# Complex Numbers

> CIE A-Level Further Mathematics · 9231 Further Pure 2
> Source: https://www.owlsprep.com/study/cie-9231-u2-complex-numbers/

This sub-topic covers advanced representations of complex numbers, de Moivre's theorem, nth roots, and geometric loci on the Argand diagram, building on basic complex number knowledge for CIE A-Level Further Mathematics.

**Prerequisites:** [Basic complex numbers (Cartesian form, arithmetic)](https://www.owlsprep.com/study/cie-9231-as-complex-numbers/); [Trigonometric identities](https://www.owlsprep.com/study/cie-9231-p1-trigonometric-identities/)

## Learning objectives

- Convert complex numbers between Cartesian, polar and exponential forms
- Apply de Moivre's theorem to powers and roots of complex numbers
- Find nth roots of unity and solve polynomial equations with complex roots
- Identify and sketch loci of complex numbers on the Argand diagram

## Representations of Complex Numbers

**Polar and Exponential Form** — For a complex number $z = x + iy$, $r = |z| = \sqrt{x^2 + y^2}$ is the modulus, and $\theta = \arg z$ is the argument, normally taken in the interval $(-\pi, \pi]$

*Notation:* $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$

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

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

**Worked example:** Convert $z = -1 + \sqrt{3}i$ 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 $\alpha$:
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. $$\text{Polar: } 2\left(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right), \quad \text{Exponential: } 2e^{2i\pi/3}$$

## de Moivre's Theorem and Applications

**de Moivre's Theorem** — For any integer $n$, $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$. For any real $n$, the identity holds for one principal value of the right-hand side.

*Example:* $(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6})^2 = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3}$

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 $\cos 3\theta$ in terms of $\cos\theta$

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 $\sin^2\theta = 1 - \cos^2\theta$ to get the result in terms of $\cos\theta$ only:
6. $$\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta$$

> **tip**
>
> Always remember to raise the modulus to the power $n$ when applying de Moivre's theorem. For $(r(\cos\theta + i\sin\theta))^n$, the result is $r^n(\cos n\theta + i\sin n\theta)$.

## nth Roots of Complex Numbers

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.

## Loci on the Argand Diagram

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 $z_2$
- $\arg(z - z_0) = \alpha$: Half-line starting at $z_0$, at angle $\alpha$ to the positive real axis
- $|z - z_0| = r$: Circle, centre $z_0$, radius $r$

## Common pitfalls

- **Wrong:** Taking $\arg z = \arctan(y/x)$ directly for points in quadrants 2 or 3
  - Why it fails: The standard arctan function only returns values between $-\pi/2$ and $\pi/2$, so this gives an incorrect angle for points not in quadrants 1 or 4
  - Correct: Calculate the reference angle, then adjust based on the quadrant: add $\pi$ for third quadrant, $\pi - \alpha$ for second quadrant, and $-\pi + \alpha$ for negative angles below $-\pi$
- **Wrong:** Stopping at $k=0$ when finding nth roots, giving only one root
  - Why it fails: Every non-zero complex number has exactly $n$ distinct nth roots, so you must generate all values
  - Correct: Use $k = 0, 1, ..., n-1$ to generate all distinct roots, then stop to avoid repeating values
- **Wrong:** Misidentifying the centre of the locus $|z - a + ib| = r$
  - Why it fails: Sign errors when rewriting the expression in standard form $|z - z_0| = r$
  - Correct: Rewrite the expression as $|z - (a - ib)| = r$ to clearly see that the centre is $z_0 = a - ib$ at coordinates $(a, -b)$
- **Wrong:** Including the starting point $z_0$ in the locus $\arg(z - z_0) = \alpha$
  - Why it fails: $\arg(z - z_0)$ is undefined when $z = z_0$, since zero has no defined angle
  - Correct: Draw the starting point of the half-line as an open circle to indicate it is not included in the locus
- **Wrong:** Applying de Moivre's theorem to complex numbers in Cartesian form
  - Why it fails: The theorem only applies to complex numbers written in polar form, direct application to $x + iy$ gives an incorrect result
  - Correct: Always convert to polar form first before applying de Moivre's theorem, then convert back to Cartesian if required by the question

## Cheatsheet

| Concept | Expression / Form | Key Property |
| --- | --- | --- |
| Cartesian | $z = x + iy$ | $\|z\| = \sqrt{x^2+y^2}, \bar{z} = x - iy$ |
| Polar | $z = r(\cos\theta + i\sin\theta)$ | $r = \|z\|, \arg z \in (-\pi, \pi]$ |
| Exponential | $z = re^{i\theta}$ | $z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}$ |
| de Moivre's Theorem | $(r(\cos\theta + i\sin\theta))^n$ | $= r^n(\cos n\theta + i\sin n\theta)$ |
| nth Roots of $w = R(\cos\phi + i\sin\phi)$ | $z_k = R^{1/n}\left(\cos \frac{\phi + 2k\pi}{n} + i\sin \frac{\phi + 2k\pi}{n}\right)$ | $k = 0, 1, ..., n-1$ 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 $r$ |
| $\arg(z - z_0) = \alpha$ | Half-line | Starts at open circle at $z_0$ |

## 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.

- [Second Order Differential Equations](https://www.owlsprep.com/study/cie-9231-u2-differential-equations/)
- [Further Mechanics](https://www.owlsprep.com/study/cie-9231-u3-overview/)

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