Complex Numbers

CIE A-Level Further Mathematics· Further Pure 2 Unit 2 Topic 4· 15 min read

1. Representations of Complex Numbers★★☆☆☆⏱ 5 min

📘 Definition

Polar and Exponential Form

$z = r (cos θ + i sin θ) = r e^{i θ}$

For a complex number $z = x + i y$ , $r = ∣ z ∣ = x^{2} + y^{2}$ is the modulus, and $θ = ar g z$ is the argument, normally taken in the interval $(- π, π]$

Example:

$z = 1 + i$ has modulus $2$ and argument $\frac{π}{4}$ , so polar form $2 (cos \frac{π}{4} + i sin \frac{π}{4})$ and exponential form $2 e^{iπ /4}$

Converting between forms uses standard coordinate geometry relationships: $x = r cos θ$ , $y = r sin θ$ , so you can always get from polar to Cartesian directly. To convert from Cartesian to polar, calculate $r$ first, then find $θ$ adjusted for the quadrant of the point.

📐 Worked Example

Convert $z = - 1 + 3 i$ to polar and exponential form

1
Step 1: Calculate the modulus $r$ :
2
$r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$
3
Step 2: Find the argument. $z$ lies in the second quadrant, so first find the reference angle $α$ :
4
$\tan\alpha = \frac{|\sqrt{3}|}{|-1|} = \sqrt{3} \implies \alpha = \frac{\pi}{3}, \quad \arg z = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
5
Step 3: Write both final forms:
6
$\text{Polar: } 2\left(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right), \quad \text{Exponential: } 2e^{2i\pi/3}$

2. de Moivre's Theorem and Applications★★★☆☆⏱ 6 min

📘 Definition

de Moivre's Theorem

For any integer $n$ , $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ . For any real $n$ , the identity holds for one principal value of the right-hand side.

Example:

$(cos \frac{π}{6} + i sin \frac{π}{6})^{2} = cos \frac{π}{3} + i sin \frac{π}{3}$

de Moivre's theorem simplifies calculating powers of complex numbers when they are in polar form, and can also be used to derive multiple-angle trigonometric identities by equating real and imaginary parts after binomial expansion.

📐 Worked Example

Use de Moivre's theorem to find $cos 3 θ$ in terms of $cos θ$

1
Step 1: Apply de Moivre's theorem and expand via binomial theorem:
2
$\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta + 3i^2\cos\theta\sin^2\theta + i^3\sin^3\theta$
3
Step 2: Simplify using $i^{2} = - 1$ , then equate real parts:
4
$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$
5
Step 3: Substitute $sin^{2} θ = 1 - cos^{2} θ$ to get the result in terms of $cos θ$ only:
6
$\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta$

3. nth Roots of Complex Numbers★★★☆☆⏱ 6 min

Any non-zero complex number has exactly $n$ distinct $n$ th roots. These roots are equally spaced around a circle of radius $r^{1/ n}$ centered at the origin on the Argand diagram. A common special case is the $n$ th roots of unity, which satisfy $z^{n} = 1$ .

📐 Worked Example

Find all cube roots of unity, in exact Cartesian form

1
Step 1: Write 1 in general polar form to capture all angles:
2
$1 = 1(\cos 2k\pi + i\sin 2k\pi), \quad k = 0, 1, 2$
3
Step 2: Apply de Moivre's theorem for roots:
4
$z^{1/3} = 1^{1/3}\left(\cos \frac{2k\pi}{3} + i\sin \frac{2k\pi}{3}\right)$
5
Step 3: Calculate each distinct root:
6
$k=0: 1(\cos 0 + i\sin 0) = 1 \\ k=1: \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \\ k=2: \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$

For polynomials with real coefficients, all complex roots occur in conjugate pairs: if $a + ib$ is a root, then $a - ib$ is also a root. This property can be used to factorise higher-degree polynomials completely.

4. Loci on the Argand Diagram★★★★☆⏱ 7 min

A locus is a set of points satisfying a given condition. Common loci for CIE exams are defined using modulus and argument, and can be sketched or described algebraically. The most common types are circles, perpendicular bisectors, and half-lines.

📐 Worked Example

Describe the locus of points $z$ satisfying $∣ z - 2 + i ∣ = 3$

1
Step 1: Rewrite the condition in the standard form $∣ z - z_{0} ∣ = r$ :
2
$|z - (2 - i)| = 3$
3
Step 2: Recall that $∣ z - z_{0} ∣$ equals the distance between point $z$ and fixed point $z_{0}$ on the Argand diagram.
4
Step 3: The condition means all points a fixed distance 3 from $z_{0} = 2 - i$ .
5
$\text{Conclusion: Locus is a circle with centre } (2, -1) \text{ and radius } 3$

$∣ z - z_{1} ∣ = ∣ z - z_{2} ∣$ : Perpendicular bisector of the segment joining $z_{1}$ and $z_{2}$
$ar g (z - z_{0}) = α$ : Half-line starting at $z_{0}$ , at angle $α$ to the positive real axis
$∣ z - z_{0} ∣ = r$ : Circle, centre $z_{0}$ , radius $r$

5. Common Pitfalls

Wrong move:

Taking $ar g z = arctan (y / x)$ directly for points in quadrants 2 or 3

Why:

The standard arctan function only returns values between $- π /2$ and $π /2$ , so this gives an incorrect angle for points not in quadrants 1 or 4

Correct move:

Calculate the reference angle, then adjust based on the quadrant: add $π$ for third quadrant, $π - α$ for second quadrant, and $- π + α$ for negative angles below $- π$

Wrong move:

Stopping at $k = 0$ when finding nth roots, giving only one root

Why:

Every non-zero complex number has exactly $n$ distinct nth roots, so you must generate all values

Correct move:

Use $k = 0, 1, ..., n - 1$ to generate all distinct roots, then stop to avoid repeating values

Wrong move:

Misidentifying the centre of the locus $∣ z - a + ib ∣ = r$

Why:

Sign errors when rewriting the expression in standard form $∣ z - z_{0} ∣ = r$

Correct move:

Rewrite the expression as $∣ z - (a - ib) ∣ = r$ to clearly see that the centre is $z_{0} = a - ib$ at coordinates $(a, - b)$

Wrong move:

Including the starting point $z_{0}$ in the locus $ar g (z - z_{0}) = α$

Why:

$ar g (z - z_{0})$ is undefined when $z = z_{0}$ , since zero has no defined angle

Correct move:

Draw the starting point of the half-line as an open circle to indicate it is not included in the locus

Wrong move:

Applying de Moivre's theorem to complex numbers in Cartesian form

Why:

The theorem only applies to complex numbers written in polar form, direct application to $x + i y$ gives an incorrect result

Correct move:

Always convert to polar form first before applying de Moivre's theorem, then convert back to Cartesian if required by the question

6. Quick Reference Cheatsheet

Concept	Expression / Form	Key Property
Cartesian	$z = x + i y$	$∣ z ∣ = x^{2} + y^{2}, \overset{z}{ˉ} = x - i y$
Polar	$z = r (cos θ + i sin θ)$	$r = ∣ z ∣, ar g z \in (- π, π]$
Exponential	$z = r e^{i θ}$	$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$
de Moivre's Theorem	$(r (cos θ + i sin θ))^{n}$	$= r^{n} (cos n θ + i sin n θ)$
nth Roots of $w = R (cos ϕ + i sin ϕ)$	$z_{k} = R^{1/ n} (cos \frac{ϕ + 2 k π}{n} + i sin \frac{ϕ + 2 k π}{n})$	$k = 0, 1, ..., n - 1$ for n distinct roots
$∣ z - z_{1} ∣ = ∣ z - z_{2} ∣$	Perpendicular bisector	Of line segment joining $z_{1}$ and $z_{2}$
$∣ z - z_{0} ∣ = r$	Circle	Centre $z_{0}$ , radius $r$
$ar g (z - z_{0}) = α$	Half-line	Starts at open circle at $z_{0}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 2
Find 5th roots of a complex number
2022 · 1
Sketch locus of complex number
2021 · 2
Use de Moivre to derive trig identity

What's Next

This sub-topic builds the foundational knowledge of complex numbers required for all further study in pure and applied mathematics. Mastery of polar form, de Moivre's theorem, and loci is essential for topics like complex integration, series, and differential equations, and complex numbers are widely used in physics and engineering applications. This topic frequently combines with polynomials and trigonometry in exam questions, so solid understanding will help you earn marks across multiple sections of the paper.