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Cambridge International AS & A Level · Topic 3.9
A-Level Maths 9709 / Topic 3.9 / AS & A Level

A square root of minus one.

Paper 3 only. Allow one new number, \(i\), with \(i^{2} = -1\), and every quadratic has two roots. Complex numbers then add, subtract, multiply and divide by the ordinary rules of algebra, with \(i^{2}\) replaced by \(-1\) whenever it appears.

Topic 3.9All complex numbers lessons →
The key idea

A complex number is \(z = x + iy\) with \(x, y\) real and \(i^{2} = -1\); \(\text{Re}\,z = x\), \(\text{Im}\,z = y\), and the conjugate is \(z^{*} = x - iy\). Two complex numbers are equal only if their real parts are equal AND their imaginary parts are equal. To divide, multiply top and bottom by the conjugate of the bottom.

1i−1−i × i× i× i× i i¹ = ii² = −1i³ = −ii⁴ = 1
Fig. 1: each multiplication by i moves one step round the cycle 1, i, −1, −i, so the powers of i repeat every four
Section 01

Meet i.

No real number squares to \(-1\), so we invent one: \(i = \sqrt{-1}\). Then the square root of any negative number is a real multiple of \(i\), and quadratics with negative discriminant have two complex roots.

Worked example Quadratics with complex roots
Solve (a) \(z^{2} + 9 = 0\) (b) \(z^{2} - 4z + 13 = 0\).
  1. (a) \(z^{2} = -9\), so \(z = \pm\sqrt{-9} = \pm\sqrt{9}\sqrt{-1} = \pm 3i\).
  2. (b) Quadratic formula: \(z = \dfrac{4 \pm \sqrt{16 - 52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2}\).
  3. \(\sqrt{-36} = 6i\), so \(z = \dfrac{4 \pm 6i}{2} = 2 \pm 3i\).
  4. Check \(2 + 3i\): \((2 + 3i)^{2} = 4 + 12i - 9 = -5 + 12i\); then \(-5 + 12i - 8 - 12i + 13 = 0\) ✓.
Answer: (a) \(z = 3i\) or \(z = -3i\) (b) \(z = 2 + 3i\) or \(z = 2 - 3i\)
Examiner trap

\(\sqrt{-36}\) is \(6i\), not \(-6\) or \(6\). And \(i^{2} = -1\) must be used every time it appears: leaving \(-9i^{2}\) as \(-9\) is the most common lost mark in the whole topic.

Section 02

Add, subtract, multiply.

Treat \(i\) like an algebraic letter: collect real and imaginary parts when adding, expand brackets when multiplying, and replace \(i^{2}\) by \(-1\) at the end.

Worked example The three operations
Given \(z = 3 + 4i\) and \(w = 1 - 2i\), find (a) \(z + w\) (b) \(z - w\) (c) \(zw\) (d) \(zz^{*}\).
  1. (a) \((3 + 1) + (4 - 2)i = 4 + 2i\).
  2. (b) \((3 - 1) + (4 - (-2))i = 2 + 6i\).
  3. (c) \((3 + 4i)(1 - 2i) = 3 - 6i + 4i - 8i^{2}\).
  4. \(-8i^{2} = +8\), so \(zw = 11 - 2i\).
  5. (d) \((3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^{2} = 9 + 16 = 25\).
Answer: (a) \(4 + 2i\) (b) \(2 + 6i\) (c) \(11 - 2i\) (d) \(25\), which is real
Your turn
Expand and simplify \((1 + i)^{4}\).
Examiner trap

\(zz^{*} = x^{2} + y^{2}\), always real and never negative. If your answer to \((3 + 4i)(3 - 4i)\) is \(-7\), you used \(i^{2} = +1\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Simplify \(i^{3}\).

Quick check+10 XP

Find \((2 + 3i)(2 - 3i)\).

Quick check+10 XP

What is the conjugate of \(-4 + 5i\)?

Quick check+10 XP

Solve \(z^{2} + 16 = 0\).

Section 03

Dividing with the conjugate.

A complex denominator is made real by multiplying top and bottom by its conjugate, since \((c + id)(c - id) = c^{2} + d^{2}\). The syllabus asks for full details of this working, so show every line.

Worked example Division
Express \(\dfrac{3 + 4i}{1 - 2i}\) in the form \(x + iy\).
  1. Multiply top and bottom by the conjugate of the denominator, \(1 + 2i\).
  2. Top: \((3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i^{2} = -5 + 10i\).
  3. Bottom: \((1 - 2i)(1 + 2i) = 1 + 4 = 5\).
  4. \(\dfrac{-5 + 10i}{5} = -1 + 2i\). Check: \((-1 + 2i)(1 - 2i) = -1 + 2i + 2i + 4 = 3 + 4i\) ✓.
Answer: \(-1 + 2i\)
Worked example Equating real and imaginary parts
Find the real numbers \(x\) and \(y\) such that \((x + iy)(2 - i) = 7 + 4i\).
  1. Expand: \(2x - xi + 2yi - y i^{2} = (2x + y) + (2y - x)i\).
  2. Real parts: \(2x + y = 7\). Imaginary parts: \(2y - x = 4\).
  3. From the second, \(x = 2y - 4\); then \(4y - 8 + y = 7\), so \(y = 3\) and \(x = 2\).
  4. Check: \((2 + 3i)(2 - i) = 4 - 2i + 6i + 3 = 7 + 4i\) ✓.
Answer: \(x = 2\), \(y = 3\)
Worked example An equation involving the conjugate
Solve \(z + 2z^{*} = 6 - 3i\).
  1. Let \(z = x + iy\), so \(z^{*} = x - iy\).
  2. \(x + iy + 2x - 2iy = 3x - iy\).
  3. Equate: \(3x = 6\) and \(-y = -3\).
Answer: \(x = 2\), \(y = 3\), so \(z = 2 + 3i\).
Your turn
Express \(\dfrac{1}{2 + i}\) in the form \(x + iy\).
Examiner trap

You cannot divide parts separately: \(\dfrac{3 + 4i}{1 - 2i}\) is NOT \(3 - 2i\). Only a REAL denominator can be shared out between the real and imaginary parts.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.

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Express \(\dfrac{5 + i}{2 - 3i}\) in the form \(x + iy\).

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Solve \(z^{2} - 6z + 34 = 0\).

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Find the real numbers \(x\) and \(y\) such that \((x + iy)(1 + 2i) = -4 + 7i\).

Exam style+20 XP

Solve \(z + 3iz^{*} = 11 + 9i\).

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Complex arithmetic, sorted.

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Go deeper · practice
Six original Cambridge-style questions
The four operations, a quadratic with complex roots, an equation with a conjugate, powers of 1 + i, and a division problem with a real-part condition.