.
\big[\left(-7 - 3i\right)-\left(2 + 2i\right)\big]w=\left(5 + i\right)-6\;\Rightarrow\;\left(-9 - 5i\right)\,w=\left(-1 + i\right)
Divide and rationalise again.
w=\dfrac{-1 + i}{-9 - 5i}=\dfrac{\left(-1 + i\right)\left(-9 + 5i\right)}{106}=\dfrac{4 - 14i}{106}=\dfrac{2}{53} - \dfrac{7}{53}i
Check in the first equation: \left(1 - i\right)\,z+\left(1 + i\right)\,w gives 3, the right side of (1), and in the second equation 2\,z+\left(-2 - 5i\right)\,w gives 2 + 3i, the right side of (2).

(z,w)=\left(\tfrac{5}{106}(29+31i),\ \tfrac{4-14i}{106}\right)

(ii) 2iz+(3-2i)\omega=1+i;\ \ (1-2i)z+(3+2i)\omega=5+6i

The coefficients are already simple. Label the equations.
2iz+(3-2i)w=1+i\quad(1)\qquad\qquad (1-2i)z+(3+2i)w=5+6i\quad(2)
**Eliminate w.** Multiply (1) by the coefficient of w in (2) and (2) by the coefficient of w in (1), so that both equations have the same w-term.
\begin{aligned}(1)\times\left(3 + 2i\right):&\ \left(-4 + 6i\right)\,z+13\,w=\left(1 + 5i\right)\\ (2)\times\left(3 - 2i\right):&\ \left(-1 - 8i\right)\,z+13\,w=\left(27 + 8i\right)\end{aligned}
Subtract the second equation from the first. The w terms cancel.
\big[\left(-4 + 6i\right)-\left(-1 - 8i\right)\big]z=\left(1 + 5i\right)-\left(27 + 8i\right)\;\Rightarrow\;\left(-3 + 14i\right)\,z=\left(-26 - 3i\right)
Divide by the coefficient of z and rationalise with its conjugate -3 - 14i.
z=\dfrac{-26 - 3i}{-3 + 14i}=\dfrac{\left(-26 - 3i\right)\left(-3 - 14i\right)}{205}=\dfrac{36 + 373i}{205}=\dfrac{36}{205} + \dfrac{373}{205}i
**Eliminate z.** Multiply (1) by the coefficient of z in (2) and (2) by the coefficient of z in (1).
\begin{aligned}(1)\times\left(1 - 2i\right):&\ \left(4 + 2i\right)\,z+\left(-1 - 8i\right)\,w=\left(3 - i\right)\\ (2)\times\left(2i\right):&\ \left(4 + 2i\right)\,z+\left(-4 + 6i\right)\,w=\left(-12 + 10i\right)\end{aligned}
Subtract the first from the second. The z terms cancel.
\big[\left(-4 + 6i\right)-\left(-1 - 8i\right)\big]w=\left(-12 + 10i\right)-\left(3 - i\right)\;\Rightarrow\;\left(-3 + 14i\right)\,w=\left(-15 + 11i\right)
Divide and rationalise again.
w=\dfrac{-15 + 11i}{-3 + 14i}=\dfrac{\left(-15 + 11i\right)\left(-3 - 14i\right)}{205}=\dfrac{199 + 177i}{205}=\dfrac{199}{205} + \dfrac{177}{205}i
Check in the first equation: \left(2i\right)\,z+\left(3 - 2i\right)\,w gives 1 + i, the right side of (1), and in the second equation \left(1 - 2i\right)\,z+\left(3 + 2i\right)\,w gives 5 + 6i, the right side of (2).

z=\tfrac{36+373i}{205},\ w=\tfrac{199+177i}{205}

(iii) \dfrac3iz-(6+2i)\omega=5;\ \ \dfrac i2z+\left(\dfrac34-\dfrac12i\right)\omega=\left(\dfrac12+2i\right)

Simplify using \dfrac3i=\dfrac{3(-i)}{i(-i)}=-3i and multiply the second equation by 4 to clear the fractions (4\cdot\tfrac i2=2i, 4\left(\tfrac34-\tfrac i2\right)=3-2i, 4\left(\tfrac12+2i\right)=2+8i).
-3iz-(6+2i)w=5\quad(1)\qquad\qquad 2iz+(3-2i)w=2+8i\quad(2)
**Eliminate w.** Multiply (1) by the coefficient of w in (2) and (2) by the coefficient of w in (1), so that both equations have the same w-term.
\begin{aligned}(1)\times\left(3 - 2i\right):&\ \left(-6 - 9i\right)\,z+\left(-22 + 6i\right)\,w=\left(15 - 10i\right)\\ (2)\times\left(-6 - 2i\right):&\ \left(4 - 12i\right)\,z+\left(-22 + 6i\right)\,w=\left(4 - 52i\right)\end{aligned}
Subtract the second equation from the first. The w terms cancel.
\big[\left(-6 - 9i\right)-\left(4 - 12i\right)\big]z=\left(15 - 10i\right)-\left(4 - 52i\right)\;\Rightarrow\;\left(-10 + 3i\right)\,z=\left(11 + 42i\right)
Divide by the coefficient of z and rationalise with its conjugate -10 - 3i.
z=\dfrac{11 + 42i}{-10 + 3i}=\dfrac{\left(11 + 42i\right)\left(-10 - 3i\right)}{109}=\dfrac{16 - 453i}{109}=\dfrac{16}{109} - \dfrac{453}{109}i
**Eliminate z.** Multiply (1) by the coefficient of z in (2) and (2) by the coefficient of z in (1).
\begin{aligned}(1)\times\left(2i\right):&\ 6\,z+\left(4 - 12i\right)\,w=\left(10i\right)\\ (2)\times\left(-3i\right):&\ 6\,z+\left(-6 - 9i\right)\,w=\left(24 - 6i\right)\end{aligned}
Subtract the first from the second. The z terms cancel.
\big[\left(-6 - 9i\right)-\left(4 - 12i\right)\big]w=\left(24 - 6i\right)-\left(10i\right)\;\Rightarrow\;\left(-10 + 3i\right)\,w=\left(24 - 16i\right)
Divide and rationalise again.
w=\dfrac{24 - 16i}{-10 + 3i}=\dfrac{\left(24 - 16i\right)\left(-10 - 3i\right)}{109}=\dfrac{-288 + 88i}{109}=-\dfrac{288}{109} + \dfrac{88}{109}i
Check in the first equation: \left(-3i\right)\,z+\left(-6 - 2i\right)\,w gives 5, the right side of (1), and in the second equation \left(2i\right)\,z+\left(3 - 2i\right)\,w gives 2 + 8i, the right side of (2).

z=\tfrac{16-453i}{109},\ w=\tfrac{-288+88i}{109}

(iv) \dfrac{1}{1-i}z+(1+i)\omega=3;\ \ \dfrac2iz-(2-3i)\omega=2+6i

Simplify using \dfrac1{1-i}=\dfrac{1+i}{2} and \dfrac2i=-2i. Multiply the first equation by 2 to clear the fraction.
(1+i)z+(2+2i)w=6\quad(1)\qquad\qquad -2iz+(-2+3i)w=2+6i\quad(2)
**Eliminate w.** Multiply (1) by the coefficient of w in (2) and (2) by the coefficient of w in (1), so that both equations have the same w-term.
\begin{aligned}(1)\times\left(-2 + 3i\right):&\ \left(-5 + i\right)\,z+\left(-10 + 2i\right)\,w=\left(-12 + 18i\right)\\ (2)\times\left(2 + 2i\right):&\ \left(4 - 4i\right)\,z+\left(-10 + 2i\right)\,w=\left(-8 + 16i\right)\end{aligned}
Subtract the second equation from the first. The w terms cancel.
\big[\left(-5 + i\right)-\left(4 - 4i\right)\big]z=\left(-12 + 18i\right)-\left(-8 + 16i\right)\;\Rightarrow\;\left(-9 + 5i\right)\,z=\left(-4 + 2i\right)
Divide by the coefficient of z and rationalise with its conjugate -9 - 5i.
z=\dfrac{-4 + 2i}{-9 + 5i}=\dfrac{\left(-4 + 2i\right)\left(-9 - 5i\right)}{106}=\dfrac{46 + 2i}{106}=\dfrac{23}{53} + \dfrac{1}{53}i
**Eliminate z.** Multiply (1) by the coefficient of z in (2) and (2) by the coefficient of z in (1).
\begin{aligned}(1)\times\left(-2i\right):&\ \left(2 - 2i\right)\,z+\left(4 - 4i\right)\,w=\left(-12i\right)\\ (2)\times\left(1 + i\right):&\ \left(2 - 2i\right)\,z+\left(-5 + i\right)\,w=\left(-4 + 8i\right)\end{aligned}
Subtract the first from the second. The z terms cancel.
\big[\left(-5 + i\right)-\left(4 - 4i\right)\big]w=\left(-4 + 8i\right)-\left(-12i\right)\;\Rightarrow\;\left(-9 + 5i\right)\,w=\left(-4 + 20i\right)
Divide and rationalise again.
w=\dfrac{-4 + 20i}{-9 + 5i}=\dfrac{\left(-4 + 20i\right)\left(-9 - 5i\right)}{106}=\dfrac{136 - 160i}{106}=\dfrac{68}{53} - \dfrac{80}{53}i
Check in the first equation: \left(1 + i\right)\,z+\left(2 + 2i\right)\,w gives 6, the right side of (1), and in the second equation \left(-2i\right)\,z+\left(-2 + 3i\right)\,w gives 2 + 6i, the right side of (2).

z=\tfrac{23+i}{53},\ w=\tfrac{68-80i}{53}

Exercise 1.4p. 28–29

Polar form r(\cos\theta+i\sin\theta) and the principal argument, products and quotients in polar form, loci from \arg and |z|, and the applications of complex numbers to oscillating particles, impedance and AC voltages (section 1.10).

Write following complex numbers in polar form.

Solution

Method (section 1.9). For z=x+iy find r=|z|=\sqrt{x^2+y^2}, the reference angle \alpha=\tan^{-1}\left|\dfrac yx\right|, and the principal argument \theta from the quadrant of z (first quadrant \theta=\alpha, second \pi-\alpha, third \alpha-\pi, fourth -\alpha). Then z=r(\cos\theta+i\sin\theta).

(i) 2+i2\sqrt3

Here x=2, y=2\sqrt3. Find the modulus.
r=\sqrt{2^2+(2\sqrt3)^2}=\sqrt{4+12}=4
Reference angle.
\alpha=\tan^{-1}\dfrac{2\sqrt3}{2}=\tan^{-1}\sqrt3=\dfrac\pi3
x>0,\ y>0: first quadrant, so \theta=\alpha=\dfrac\pi3.
2+i2\sqrt3=4\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)

z=2+2\sqrt3\,i in the first quadrant, r=4, \theta=\pi/3.

4\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)

(ii) 3-i\sqrt3

Here x=3, y=-\sqrt3. Find the modulus.
r=\sqrt{9+3}=\sqrt{12}=2\sqrt3
Reference angle (use |y/x|).
\alpha=\tan^{-1}\dfrac{\sqrt3}{3}=\tan^{-1}\dfrac1{\sqrt3}=\dfrac\pi6
x>0,\ y<0: fourth quadrant, so \theta=-\alpha=-\dfrac\pi6.
3-i\sqrt3=2\sqrt3\left[\cos\left(-\dfrac\pi6\right)+i\sin\left(-\dfrac\pi6\right)\right]
Since \cos(-\theta)=\cos\theta and \sin(-\theta)=-\sin\theta this is also
=2\sqrt3\left(\cos\dfrac\pi6-i\sin\dfrac\pi6\right)

z=3-\sqrt3\,i in the fourth quadrant, r=2\sqrt3, \theta=-\pi/6.

2\sqrt3\left(\cos\dfrac\pi6-i\sin\dfrac\pi6\right)

(iii) -2-i2

Here x=-2, y=-2. Find the modulus.
r=\sqrt{4+4}=2\sqrt2
Reference angle.
\alpha=\tan^{-1}\left|\dfrac{-2}{-2}\right|=\tan^{-1}1=\dfrac\pi4
x<0,\ y<0: third quadrant, so \theta=\alpha-\pi=-\dfrac{3\pi}4.
-2-2i=2\sqrt2\left[\cos\left(-\dfrac{3\pi}4\right)+i\sin\left(-\dfrac{3\pi}4\right)\right]
Using \cos(-\theta)=\cos\theta, \sin(-\theta)=-\sin\theta:
=2\sqrt2\left(\cos\dfrac{3\pi}4-i\sin\dfrac{3\pi}4\right)

z=-2-2i in the third quadrant, r=2\sqrt2, \theta=-3\pi/4.

2\sqrt2\left(\cos\dfrac{3\pi}4-i\sin\dfrac{3\pi}4\right)

Write the complex numbers in rectangular form

Solution

Method. Multiply and divide in polar form (section 1.9.3): r_1(\cos\theta_1+i\sin\theta_1)\cdot r_2(\cos\theta_2+i\sin\theta_2)=r_1r_2[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)] and the quotient has r_1/r_2 and \theta_1-\theta_2. Then evaluate the cosine and sine.

(i) \left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)

Both moduli are 1. Add the angles.
\dfrac\pi6+\dfrac\pi3=\dfrac{\pi+2\pi}{6}=\dfrac\pi2
Use the product rule.
=\cos\dfrac\pi2+i\sin\dfrac\pi2=0+i(1)=i

(ii) \dfrac{\cos\frac\pi6-i\sin\frac\pi6}{2\left(\cos\frac\pi3+i\sin\frac\pi3\right)}

Write the numerator in the form \cos\theta+i\sin\theta: \cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta) with \theta=\dfrac\pi6.
\cos\dfrac\pi6-i\sin\dfrac\pi6=\cos\left(-\dfrac\pi6\right)+i\sin\left(-\dfrac\pi6\right)
Use the quotient rule: divide the moduli (\tfrac12) and subtract the angles.
\dfrac12\left[\cos\left(-\dfrac\pi6-\dfrac\pi3\right)+i\sin\left(-\dfrac\pi6-\dfrac\pi3\right)\right]=\dfrac12\left[\cos\left(-\dfrac\pi2\right)+i\sin\left(-\dfrac\pi2\right)\right]
Evaluate: \cos\left(-\tfrac\pi2\right)=0 and \sin\left(-\tfrac\pi2\right)=-1.
=\dfrac12(0-i)=-\dfrac12i

-\dfrac12i

If (x_1+iy_1)(x_2+iy_2)(x_3+iy_3)\dots(x_n+iy_n)=a+ib, show that:

Solution

Setting up. Put z_r=x_r+iy_r for r=1,2,\dots,n, so that z_1z_2\cdots z_n=a+ib. We use two facts: the modulus of a product is the product of the moduli (|z_1z_2|=|z_1||z_2|, Key Facts section 1.4, repeated n-1 times), and in polar form the arguments add (\mathrm{Arg}(z_1z_2)=\mathrm{Arg}\,z_1+\mathrm{Arg}\,z_2, Key Facts section 1.9).

(i) (x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2)\dots(x_n^2+y_n^2)=a^2+b^2

Take the modulus of both sides of the given equation.
|z_1z_2\cdots z_n|=|a+ib|
The modulus of a product is the product of the moduli.
|z_1||z_2|\cdots|z_n|=\sqrt{a^2+b^2}
Write each modulus as |z_r|=\sqrt{x_r^2+y_r^2}.
\sqrt{x_1^2+y_1^2}\,\sqrt{x_2^2+y_2^2}\cdots\sqrt{x_n^2+y_n^2}=\sqrt{a^2+b^2}
Square both sides.
(x_1^2+y_1^2)(x_2^2+y_2^2)\cdots(x_n^2+y_n^2)=a^2+b^2

(ii) \displaystyle\sum_{r=1}^{n}\tan^{-1}\left(\dfrac{y_r}{x_r}\right)=\tan^{-1}\left(\dfrac ba\right)+2k\pi,\ k\in\mathbb{Z}

Write each factor in polar form. Here \theta_r=\tan^{-1}\dfrac{y_r}{x_r} is the argument of z_r and r_r=|z_r|.
z_r=r_r(\cos\theta_r+i\sin\theta_r)
Multiply. By the product rule (Property 4, section 1.9.3, repeated) the moduli multiply and the arguments add.
z_1z_2\cdots z_n=r_1r_2\cdots r_n\left[\cos(\theta_1+\dots+\theta_n)+i\sin(\theta_1+\dots+\theta_n)\right]
Let R=r_1r_2\cdots r_n and \Theta=\theta_1+\theta_2+\dots+\theta_n. Compare with a+ib (real and imaginary parts).
a=R\cos\Theta,\qquad b=R\sin\Theta
Divide the second equation by the first (the factor R cancels).
\dfrac ba=\tan\Theta
Take the inverse tangent. The argument of a complex number is fixed only up to a multiple of 2\pi, so we add 2k\pi.
\Theta=\tan^{-1}\dfrac ba+2k\pi,\quad k\in\mathbb{Z}
Replace \Theta by the sum.
\sum_{r=1}^{n}\tan^{-1}\left(\dfrac{y_r}{x_r}\right)=\tan^{-1}\left(\dfrac ba\right)+2k\pi

Write a given complex number in the algebraic form:

Solution

Method. Evaluate \cos\theta and \sin\theta using the reference angle and the quadrant, multiply by r and write the result as x+iy.

(i) \sqrt2(\cos315^\circ+i\sin315^\circ)

315^\circ is in the fourth quadrant, reference angle 45^\circ: cosine positive, sine negative.
\cos315^\circ=\dfrac{1}{\sqrt2},\quad\sin315^\circ=-\dfrac1{\sqrt2}
Multiply by \sqrt2.
\sqrt2\left(\dfrac1{\sqrt2}-\dfrac{i}{\sqrt2}\right)=1-i

1-i

(ii) 5(\cos210^\circ+i\sin210^\circ)

210^\circ is in the third quadrant, reference angle 30^\circ: both negative.
\cos210^\circ=-\dfrac{\sqrt3}{2},\quad\sin210^\circ=-\dfrac12
Multiply by 5.
5\left(-\dfrac{\sqrt3}{2}-\dfrac12i\right)=-\dfrac{5\sqrt3}{2}-\dfrac52i

-\dfrac{5\sqrt3}{2}-\dfrac52i

(iii) 2\left(\cos\dfrac{3\pi}2+i\sin\dfrac{3\pi}2\right)

At \dfrac{3\pi}{2} (that is 270^\circ) the cosine is 0 and the sine is -1.
2(0+i(-1))=-2i

-2i

(iv) 4\left(\cos\dfrac{5\pi}6+i\sin\dfrac{5\pi}6\right)

\dfrac{5\pi}6=150^\circ is in the second quadrant, reference angle \dfrac\pi6: cosine negative, sine positive.
\cos\dfrac{5\pi}{6}=-\dfrac{\sqrt3}2,\quad\sin\dfrac{5\pi}{6}=\dfrac12
Multiply by 4.
4\left(-\dfrac{\sqrt3}2+\dfrac12i\right)=-2\sqrt3+2i

-2\sqrt3+2i

(v) 2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)

First quadrant values.
\cos\dfrac\pi6=\dfrac{\sqrt3}2,\quad\sin\dfrac\pi6=\dfrac12
Multiply by 2.
2\left(\dfrac{\sqrt3}2+\dfrac12i\right)=\sqrt3+i

\sqrt3+i

(vi) \cos135^\circ+i\sin135^\circ

135^\circ is in the second quadrant, reference angle 45^\circ.
\cos135^\circ=-\dfrac1{\sqrt2},\quad\sin135^\circ=\dfrac1{\sqrt2}
Write the number.
=-\dfrac1{\sqrt2}+\dfrac1{\sqrt2}i

-\dfrac{1}{\sqrt2}+\dfrac1{\sqrt2}i

(vii) 10(\cos50^\circ+i\sin50^\circ)

50^\circ is not a standard angle, so use a calculator (degree mode).
\cos50^\circ\approx0.6428,\quad\sin50^\circ\approx0.7660
Multiply by 10.
10(0.6428+0.7660i)\approx6.43+7.66i

\approx6.43+7.66i

(viii) \sqrt2\left(\cos\dfrac{3\pi}4+i\sin\dfrac{3\pi}4\right)

\dfrac{3\pi}4=135^\circ: second quadrant.
\cos\dfrac{3\pi}4=-\dfrac1{\sqrt2},\quad\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}
Multiply by \sqrt2.
\sqrt2\left(-\dfrac1{\sqrt2}+\dfrac1{\sqrt2}i\right)=-1+i

-1+i

Convert the following equations and inequations in Cartesian form:

Solution

Method. Put z=x+iy. Write the complex number inside \arg in the form u+iv. Then \arg(u+iv)=\theta means \tan\theta=\dfrac vu (in the correct quadrant). For |z| use \sqrt{x^2+y^2}.

(i) \arg(z-1)=-\dfrac\pi4

Write z-1=(x-1)+iy.
The argument is the angle \theta with \tan\theta=\dfrac{y}{x-1}. Take \tan of both sides.
\tan\left[\arg(z-1)\right]=\dfrac{y}{x-1}=\tan\left(-\dfrac\pi4\right)=-1
Clear the fraction.
y=-(x-1)\;\Rightarrow\;x+y=1
An argument of -\dfrac\pi4 lies in the fourth quadrant, so the points have x>1 and y<0: only that part of the line is included.

The solid ray from (1,0) downwards at -45^\circ is the locus; the dashed line is x+y=1.

x+y=1\quad(x>1)

(ii) z\bar z=4\left|e^{i\theta}\right|

Use z\bar z=(x+iy)(x-iy)=x^2+y^2.
z\bar z=x^2+y^2
Use e^{i\theta}=\cos\theta+i\sin\theta, so its modulus is \sqrt{\cos^2\theta+\sin^2\theta}=1.
\left|e^{i\theta}\right|=1
Substitute.
x^2+y^2=4(1)=4

A circle with centre the origin and radius 2.

x^2+y^2=4

(iii) -\dfrac\pi3\le\arg(z-4)\le\dfrac\pi3

Write z-4=(x-4)+iy, so \tan\left[\arg(z-4)\right]=\dfrac{y}{x-4}.
Take the tangent of each part of the inequality. Tangent increases on \left(-\tfrac\pi2,\tfrac\pi2\right), so the order is kept.
\tan\left(-\dfrac\pi3\right)\le\dfrac{y}{x-4}\le\tan\dfrac\pi3
Evaluate the tangents.
-\sqrt3\le\dfrac{y}{x-4}\le\sqrt3
Since the argument is between -\dfrac\pi3 and \dfrac\pi3 (a right-hand angle) x-4>0, so this is the wedge |y|\le\sqrt3(x-4).

The shaded wedge with vertex (4,0) between the rays at \pm60^\circ.

-\sqrt3\le\dfrac{y}{x-4}\le\sqrt3

(iv) 0\le\arg\left(\dfrac{z-4}{1+i}\right)\le\dfrac\pi6

Rationalise the quotient with the conjugate 1-i of the denominator ((1+i)(1-i)=2).
\dfrac{z-4}{1+i}=\dfrac{[(x-4)+iy](1-i)}{2}=\dfrac{(x-4+y)+i(y-x+4)}{2}
Call the real part u=\dfrac{x+y-4}{2} and the imaginary part v=\dfrac{-x+y+4}{2}.
An argument between 0 and \dfrac\pi6 means the point is in the first quadrant and the angle is at most 30^\circ, so 0\le v and v\le u\tan\dfrac\pi6=\dfrac{u}{\sqrt3} (with u>0).
0\le v,\qquad \sqrt3\,v\le u
First inequality: v\ge0 means
-x+y+4\ge0
Second inequality: \sqrt3(-x+y+4)\le x+y-4. Bring everything to the right side.
0\le x+y-4+\sqrt3x-\sqrt3y-4\sqrt3=(1+\sqrt3)x+(1-\sqrt3)y-4(1+\sqrt3)
Multiply by -1 (reversing the inequality) to get the form in the key, and join the two results.
(-\sqrt3-1)x+(\sqrt3-1)y+4(\sqrt3+1)\le0\le-x+y+4
Check on the boundary: a point on the ray at 45^\circ from (4,0), say (5,1), has -x+y+4=0; a point on the ray at 75^\circ satisfies the other expression =0.

The region lies between the rays at 45^\circ and 75^\circ from (4,0) (that is \arg(z-4) from \tfrac\pi4 to \tfrac{5\pi}{12}).

(-\sqrt3-1)x+(\sqrt3-1)y+4(\sqrt3+1)\le0\le-x+y+4

(v) \arg\left(\dfrac{1-iz}{1-z}\right)=\dfrac\pi4;\ z\ne i

Write 1-iz=1-i(x+iy)=(1+y)-ix and 1-z=(1-x)-iy. Multiply the top and bottom by the conjugate (1-x)+iy.
\dfrac{(1+y)-ix}{(1-x)-iy}=\dfrac{[(1+y)-ix][(1-x)+iy]}{(1-x)^2+y^2}
Expand the numerator: real part and imaginary part.
\begin{aligned}\text{Re}&=(1+y)(1-x)+xy=1-x+y\\ \text{Im}&=y(1+y)-x(1-x)=x^2+y^2+y-x\end{aligned}
An argument of \dfrac\pi4 means \tan\dfrac\pi4=1=\dfrac{\text{Im}}{\text{Re}}, so Im = Re (and both positive).
x^2+y^2+y-x=1-x+y
Cancel y-x on both sides.
x^2+y^2=1

The locus is the unit circle without the dashed arc (where Re would be negative) and without z=1 and z=-i.

x^2+y^2=1

(vi) \dfrac12\arg(z-i)=\dfrac\pi3-\dfrac12\arg(z+i)

Multiply by 2 and bring the arguments together.
\arg(z-i)+\arg(z+i)=\dfrac{2\pi}3
The argument of a product is the sum of the arguments, so the left side is \arg\left[(z-i)(z+i)\right]. Use (z-i)(z+i)=z^2-i^2=z^2+1.
\arg(z^2+1)=\dfrac{2\pi}3
With z=x+iy, z^2+1=(x^2-y^2+1)+i(2xy). Take the tangent of both sides.
\dfrac{2xy}{x^2-y^2+1}=\tan\dfrac{2\pi}3=-\sqrt3
Clear the fraction and collect terms on one side.
2xy=-\sqrt3(x^2-y^2+1)\;\Rightarrow\;\sqrt3(x^2-y^2+1)+2xy=0
Check the sign with a point: for x\approx0.93,\ y=2 we get \arg(z-i)+\arg(z+i)\approx47.1^\circ+72.8^\circ=119.9^\circ\approx120^\circ, and \sqrt3(0.865-4+1)+2(0.93)(2)\approx-3.70+3.72\approx0.

\sqrt3\,(x^2-y^2+1)+2xy=0

Calculate the position of a particle from mean position when amplitude is 0.004mm and angle is:

Solution

Method (section 1.10, equation (1)). The position from the mean position is x=x_{max}e^{i\theta}=x_{max}(\cos\theta+i\sin\theta) with x_{max}=0.004\,\mathrm{mm} the amplitude.

(i) \dfrac\pi4

Substitute the amplitude and \theta=\dfrac\pi4.
x=0.004\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)
Both \cos\dfrac\pi4 and \sin\dfrac\pi4 equal \dfrac{\sqrt2}{2}.
x=0.004\cdot\dfrac{\sqrt2}2(1+i)=0.002\sqrt2\,(1+i)=\dfrac{\sqrt2}{500}(1+i)\ \mathrm{mm}

x=\dfrac{\sqrt2}{500}(1+i)\,\mathrm{mm}\approx0.00283(1+i)\,\mathrm{mm}

(ii) \dfrac\pi3

Substitute \theta=\dfrac\pi3: \cos\dfrac\pi3=\dfrac12, \sin\dfrac\pi3=\dfrac{\sqrt3}2.
x=0.004\left(\dfrac12+i\dfrac{\sqrt3}2\right)=0.002\left(1+\sqrt3\,i\right)=\dfrac1{500}\left(1+\sqrt3\,i\right)\mathrm{mm}

x=\dfrac{1}{500}(1+\sqrt3\,i)\,\mathrm{mm}

(iii) \dfrac\pi6

Substitute \theta=\dfrac\pi6: \cos\dfrac\pi6=\dfrac{\sqrt3}2, \sin\dfrac\pi6=\dfrac12.
x=0.004\left(\dfrac{\sqrt3}2+\dfrac i2\right)=0.002\left(\sqrt3+i\right)=\dfrac{1}{500}\left(\sqrt3+i\right)\mathrm{mm}

x=\dfrac1{500}(\sqrt3+i)\,\mathrm{mm}

When particle is at a position of x=2+3i from its mean position and x_{max}=1+2\sqrt3\,i is the position at maximum distance from mean position as it can be seen under microscope at this point.

Solution

Method. The position is x=x_{max}e^{i\omega t} (section 1.10), so the angle at time t is \theta=\omega t and e^{i\theta}=\dfrac{x}{x_{max}}. The frequency comes from \omega=2\pi f.

(i) Calculate the angle at time t=0 and find the position of the particle.

At t=0 the angle is \theta=\omega t=\omega\cdot0=0.
\theta=0
Substitute into x=x_{max}e^{i\theta} with e^{0}=1. The particle is at its maximum displacement.
x=x_{max}e^{0}=x_{max}=1+2\sqrt3\,i

\theta=0,\quad x=1+2\sqrt3\,i

(ii) If x=2+3i and x_{max}=1+2\sqrt3\,i. Calculate the frequency when t=2.

Find e^{i\theta}=\dfrac{x}{x_{max}} by rationalising with the conjugate 1-2\sqrt3\,i (note |1+2\sqrt3\,i|^2=1+12=13).
\dfrac{2+3i}{1+2\sqrt3\,i}=\dfrac{(2+3i)(1-2\sqrt3\,i)}{13}=\dfrac{2-4\sqrt3\,i+3i+6\sqrt3}{13}=\dfrac{(2+6\sqrt3)+(3-4\sqrt3)i}{13}
Compare with \cos\theta+i\sin\theta.
\cos\theta=\dfrac{2+6\sqrt3}{13}\approx0.9533,\qquad\sin\theta=\dfrac{3-4\sqrt3}{13}\approx-0.3022
Cosine positive and sine negative: the angle is in the fourth quadrant. Its principal value is
\theta=\tan^{-1}\left(\dfrac{-0.3022}{0.9533}\right)\approx-0.3070\ \mathrm{rad}
The angle \omega t must be positive for t>0 and \omega>0. Add 2\pi to take the first positive angle with the same e^{i\theta}.
\omega t=-0.3070+2\pi\approx5.9762\ \mathrm{rad}
Put t=2 to find \omega.
\omega=\dfrac{5.9762}{2}\approx2.9881\ \mathrm{rad\,s^{-1}}
Use \omega=2\pi f.
f=\dfrac{\omega}{2\pi}=\dfrac{2.9881}{6.2832}\approx0.4756\ \mathrm{Hz}

f\approx0.4756\ \mathrm{Hz}

Find the impedance Z for the following values:

Solution

Method (section 1.10). E=I\times Z, so Z=\dfrac EI. Divide by multiplying the top and bottom by the conjugate of I.

(i) E=(-50+100i) volts, I=(-6-2i) amps

Write Z=\dfrac EI and rationalise with the conjugate -6+2i.
Z=\dfrac{-50+100i}{-6-2i}=\dfrac{(-50+100i)(-6+2i)}{(-6)^2+2^2}
Expand the numerator and use i^2=-1: (-50)(-6)=300, (-50)(2i)=-100i, (100i)(-6)=-600i, (100i)(2i)=-200.
=\dfrac{300-100i-600i-200}{40}=\dfrac{100-700i}{40}
Divide by 40.
Z=\dfrac52-\dfrac{35}2i\ \Omega

Z=\dfrac52-\dfrac{35}2i

(ii) E=(100+10i) volts, I=(-8+3i) amps

Rationalise with the conjugate -8-3i.
Z=\dfrac{100+10i}{-8+3i}=\dfrac{(100+10i)(-8-3i)}{64+9}
Expand: -800-300i-80i-30i^2=-800-380i+30.
=\dfrac{-770-380i}{73}
Take out the common factor -10.
Z=-\dfrac{10}{73}(77+38i)\ \Omega

Z=-\dfrac{10}{73}(77+38i)

Find the impedance in an AC circuit if v(t)=3\cos(t) and i=10 amperes.

Solution

Use the Ohm's-law form of section 1.10: the impedance is the voltage divided by the current, Z=\dfrac{v}{i}. Write the voltage as the real part of a complex voltage.
v(t)=3\cos t=\mathrm{Re}\left(3e^{it}\right)
Divide by the current i=10 (a real number).
Z=\dfrac{v(t)}{i}=\dfrac{3\cos t}{10}
Simplify.
Z=0.3\cos t\ \Omega

Z=0.3\cos t

Book answer. 0.3\cos t

Determine the current flowing through an AC circuit having an impedance of 5 ohm where voltage source is given as v(t)=4\cos(3t).

Solution

For a resistor the impedance is the real number Z_R=R=5\ \Omega, and i=\mathrm{Re}\left(\dfrac{V_R}{Z_R}\right) with V_R=V_0e^{i\omega t} (section 1.10). Here V_0=4 and \omega=3.
v(t)=4\cos(3t)=\mathrm{Re}\left(4e^{3it}\right)
Divide by the impedance 5 (a real number, so taking the real part afterwards gives the same result).
i=\mathrm{Re}\left(\dfrac{4e^{3it}}{5}\right)=\dfrac{4\cos3t}{5}
Simplify.
i=0.8\cos(3t)\ \text{amperes}

i=0.8\cos(3t)

Book answer. 0.8\cos t

Note. The key prints 0.8\cos t, which leaves out the factor 3 of the angle. Since v(t)=4\cos(3t) and Z=5, the current is i=0.8\cos(3t); the angular frequency \omega=3 is unchanged by dividing by a real impedance.

Compute the potential difference across two AC power supplies with respect to time. Where V_1=5\cos(t+45^\circ) and V_2=3\sin(t+30^\circ).

Solution

Method of Example 27. Write each voltage as the real part of a complex number Ze^{it} and add the complex numbers; the potential difference is the real part of the sum. In Example 27 the book treats both supplies as r\cos(t+\phi) with the angle \phi given, and we follow it, because the key is computed that way.

Write the two complex voltages.
Z_1=5e^{i(t+45^\circ)},\qquad Z_2=3e^{i(t+30^\circ)}
Add and take out the common factor e^{it}.
Z_{total}=Z_1+Z_2=e^{it}\left(5e^{i45^\circ}+3e^{i30^\circ}\right)
Expand with e^{i\phi}=\cos\phi+i\sin\phi and use \cos45^\circ=\sin45^\circ=0.7071, \cos30^\circ=0.8660, \sin30^\circ=0.5.
=e^{it}\left\{5(0.7071+0.7071i)+3(0.8660+0.5i)\right\}
Collect real and imaginary parts.
=e^{it}\{(3.5355+2.5981)+i(3.5355+1.5)\}=e^{it}\{6.1336+5.0355i\}
Find the modulus and the argument of the bracket.
r=\sqrt{6.1336^2+5.0355^2}=\sqrt{62.98}\approx7.936,\qquad\phi=\tan^{-1}\dfrac{5.0355}{6.1336}\approx39.39^\circ
Write the bracket as re^{i\phi}, so Z_{total}=7.936\,e^{i(t+39.39^\circ)}, and take the real part.
V_{total}=\mathrm{Re}(Z_{total})=7.936\cos(t+39.39^\circ)

V_{total}=7.936\cos\left(t+39.39^\circ\right)

Book answer. 7.936\cos(t+39.39^\circ)

Note. The key (and Example 27) read \sin(t+\phi) as \cos(t+\phi) when forming the complex voltage. Strictly, 3\sin(t+30^\circ)=3\cos(t-60^\circ), and adding 5\cos(t+45^\circ)+3\cos(t-60^\circ) with the same method gives 5.122\cos(t+10.55^\circ). We give the answer the book's method and key give, 7.936\cos(t+39.39^\circ).

Given that V_A=4\cos(t+120^\circ) and V_B=5\sin(t+60^\circ). What will be the total potential difference of these AC power supplies?

Solution

Method of Example 27, as in question 11: write each voltage as r\,e^{i(t+\phi)}, add, and take the real part.

Write the two complex voltages.
Z_A=4e^{i(t+120^\circ)},\qquad Z_B=5e^{i(t+60^\circ)}
Add and take out the common factor e^{it}.
Z_{total}=e^{it}\left(4e^{i120^\circ}+5e^{i60^\circ}\right)
Expand with \cos120^\circ=-\tfrac12, \sin120^\circ=\tfrac{\sqrt3}2, \cos60^\circ=\tfrac12, \sin60^\circ=\tfrac{\sqrt3}2.
=e^{it}\left\{4\left(-\dfrac12+\dfrac{\sqrt3}2i\right)+5\left(\dfrac12+\dfrac{\sqrt3}2i\right)\right\}
Collect real and imaginary parts.
=e^{it}\left\{\dfrac12+\dfrac{9\sqrt3}{2}i\right\}
Find the modulus.
r=\sqrt{\dfrac14+\dfrac{243}{4}}=\sqrt{\dfrac{244}{4}}=\sqrt{61}
Find the argument. The real part is positive and the imaginary part is positive, so the angle is in the first quadrant.
\phi=\tan^{-1}\left(9\sqrt3\right)\approx86.33^\circ
Write Z_{total}=\sqrt{61}\,e^{i(t+86.33^\circ)} and take the real part.
V_{total}=\sqrt{61}\cos\left(t+86.33^\circ\right)

V_{total}=\sqrt{61}\cos\left(t+86.33^\circ\right)

Book answer. \sqrt{61}\cos(t+86.33^\circ)

Note. As in question 11 the key reads \sin(t+60^\circ) as \cos(t+60^\circ) in the complex form. Strictly 5\sin(t+60^\circ)=5\cos(t-30^\circ) and the sum would be 2.522\cos(t+22.48^\circ). We give the book's answer.

Review Exercise 1p. 29

Revision of the whole unit: definitions and the Argand plane (the multiple choice questions), powers of i, moduli, conjugates, factorising z^2+a^2, locus of a modulus condition, completing the square and the amplitude of a particle.

1(i)

Choose the correct option: Every real number is also a ______ number.

Anatural
Binteger
CcomplexAnswer
Drational

Why

Every real number x can be written as x+i0, so it is a complex number: \mathbb{R}\subset\mathbb{C} (section 1.1.1).

Book answer. (c)

1(ii)

Choose the correct option: Every complex number has ______ part(s).

Aone
BtwoAnswer
Cthree
Dno

Why

A complex number x+iy has two parts: the real part x and the imaginary part y (section 1.1.1).

Book answer. (b)

1(iii)

Choose the correct option: Magnitude of a complex number z is the distance of z from ______.

A(0, 0)Answer
B(1, 0)
C(0, 1)
D(1, 1)

Why

|z|=\sqrt{x^2+y^2} is the distance of the point (x,y) from the origin (0,0) (section 1.4).

Book answer. (a)

1(iv)

Choose the correct option: If z is a complex number then its mirror image is ______.

A|z|
B1/z
C-z
D\bar zAnswer

Why

The conjugate \bar z=x-iy is the mirror image of z=x+iy in the x-axis (section 1.3).

Book answer. (d)

1(v)

Choose the correct option: In complex plane imaginary part is drawn along ______.

Ax-axis
By-axisAnswer
Corigin
Dxy-plane

Why

In the complex (Argand) plane the real part x is measured along the x-axis (real axis) and the imaginary part y along the y-axis (imaginary axis).

Book answer. (b)

1(vi)

Choose the correct option: If z_1=3+2i and z_2=5+6i then

Az_1>z_2
Bz_1<z_2
C|z_1|>|z_2|
D|z_1|<|z_2|Answer

Why

Complex numbers cannot be compared with > or < (section 1.1.2), so (a) and (b) are meaningless.
The moduli are real numbers and can be compared.
|z_1|=\sqrt{9+4}=\sqrt{13},\qquad|z_2|=\sqrt{25+36}=\sqrt{61}
Since \sqrt{13}<\sqrt{61}, option (d) is true.
|z_1|<|z_2|

Book answer. (d)

1(vii)

Choose the correct option: Diagram representing a complex number is called ______ diagram.

Avector
BVenn
CArgandAnswer
Dordered pair

Why

The plane in which a complex number x+iy is shown as the point (x,y) is the Argand diagram (complex plane).

Book answer. (c)

1(viii)

Choose the correct option: If z=3+4i then z^{-1} is

A\left(\frac13,\frac14\right)
B\left(-\frac13,-\frac14\right)
C\left(\frac{3}{25},\frac{-4}{25}\right)
D\left(\frac{3}{25},\frac{-4}{25}\right)Answer

Why

Rationalise with the conjugate 3-4i.
z^{-1}=\dfrac{1}{3+4i}=\dfrac{3-4i}{9+16}=\dfrac{3-4i}{25}
Write as an ordered pair (real part, imaginary part). Check with the formulas of section 1.5: \mathrm{Re}=\dfrac{3}{25} and \mathrm{Im}=-\dfrac4{25}.
z^{-1}=\left(\dfrac{3}{25},\dfrac{-4}{25}\right)

Book answer. (d)

Note. Options (c) and (d) are printed identically in the book, \left(\tfrac3{25},\tfrac{-4}{25}\right). The value is correct for both; the key chooses (d).

1(ix)

Choose the correct option: The value of (\sqrt{-25})(\sqrt{-4}) is

A10
B-10Answer
C10i
D-10i

Why

Write each root with i.
\sqrt{-25}=5i,\qquad\sqrt{-4}=2i
Multiply and use i^2=-1.
(5i)(2i)=10i^2=-10

Book answer. (b)

1(x)

Choose the correct option: If \left(\dfrac{1+i}{1-i}\right)^n=1 then least positive value of n is

A1
B2
C3
D4Answer

Why

Simplify the base: multiply the top and bottom by 1+i.
\dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{(1-i)(1+i)}=\dfrac{2i}{2}=i
We need i^n=1. The powers are i,\,-1,\,-i,\,1 for n=1,2,3,4.
i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1
The least positive n is 4.
n=4

Book answer. (b)

Note. The key gives (b), but i^2=-1\ne1. The least positive n with i^n=1 is n=4, option (d). The key entry is a misprint.

Find the values of the following:

Solution

(i) i^2+i^4+i^6+\cdots+i^{100}

The terms are i^{2k}=(i^2)^k=(-1)^k for k=1,2,\dots,50. So the terms alternate -1,\,+1,\,-1,\,+1,\dots
i^2=-1,\ i^4=1,\ i^6=-1,\ i^8=1,\dots,\ i^{100}=1
There are 50 terms, which makes 25 pairs, and each pair (-1)+1=0.
25(0)=0

(ii) \left|\dfrac{(3-2i)(1+i)}{2-3i}\right|

Use |z_1z_2|=|z_1||z_2| and \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}.
\dfrac{|3-2i|\,|1+i|}{|2-3i|}=\dfrac{\sqrt{13}\cdot\sqrt2}{\sqrt{13}}
Cancel \sqrt{13}.
=\sqrt2

\sqrt2

(iii) \left|(3-2i)(4-i)\right|

Use |z_1z_2|=|z_1||z_2|.
|3-2i|=\sqrt{13},\quad|4-i|=\sqrt{17}
Multiply.
\sqrt{13}\sqrt{17}=\sqrt{221}
Check: (3-2i)(4-i)=12-3i-8i+2i^2=10-11i and \sqrt{100+121}=\sqrt{221}.

\sqrt{221}

(iv) \left(\dfrac{3+5i}{2-3i}\right)^{-1}

Invert the fraction.
\left(\dfrac{3+5i}{2-3i}\right)^{-1}=\dfrac{2-3i}{3+5i}
Rationalise with the conjugate 3-5i.
\dfrac{(2-3i)(3-5i)}{9+25}=\dfrac{6-10i-9i+15i^2}{34}
Simplify.
=\dfrac{-9-19i}{34}=-\dfrac9{34}-\dfrac{19}{34}i

-\dfrac{9}{34}-\dfrac{19}{34}i

Factorize the following:

Solution

Method. Take out the common factor, then use z^2+a^2=(z-ia)(z+ia) (section 1.7.1).

(i) 3x^2+108

Take out 3.
3x^2+108=3(x^2+36)
Write 36=6^2 and use the formula with a=6.
=3(x-6i)(x+6i)

3(x-6i)(x+6i)

(ii) 4x^2+40

Take out 4.
4x^2+40=4(x^2+10)
Write 10=(\sqrt{10})^2 and use the formula with a=\sqrt{10}.
=4(x-\sqrt{10}\,i)(x+\sqrt{10}\,i)

4(x-\sqrt{10}\,i)(x+\sqrt{10}\,i)

Locate the complex number z=x+iy on the complex plane if \left|\dfrac{z+2i}{z-2i}\right|=1.

Solution

Use \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}. The equation says the two moduli are equal.
|z+2i|=|z-2i|\quad(z\ne2i)
Write z+2i=x+i(y+2) and z-2i=x+i(y-2) and use the modulus formula.
\sqrt{x^2+(y+2)^2}=\sqrt{x^2+(y-2)^2}
Square both sides and expand.
x^2+y^2+4y+4=x^2+y^2-4y+4
Cancel the common terms.
8y=0\;\Rightarrow\;y=0
So z=x+0i=x is real: z lies on the real axis (the x-axis). Geometrically, z is equally distant from 2i and -2i, so it lies on the perpendicular bisector of the segment joining them, which is the real axis.

The points z are equidistant from 2i and -2i: the real axis.

y=0\ \ (z\text{ lies on the real axis})

Book answer. z=x

Find z when (z-3i)(2+5i)=3-4i.

Solution

Divide both sides by (2+5i).
z-3i=\dfrac{3-4i}{2+5i}
Rationalise with the conjugate 2-5i.
\dfrac{(3-4i)(2-5i)}{4+25}=\dfrac{6-15i-8i+20i^2}{29}=\dfrac{-14-23i}{29}
Add 3i to both sides. Write 3i=\dfrac{87}{29}i.
z=\dfrac{-14-23i}{29}+\dfrac{87i}{29}=\dfrac{-14+64i}{29}
Check: \left(z-3i\right)=\dfrac{-14-23i}{29}, and \dfrac{(-14-23i)(2+5i)}{29}=\dfrac{-28-70i-46i-115i^2}{29}=\dfrac{87-116i}{29}=3-4i.

z=-\dfrac{14}{29}+\dfrac{64}{29}i

Book answer. z=\dfrac{-14}{29}+\dfrac{64}{29}i

Evaluate \left[\dfrac{1}{i^{10}}+(2-i)^2+\sqrt{-25}\right]^3.

Solution

Evaluate each term. i^{10}=i^8i^2=-1.
\dfrac{1}{i^{10}}=\dfrac1{-1}=-1
Expand the square.
(2-i)^2=4-4i+i^2=3-4i
Write the root with i.
\sqrt{-25}=5i
Add the three terms.
-1+(3-4i)+5i=2+i
Cube: first the square, then multiply again by (2+i).
(2+i)^2=3+4i,\qquad(2+i)^3=(3+4i)(2+i)=6+3i+8i+4i^2=2+11i

2+11i

Book answer. 2+11i

Solve by completing square method 2z^2-11z+16=0.

Solution

Divide by 2 (the coefficient of z^2) and move the constant to the right side.
z^2-\dfrac{11}2z=-8
Add the square of half the coefficient of z, \left(\dfrac{11}{4}\right)^2=\dfrac{121}{16}, to both sides.
z^2-\dfrac{11}2z+\dfrac{121}{16}=-8+\dfrac{121}{16}\;\Rightarrow\;\left(z-\dfrac{11}4\right)^2=\dfrac{-128+121}{16}=-\dfrac7{16}
Take the square root of both sides; \sqrt{-7}=i\sqrt7.
z-\dfrac{11}4=\pm\dfrac{i\sqrt7}{4}
Solve for z.
z=\dfrac{11\pm i\sqrt7}{4}
Check with the discriminant: b^2-4ac=121-128=-7, so z=\dfrac{11\pm\sqrt{-7}}{4}, the same result.

z=\dfrac{11\pm i\sqrt7}{4}

Book answer. z=\dfrac{11\pm i\sqrt{71}}{4}

Note. The key prints \sqrt{71} under the i. The discriminant is b^2-4ac=(-11)^2-4(2)(16)=121-128=-7, so the roots contain i\sqrt7. Substituting z=\dfrac{11+i\sqrt7}{4} gives 2z^2-11z+16=\dfrac{57+11i\sqrt7}{4}-\dfrac{121+11i\sqrt7}{4}+16=0, which confirms \sqrt7.

When particle is at a position of \sqrt2+i\sqrt2 nm from its mean position calculate its amplitude when \theta=45^\circ.

Solution

Use x=x_{max}e^{i\theta} (section 1.10), so x_{max}=\dfrac{x}{e^{i\theta}}=x\,e^{-i\theta}, where x is the position and x_{max} the amplitude.
x_{max}=(\sqrt2+i\sqrt2)\left(\cos45^\circ-i\sin45^\circ\right)
Substitute \cos45^\circ=\sin45^\circ=\dfrac{\sqrt2}2 and take out \sqrt2.
x_{max}=\sqrt2(1+i)\cdot\dfrac{\sqrt2}{2}(1-i)=(1+i)(1-i)
Multiply: (1+i)(1-i)=1-i^2=2.
x_{max}=2\ \mathrm{nm}
Check using moduli: |x|=\sqrt{2+2}=2 and |e^{i\theta}|=1, so |x_{max}|=2. Also 2e^{i45^\circ}=\sqrt2+i\sqrt2, the given position.

x_{max}=2\ \mathrm{nm}

Book answer. 2

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 11 mark

What are the multiplicative factors of z^2+4z+5?

A(z+2-i)(z+2+i)Answer
B(z-2-i)(z-2+i)
C(z+2-2i)(z+2+2i)
D(z-2-2i)(z-2+2i)

Why

Complete the square: z^2+4z+5=(z+2)^2+1=(z+2)^2-i^2, because 1=-i^2. A difference of squares gives (z+2-i)(z+2+i).

Check by expanding: (z+2)^2-i^2=z^2+4z+4+1. Options B and D have the wrong sign of the 2 (they factorise z^2-4z+\dots). Options C and D use 2i instead of i: (z+2)^2-(2i)^2=z^2+4z+8, which is not the given expression.

2025 Annual I, Set B, MCQ 21 mark

The principal argument (\theta) of the complex number z=-1+i\sqrt{3} is:

A-\dfrac{\pi}{3}
B\dfrac{\pi}{3}
C\dfrac{2\pi}{3}Answer
D\dfrac{5\pi}{3}

Why

Here x=-1<0 and y=\sqrt3>0, so z is in the second quadrant. The acute angle is \alpha=\tan^{-1}\left|\dfrac{\sqrt3}{-1}\right|=\dfrac{\pi}{3}, and in the second quadrant \theta=\pi-\alpha=\dfrac{2\pi}{3}.

Option B is only the acute angle \alpha; option A is an angle in the fourth quadrant. Option D, \dfrac{5\pi}{3}, points in the fourth quadrant and is outside the principal range -\pi<\theta\le\pi. Check: 2\left(\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3\right)=2\left(-\dfrac12+\dfrac{\sqrt3}{2}i\right)=-1+i\sqrt3.

2026 Annual I, Set B, MCQ 61 mark

Factorize z^2+289:

A(z+17)(z-17)
B(z+17i)(z+17i)
C(z+17i)(z-17i)Answer
D(z-17i)(z-17i)

Why

Since i^2=-1, we have 289=-(17i)^2, so z^2+289=z^2-(17i)^2. This is a difference of squares, a^2-b^2=(a-b)(a+b) with a=z and b=17i: z^2+289=(z-17i)(z+17i).

Option A is the factorisation of z^2-289, a different expression. Options B and D are perfect squares: (z+17i)^2=z^2+34iz-289 has a middle term, so neither can equal z^2+289. Check of C: (z+17i)(z-17i)=z^2-(17i)^2=z^2+289.

2026 Annual I, Set B, MCQ 71 mark

Argument of z=\sqrt3+i is:

A\dfrac{\pi}{3}
B\dfrac{\pi}{4}
C\dfrac{\pi}{9}
D\dfrac{\pi}{6}Answer

Why

Here x=\sqrt3>0 and y=1>0, so z lies in the first quadrant and \operatorname{Arg}z=\alpha=\tan^{-1}\left|\dfrac yx\right|=\tan^{-1}\dfrac{1}{\sqrt3}=\dfrac{\pi}{6}.

Option A, \dfrac{\pi}{3}, comes from inverting the ratio, \tan^{-1}\dfrac{\sqrt3}{1}. Remember that \tan\theta=\dfrac yx, imaginary part over real part. As a check, |z|=2 and 2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)=2\left(\dfrac{\sqrt3}{2}+\dfrac i2\right)=\sqrt3+i.

2025 Annual I, Set B, Q.2 (i)4 marks

Solve the quadratic equation 3z^2+2z+2=0, z\in C by completing the square method.

Solution

Divide by the leading coefficient 3 so that the coefficient of z^2 is 1.
z^2+\dfrac23z+\dfrac23=0
Take the constant to the right side.
z^2+\dfrac23z=-\dfrac23
Add the square of half the coefficient of z, \left(\dfrac13\right)^2=\dfrac19, to both sides.
z^2+\dfrac23z+\dfrac19=-\dfrac23+\dfrac19\;\Rightarrow\;\left(z+\dfrac13\right)^2=-\dfrac59
Take square roots. The right side is negative, and \sqrt{-1}=i.
z+\dfrac13=\pm\dfrac{\sqrt5}{3}\,i
Solve for z.
z=-\dfrac13\pm\dfrac{\sqrt5}{3}\,i=\dfrac{-1\pm i\sqrt5}{3}
Check with the quadratic formula: the discriminant is 2^2-4(3)(2)=-20, so z=\dfrac{-2\pm i\sqrt{20}}{6}=\dfrac{-1\pm i\sqrt5}{3}\;\checkmark.

z=\dfrac{-1+i\sqrt5}{3}\quad\text{or}\quad z=\dfrac{-1-i\sqrt5}{3}

Note. The two roots are complex conjugates, as always for a quadratic with real coefficients and a negative discriminant.

2026 Annual I, Set B, Q.2 (iii)4 marks

Given z_1=2(\cos 30^\circ+i\sin 30^\circ) and z_2=3(\cos 60^\circ+i\sin 60^\circ), find z_1z_2 in polar form.

Solution

Both numbers are already in polar form r(\cos\theta+i\sin\theta), with r_1=2,\ \theta_1=30^\circ and r_2=3,\ \theta_2=60^\circ. For a product, multiply the moduli and add the arguments (Property 4).
z_1z_2=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]
Substitute the values.
z_1z_2=(2)(3)\left[\cos(30^\circ+60^\circ)+i\sin(30^\circ+60^\circ)\right]=6\left(\cos 90^\circ+i\sin 90^\circ\right)
Check in rectangular form: z_1=\sqrt3+i and z_2=\dfrac32+\dfrac{3\sqrt3}{2}i, so
z_1z_2=\left(\sqrt3+i\right)\left(\dfrac32+\dfrac{3\sqrt3}{2}i\right)=\dfrac{3\sqrt3}{2}+\dfrac92i+\dfrac32i-\dfrac{3\sqrt3}{2}=6i\;\checkmark

z_1z_2=6\left(\cos 90^\circ+i\sin 90^\circ\right)\ (=6i)

Note. Marking: the rule (multiply moduli, add arguments) and the substitution earn most of the marks. Leaving the answer as 6(\cos90^\circ+i\sin90^\circ) is enough; 6i is the rectangular form.

2026 Annual I, Set B, Q.2 (vi)4 marks

Solve the following equations: (2+i)\omega=1; (3-i)z+5\omega=2-i.

Solution

The first equation contains only \omega, so solve it first (substitution method). Divide by 2+i and multiply top and bottom by the conjugate 2-i.
\omega=\dfrac{1}{2+i}=\dfrac{2-i}{(2+i)(2-i)}=\dfrac{2-i}{4+1}=\dfrac{2-i}{5}
Substitute in the second equation. Note that 5\omega=2-i.
(3-i)z+(2-i)=2-i
Subtract 2-i from both sides.
(3-i)z=0
The factor 3-i\neq0, so the other factor must be zero.
z=0
Check in both equations.
(2+i)\dfrac{2-i}{5}=\dfrac{4+1}{5}=1,\qquad (3-i)(0)+5\cdot\dfrac{2-i}{5}=2-i\;\checkmark

z=0,\qquad \omega=\dfrac{2-i}{5}=\dfrac25-\dfrac15i

Note. This is a system of simultaneous linear equations with complex coefficients (Exercise 1.3, Q.4). The elimination method works too: multiply the first equation by 5 and subtract 5\omega from the second.

2025 Annual I, Set B, Q.2 (x) (a)4 for parts (a) and (b) together = 4 marks

If z_1=10(\cos 100^\circ+i\sin 100^\circ) and z_2=5(\cos 40^\circ+i\sin 40^\circ), find z_1\cdot z_2 in polar form.

Solution

Product rule (Property 4): multiply the moduli and add the arguments. Here r_1=10,\ \theta_1=100^\circ and r_2=5,\ \theta_2=40^\circ.
z_1z_2=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]
Substitute.
z_1z_2=(10)(5)\left[\cos(100^\circ+40^\circ)+i\sin(100^\circ+40^\circ)\right]=50\left(\cos140^\circ+i\sin140^\circ\right)

z_1z_2=50\left(\cos140^\circ+i\sin140^\circ\right)

Note. The paper prints 4 marks for the whole of part (x), which has parts (a) and (b); no split is printed.

2025 Annual I, Set B, Q.2 (x) (b)4 for parts (a) and (b) together = 4 marks

If z_1=10(\cos 100^\circ+i\sin 100^\circ) and z_2=5(\cos 40^\circ+i\sin 40^\circ), find \dfrac{z_1}{z_2} in polar form.

Solution

Division rule (Property 5): divide the moduli and subtract the arguments.
\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\left[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right]
Substitute r_1=10,\ \theta_1=100^\circ,\ r_2=5,\ \theta_2=40^\circ.
\dfrac{z_1}{z_2}=\dfrac{10}{5}\left[\cos(100^\circ-40^\circ)+i\sin(100^\circ-40^\circ)\right]=2\left(\cos60^\circ+i\sin60^\circ\right)

\dfrac{z_1}{z_2}=2\left(\cos60^\circ+i\sin60^\circ\right)

Note. The paper prints 4 marks for the whole of part (x). In rectangular form the answer is 1+i\sqrt3.

2026 Annual I, Set D, MCQ 11 mark

Factorize z^2+169:

A(z+13)(z-13)
B(z+13i)(z-13i)Answer
C(z-13i)(z-13i)
D(z+13i)(z+13i)

Why

Since 169=-(13i)^2, we get z^2+169=z^2-(13i)^2=(z-13i)(z+13i).

Option A is the factorisation of z^2-169. Options C and D are perfect squares whose expansion has a middle term \mp26iz, so they cannot equal z^2+169.

2025 Annual I, Set D, MCQ 11 mark

What are the multiplicative factors of z^2+9?

A(z+3i)(z-3i)Answer
B(z-3)(z+3)
C(z+3)^2
D(z-3i)^2

Why

9=-(3i)^2, so z^2+9=z^2-(3i)^2=(z-3i)(z+3i).

Option B factorises z^2-9. Options C and D are perfect squares and have a middle term when expanded, so they cannot equal z^2+9.

2026 Annual I, Set D, MCQ 21 mark

Argument of z=1+i is:

A\dfrac{\pi}{6}
B\dfrac{\pi}{4}Answer
C\dfrac{\pi}{3}
D\dfrac{\pi}{2}

Why

z=1+i lies in the first quadrant (x=1>0,\ y=1>0), so \operatorname{Arg}z=\tan^{-1}\left|\dfrac11\right|=\tan^{-1}1=\dfrac{\pi}{4}.

Option D, \dfrac\pi2, is the argument of the pure imaginary number i (when x=0), not of 1+i. Here the real and imaginary parts are equal, so the point lies on the line y=x at 45^\circ.

2025 Annual I, Set D, MCQ 21 mark

The rectangular form of 4\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right) is:

A-2+2i\sqrt3Answer
B-2-2i\sqrt3
C2+2i\sqrt3
D2-2i\sqrt3

Why

The angle \dfrac{2\pi}{3}=120^\circ is in the second quadrant, where cosine is negative and sine is positive: \cos\dfrac{2\pi}{3}=-\dfrac12 and \sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}. So x=4\left(-\dfrac12\right)=-2 and y=4\cdot\dfrac{\sqrt3}{2}=2\sqrt3.

Option C would need \cos\theta>0 (first quadrant, \theta=\dfrac\pi3), and option B would need both negative (third quadrant). The check |z|=\sqrt{4+12}=4 holds for all four options, so the signs must be decided from the quadrant.

2026 Annual I, Set D, Q.2 (i)4 marks

Solve the system: z+2\omega=4i; 2z+5\omega=3.

Solution

Use the elimination method. Multiply the first equation by 2 so that the coefficients of z match.
2z+4\omega=8i
Subtract this from the second equation to eliminate z.
(2z+5\omega)-(2z+4\omega)=3-8i\;\Rightarrow\;\omega=3-8i
Put \omega in the first equation to find z.
z=4i-2\omega=4i-2(3-8i)=4i-6+16i=-6+20i
Check in the second equation.
2z+5\omega=2(-6+20i)+5(3-8i)=-12+40i+15-40i=3\;\checkmark

z=-6+20i,\qquad \omega=3-8i

Note. The constant on the right of the first equation is the complex number 4i; the method is the same as for real constants.

2025 Annual I, Set D, Q.2 (i) (a)4 for parts (a) and (b) together = 4 marks

Express in rectangular form: \left[5(\cos 45^\circ+i\sin 45^\circ)\right]\left[4(\cos 15^\circ+i\sin 15^\circ)\right].

Solution

Multiply the moduli and add the arguments (Property 4).
\left[5(\cos45^\circ+i\sin45^\circ)\right]\left[4(\cos15^\circ+i\sin15^\circ)\right]=(5)(4)\left[\cos(45^\circ+15^\circ)+i\sin(45^\circ+15^\circ)\right]
Simplify to polar form.
=20\left(\cos60^\circ+i\sin60^\circ\right)
Use \cos60^\circ=\dfrac12 and \sin60^\circ=\dfrac{\sqrt3}{2} to go to rectangular form x+iy.
=20\left(\dfrac12+\dfrac{\sqrt3}{2}i\right)=10+10\sqrt3\,i

10+10\sqrt3\,i

Note. The paper prints 4 marks for the whole of part (i). This is the same method as Exercise 1.4, Q.2.

2025 Annual I, Set D, Q.2 (i) (b)4 for parts (a) and (b) together = 4 marks

Express in rectangular form: \dfrac{\left[27(\cos 120^\circ+i\sin 120^\circ)\right]}{\left[3(\cos 60^\circ+i\sin 60^\circ)\right]}.

Solution

Divide the moduli and subtract the arguments (Property 5).
\dfrac{27(\cos120^\circ+i\sin120^\circ)}{3(\cos60^\circ+i\sin60^\circ)}=\dfrac{27}{3}\left[\cos(120^\circ-60^\circ)+i\sin(120^\circ-60^\circ)\right]
Simplify.
=9\left(\cos60^\circ+i\sin60^\circ\right)
Convert to rectangular form.
=9\left(\dfrac12+\dfrac{\sqrt3}{2}i\right)=\dfrac92+\dfrac{9\sqrt3}{2}\,i

\dfrac92+\dfrac{9\sqrt3}{2}\,i

Note. The paper prints 4 marks for the whole of part (i).

2026 Annual I, Set D, Q.2 (x)4 marks

Convert z=1+i into polar form and then find z^2 using polar representation.

Solution

Find the modulus.
r=|z|=\sqrt{1^2+1^2}=\sqrt2
Find the argument. z is in the first quadrant, so \theta=\tan^{-1}\left|\dfrac11\right|.
\theta=\tan^{-1}1=\dfrac\pi4
Write the polar form.
1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)
For z^2=z\cdot z multiply the moduli and add the arguments (Property 4): the modulus is squared and the argument doubled.
z^2=(\sqrt2)^2\left[\cos\left(2\cdot\dfrac\pi4\right)+i\sin\left(2\cdot\dfrac\pi4\right)\right]=2\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)
Evaluate and check with direct squaring: \cos\dfrac\pi2=0 and \sin\dfrac\pi2=1.
z^2=2(0+i)=2i,\qquad (1+i)^2=1+2i+i^2=2i\;\checkmark

1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right),\qquad z^2=2\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)=2i

2025 Annual I, Set D, Q.2 (x)4 marks

Solve the simultaneous linear equations 2iz-3i\omega=4 and -3iz+5i\omega=9 carrying complex coefficients (i=\sqrt{-1}).

Solution

Divide both equations by i to remove the common factor. Use \dfrac1i=-i (because i\cdot(-i)=1).
2z-3\omega=\dfrac4i=-4i,\qquad -3z+5\omega=\dfrac9i=-9i
Eliminate z: multiply the first by 3 and the second by 2, then add.
6z-9\omega=-12i,\qquad -6z+10\omega=-18i\;\Rightarrow\;\omega=-30i
Put \omega in 2z-3\omega=-4i.
2z=-4i+3\omega=-4i-90i=-94i\;\Rightarrow\;z=-47i
Check in the original equations, using i^2=-1.
\begin{aligned}2i(-47i)-3i(-30i)&=94-90=4\\ -3i(-47i)+5i(-30i)&=-141+150=9\;\checkmark\end{aligned}

z=-47i,\qquad \omega=-30i

Note. Every coefficient is a multiple of i, so dividing by i first makes the numbers real-looking and the working shorter. The elimination method of Exercise 1.3, Q.4 gives the same answer.

Exam focus

Solve simultaneous linear equations with complex coefficients, by elimination or substitution (for example (2+i)\omega=1 with (3-i)z+5\omega=2-i). (problem)
Asked in all three recent new-syllabus sets that carry this topic: 2026 Set B Q.2 (vi), 2026 Set D Q.2 (i) and 2025 Set D Q.2 (x), each for 4 marks. Exercise 1.3, Q.4 is the practice set.
Go to answer
Use the product and division rules in polar form: find z_1z_2 and \dfrac{z_1}{z_2}, or convert to rectangular form. (problem)
2026 Set B Q.2 (iii), 2025 Set B Q.2 (x) (a) and (b), and 2025 Set D Q.2 (i) (a) and (b). Multiply (divide) the moduli and add (subtract) the arguments.
Go to answer
Find the principal argument of a complex number (check the quadrant first), e.g. \sqrt3+i, 1+i and -1+i\sqrt3. (mcq)
MCQ in every paper: 2026 Set B MCQ 7, 2026 Set D MCQ 2 and 2025 Set B MCQ 2. The acute angle \tan^{-1}|y/x| then the quadrant rule gives the answer in seconds.
Go to answer
Factorize z^2+a^2 as (z-ia)(z+ia), and factorize a quadratic such as z^2+4z+5 into complex linear factors. (mcq)
One MCQ in each of the four papers: 2026 Set B MCQ 6 (z^2+289), 2026 Set D MCQ 1 (z^2+169, identical to Exercise 1.3 Q.1 (i)), 2025 Set B MCQ 1 and 2025 Set D MCQ 1.
Go to answer
Convert a complex number to polar form and use the polar form to find a power, e.g. z=1+i and z^2. (problem)
2026 Set D Q.2 (x), 4 marks. Modulus, principal argument, then square the modulus and double the argument.
Go to answer
Solve a quadratic equation with real coefficients and negative discriminant by completing the square, e.g. 3z^2+2z+2=0. (problem)
2025 Set B Q.2 (i), 4 marks. Practice from Exercise 1.3, Q.2 and Q.3. Divide by the leading coefficient first, then add the square of half the coefficient of z.
Go to answer
Convert between polar and rectangular form with exact values of \cos and \sin, e.g. 4\left(\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3\right). (mcq)
2025 Set D MCQ 2 and the rectangular-form answers of 2025 Set D Q.2 (i). Exercise 1.4, Q.1, Q.2 and Q.4 cover it.
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Complex numbers in applications (simple harmonic motion, impedance, AC voltage and current). (problem)
Not in the 2025 or 2026 new-syllabus papers, but the book gives Exercise 1.4, Q.6 to Q.12 on these, so a short numerical is possible. Learn Examples 25 to 27.
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