Contents · 0% complete

Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9709Paper 3 · Pure Mathematics 3Complex numbersArithmetic and conjugates

Complex numbers · Subtopic 1 of 5

Complex Numbers: Arithmetic and Conjugates

Meet $i$ , where $i^2=-1$ . Add, subtract and multiply complex numbers like brackets, divide using the conjugate, and solve equations by equating real and imaginary parts.

Cambridge 9709 Paper 3Edexcel YFM01 WFM01Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Practise this topicTimed test Ask Salam TutorSwitch board

Section 1 of 14

Learning objectives

Use $i^2=-1$ and simplify powers of $i$ .
Add, subtract and multiply complex numbers in the form $x+iy$ .
Find the conjugate $z^*$ and use it to divide one complex number by another.
Solve equations by equating real parts and imaginary parts.

Pause and think

$x^2+1=0$ has no real solutions. Suppose we simply invent a number $i$ with $i^2=-1$ . What would $i^3$ and $i^4$ have to be?

Section 2 of 14

Core idea

Start hereThe square root of minus one

Solve $x^2=-1$ . No real number works: squares are never negative. For centuries mathematicians called such solutions "imaginary" and avoided them. Then they found that if you simply invent a number $i$ with $i^2=-1$ , all the usual algebra still works, every polynomial gets its full set of roots, and electrical engineering and quantum physics become far easier.

The big idea

A complex number is $z=x+iy$ ( $x,y$ real, $i^2=-1$ ). Add, subtract and multiply like algebra, replacing $i^2$ by $-1$ . Divide by multiplying top and bottom by the conjugate $z^*=x-iy$ , since $zz^*=x^2+y^2$ is real.

1. Definitions

z=x+iy

$i$: a number with $i^2=-1$ .
$x=\operatorname{Re}(z)$: the real part.
$y=\operatorname{Im}(z)$: the imaginary part (a real number!).
$z^*$: the conjugate $x-iy$ .

z, its conjugate z*, and z + w on the Argand diagram

Each complex number is a point (or arrow). The conjugate is a reflection in the real axis. Adding is like adding vectors: the parallelogram rule.

Key takeawayEqual complex numbers

a+ib=c+id

means

a=c

and

b=d

. One complex equation gives two real equations.

2. Arithmetic

Worked example $z=3+2i$ , $w=1-4i$

Find $z+w$ , $zw$ , $z^2$ .

Goal: Simplified forms.
Strategy: Expand; $i^2=-1$ .

$z+w=4-2i$
Add parts.
$zw=3-12i+2i-8i^2=3-10i+8=11-10i$
Expand; $-8i^2=+8$ .
$z^2=9+12i+4i^2=5+12i$
$(a+b)^2$ .

Answer

4-2i

11-10i

5+12i

✓ Check

|z|^2|w|^2=13\times17=221

and

|zw|^2=121+100=221

✓.

3. Division

zz^*

is real

$(x+iy)(x-iy)=x^2-ixy+ixy-i^2y^2$
Expand.
$=x^2+y^2$
$-i^2=1$ ; imaginary parts cancel.

Worked example $\dfrac{3+2i}{1-4i}$

Write as $a+bi$ .

Goal: Real and imaginary parts.
Strategy: Multiply top and bottom by $1+4i$ .

$\frac{(3+2i)(1+4i)}{(1-4i)(1+4i)}$
Conjugate of the bottom.
$=\frac{3+12i+2i+8i^2}{1+16}=\frac{-5+14i}{17}$
Expand.

Answer

-\frac5{17}+\frac{14}{17}i

✓ CheckMultiply back by

1-4i

\frac{(-5+14i)(1-4i)}{17}=\frac{-5+20i+14i+56}{17}=3+2i

✓.

Worked exampleSolve $z+2z^*=6-i$

Find $z$ .

Goal: $z=x+iy$ .
Strategy: Substitute and equate parts.

$x+iy+2x-2iy=6-i$
Substitute.
$3x=6,\ -y=-1$
Real and imaginary parts.

Answer

z=2+i

✓ Check

2+i+4-2i=6-i

✓.

Pause and think

Find $i^3$ , $i^4$ , $i^{10}$ .

Summary

$i^2=-1$ .
Conjugate for division.
Equate parts.

Conjugate: $x-iy$ .
Argand diagram: Plane with real and imaginary axes.

HistoryNumbers nobody wanted. In 1545 Gerolamo Cardano published a method for solving cubic equations and found it sometimes needed square roots of negative numbers, even when the final answer was real. He called them "as subtle as they are useless". In 1572 Rafael Bombelli wrote down the rules for calculating with them, and in 1637 René Descartes gave them the dismissive name "imaginary". The symbol

i

was introduced by Leonhard Euler in 1777.

Section 3 of 14

Key formulae

Powers of i

i^2=-1\qquad i^3=-i\qquad i^4=1\qquad i^{4k+r}=i^r

Multiplying and dividing

\begin{gathered}(a+bi)(c+di)=(ac-bd)+(ad+bc)i \\[4pt] \frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(a+bi)(c-di)}{c^2+d^2}\end{gathered}

Conjugates and equal numbers

\begin{gathered}z=x+iy\ \Rightarrow\ z^*=x-iy,\quad z+z^*=2x,\quad zz^*=x^2+y^2 \\[4pt] a+bi=c+di\iff a=c\ \text{and}\ b=d\end{gathered}

TrickThe conjugate is the "difference of two squares" trick.

(c+di)(c-di)=c^2-(di)^2=c^2+d^2

. You already rationalise

\dfrac{1}{3+\sqrt2}

by multiplying by

3-\sqrt2

; dividing complex numbers is exactly the same move, with

i

in place of the surd.

Section 4 of 14

Worked examples

Example 1 · Multiplying0 / 2 steps

Express

(3+2i)(4-i)

in the form

x+iy

Example 2 · Dividing0 / 3 steps

Express

\dfrac{5+i}{2-3i}

in the form

x+iy

Example 3 · Equating parts (exam style)0 / 3 steps

Find the complex number

z

such that

z+2z^*=6-3i

Section 5 of 14

Explore the graph

Multiplying by i rotates by a right angle

Loading the interactive graph…

Try this

The first point is $z=a+bi$ (horizontal axis real, vertical axis imaginary). The others are $iz=-b+ai$ , $i^2z=-a-bi$ and $i^3z=b-ai$ . What shape do the four points make?
Move $a$ and $b$ . Each multiplication by $i$ turns the point a quarter-turn anticlockwise about the origin.
Which point is the conjugate $z^*=a-bi$ ? (None of these: it is the reflection of $z$ in the real axis.)

Multiplying by $i$ is a rotation of $90^\circ$ . Doing it twice turns through $180^\circ$ , which is multiplying by $-1$ : that is why $i^2=-1$ .

In the real worldElectricity runs on complex numbers. Electrical engineers describe alternating currents and voltages with complex numbers, because a phase shift is a rotation. They write

j

instead of

i

, since

i

already means current. Dividing complex numbers is how they find the current in a circuit with resistors, capacitors and coils.

Fun factWhat is i to the power i? Surprisingly,

i^i

is a real number: about

0.2079

. You will see why once you meet

e^{i\theta}

in the polar-form subtopic:

i=e^{i\pi/2}

, so

i^i=e^{i\cdot i\pi/2}=e^{-\pi/2}

Section 6 of 14

Common mistakes

Incorrect

(3+2i)(4-i)=12-2i^2=14

Correct

14+5i

Expand all four terms; the "inside" and "outside" products give the imaginary part.

Incorrect

i^2=1

Correct

i^2=-1

That is the definition of

i

. Replace every

i^2

-1

Incorrect

\operatorname{Im}(3-4i)=-4i

Correct

-4

The imaginary part is the real number multiplying

i

Incorrect

\dfrac{5+i}{2-3i}=\dfrac{5+i}{2}-\dfrac{5+i}{3i}

CorrectMultiply by the conjugate

2+3i

You cannot split a fraction over a sum in the denominator.

$\sqrt{-1}$ doesn't exist

i

I'm about to ruin your whole career

Electrical engineers

we use $j$ and we love it

Section 7 of 14

Exam tip

Exam tipShow the conjugate step. For a division, write the multiplication by the conjugate explicitly, for example

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

. That line earns the method mark even if the arithmetic after it slips. Give the final answer in the form asked, usually

x+iy

with fractions, not decimals.

Section 8 of 14

Guided practice

SA-9709-P3-CX-G014 marksL1 FoundationCalculatorArithmetic

Given

z=3+2i

and

w=1-i

, find

zw

and

\dfrac zw

, each in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P3-CX1-0011 markL1 FoundationCalculatorArithmetic

Express

(4+3i)+(2-5i)-(1-i)

in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX1-0022 marksL1 FoundationCalculatorArithmetic

Express

(3+2i)(4-i)

in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX-0012 marksL1 FoundationCalculatorArithmetic

Express

(2-3i)^2

in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX1-0033 marksL2 StandardCalculatorArithmetic

Express

\dfrac{5+i}{2-3i}

in the form

x+iy

, giving

x

and

y

as fractions.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX-0123 marksL2 StandardCalculatorEquating parts

Find the complex number

z

such that

z(2-i)=7+i

, giving your answer in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX1-0053 marksL2 StandardCalculatorEquating parts

The real numbers

x

and

y

satisfy

(x+iy)(1+2i)=7+4i

. Find

x

and

y

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX1-0063 marksL2 StandardCalculatorArithmetic

Given

z=2-i

, express

z^2+z^*

in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P3-CX1-0072 marksL2 StandardCalculatorArithmetic

Find the value of

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

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P3-CX1-0084 marksL4 ChallengingCalculatorEquating parts

Find the complex number

z

such that

z+2z^*=6-3i

, giving your answer in the form

x+iy

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$i^2=-1$ ; powers of $i$ repeat every $4$ .
Add and multiply like brackets, then replace $i^2$ by $-1$ .
Divide by multiplying top and bottom by the conjugate of the bottom.
Two complex numbers are equal exactly when their real parts and their imaginary parts are equal.

Section 12 of 14

Quick quiz

Q1 · SA-9709-P3-CX-Q011 markL1 FoundationCalculatorArithmetic

i^2

equals

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P3-CX-Q021 markL1 FoundationCalculatorArithmetic

The conjugate of

3-4i

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P3-CX1-0041 markL1 FoundationCalculatorArithmetic

i^7

equals

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 9709 AS & A Level Mathematics
Where: Paper 3, Pure Mathematics 3 · 3.9 Complex numbers
Expected: Understand the idea of a complex number, recall the meaning of the terms real part, imaginary part, modulus, argument and conjugate, and use the fact that two complex numbers are equal if and only if both real and imaginary parts are equal; carry out operations of addition, subtraction, multiplication and division of two complex numbers expressed in Cartesian form x + iy.
Calculator: Permitted. Arguments are in radians, in the range −π < θ ≤ π unless the question says otherwise.
Typical style: Usually the first 2–3 marks of a longer question: "find zw and z/w in the form x + iy".
Open in: 9709 · Paper 3

Section 14 of 14

Past paper questions

6 questions (38 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

9709/32 · May/June 2025Question 55 marks
Find the exact square roots of a complex number by forming a quartic.
Question paper, page 7 Mark scheme
9709/32 · Feb/March 2025Question 55 marks
Exact square roots of a complex number by forming a quartic.
Question paper, page 8 Mark scheme
9709/32 · Oct/Nov 2024Question 35 marks
Square roots of a complex number.
Question paper, page 4 Mark scheme
9709/32 · May/June 2024Question 910 marks
Complex products, modulus–argument form and a geometric result.
Question paper, page 14 Mark scheme
9709/32 · Oct/Nov 2023Question 87 marks
Equate real and imaginary parts to form and solve an equation.
Question paper, page 12 Mark scheme
9709/32 · May/June 2023Question 56 marks
Simplify a complex expression and find its argument.
Question paper, page 7 Mark scheme

All 9709 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9709, 2026–2027 (official specification).