VCE Specialist Mathematics (Year 12) · Complex Numbers · 22m read

Complex Numbers — Introduction and the Complex Plane

Ever wondered how engineers describe waves and electrical circuits mathematically? Complex numbers combine real and imaginary parts (like 3 + 2i) to do just that. You'll learn to visualize them on the complex plane, switch between Cartesian and polar forms, and perform calculations with them—skills that unlock careers in engineering, physics, and digital technology.

Learning objective: Define complex numbers, represent them in Cartesian and polar forms, and perform arithmetic operations in both forms.

By the end of this lesson you can:

• I define $i = \sqrt{-1}$ and represent $z = a + bi$ in the Argand diagram.
• I find modulus $|z| = \sqrt{a^2+b^2}$ and argument $\arg(z)$.
• I convert between Cartesian $z = a+bi$ and polar $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ forms.
• I perform $+$, $-$, $\times$, $\div$ and conjugation in Cartesian form.
• I use de Moivre's theorem to find powers and roots.

Complex numbers add a second dimension to the number line. Real numbers live on a line; complex numbers live on a map — the Argand plane — with real and imaginary directions.

The imaginary unit and complex number arithmetic

### The problem with real numbers

The equation $x^2 + 1 = 0$ has no real solution because $x^2 \geq 0$ for all real $x$. Complex numbers extend the reals by introducing:

$$i = \sqrt{-1} \ \implies \ i^2 = -1$$

### Definition

A complex number $z$ has the form:

$$z = a + bi \qquad a, b \in \mathbb{R}$$

where $a = \text{Re}(z)$ is the real part and $b = \text{Im}(z)$ is the imaginary part.

> Note: $b$ is real — it's the coefficient of $i$, not itself imaginary.

### Powers of $i$

$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \ldots$$

The powers cycle with period 4: $i^{4k} = 1$, $i^{4k+1} = i$, $i^{4k+2} = -1$, $i^{4k+3} = -i$.

### Arithmetic in Cartesian form

Addition/Subtraction: Real parts with real, imaginary parts with imaginary.

$$(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i$$

Multiplication: FOIL and use $i^2 = -1$.

$$(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$$

Complex conjugate: $\bar{z} = a - bi$

Properties: $z \bar{z} = a^2 + b^2 = |z|^2 \geq 0$ (always real and non-negative).

Division: Multiply numerator and denominator by the conjugate of the denominator.

$$\frac{a + bi}{c + di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd) + (bc-ad)i}{c^2+d^2}$$

The Argand diagram and polar form

### Argand diagram

Every complex number $z = a + bi$ maps to the point $(a, b)$ in the Argand plane (complex plane):
- Horizontal axis: real part ($a$)
- Vertical axis: imaginary part ($b$)

### Modulus and argument

The modulus $|z|$ is the distance from the origin:
$$|z| = r = \sqrt{a^2 + b^2}$$

The argument $\arg(z) = \theta$ is the angle from the positive real axis:
$$\theta = \tan^{-1}\!\left(\frac{b}{a}\right) \quad \text{(adjust for quadrant)}$$

The principal argument $\text{Arg}(z) \in (-\pi, \pi]$.

### Polar form

$$z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta$$

where $\text{cis}\,\theta$ is shorthand for $\cos\theta + i\sin\theta$.

### Euler's formula

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

So: $z = re^{i\theta}$ (exponential form — used in advanced work).

### Conversion

Cartesian to Polar:
$$r = \sqrt{a^2+b^2}, \quad \theta = \text{Arg}(a + bi) \quad \text{(use quadrant)}$$

Polar to Cartesian:
$$a = r\cos\theta, \quad b = r\sin\theta$$

### Multiplication and division in polar form

$$z_1 z_2 = r_1 r_2 \, \text{cis}(\theta_1 + \theta_2)$$
$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \, \text{cis}(\theta_1 - \theta_2)$$

Geometrically: multiplication scales by $r_2$ and rotates by $\theta_2$.

De Moivre's theorem — powers and roots

### De Moivre's theorem

For any real $n$ and $z = r\,\text{cis}\,\theta$:

$$z^n = r^n\,\text{cis}(n\theta) = r^n(\cos n\theta + i\sin n\theta)$$

This is a direct consequence of the polar multiplication rule applied $n$ times.

### Roots of complex numbers

The $n$-th roots of $w = R\,\text{cis}\,\phi$ are the $n$ complex numbers:

$$z_k = R^{1/n}\,\text{cis}\!\left(\frac{\phi + 2k\pi}{n}\right), \quad k = 0, 1, 2, \ldots, n-1$$

Geometrically, the $n$ roots are equally spaced around a circle of radius $R^{1/n}$.

### Roots of unity

The $n$ solutions to $z^n = 1$ are:

$$z_k = \text{cis}\!\left(\frac{2k\pi}{n}\right), \quad k = 0, 1, \ldots, n-1$$

These are equally spaced at $\frac{360°}{n}$ around the unit circle.

Complex Numbers Formula Sheet

• Modulus: |z| = \sqrt{a^2 + b^2} (Distance from origin in Argand plane)
• Conjugate: \bar{z} = a - bi (Flip sign of imaginary part)
• Conjugate product: z\bar{z} = |z|^2 = a^2+b^2 (Always real and $\geq 0$)
• Division: \dfrac{z_1}{z_2} = \dfrac{z_1 \bar{z}_2}{|z_2|^2} (Multiply by conjugate/conjugate)
• Polar form: z = r\,\text{cis}\,\theta = re^{i\theta} ($r = |z|$, $\theta = \arg(z)$)
• Polar multiply: z_1 z_2 = r_1 r_2\,\text{cis}(\theta_1+\theta_2) (Multiply moduli, add arguments)
• De Moivre: z^n = r^n\,\text{cis}(n\theta) (For any real $n$)
• $n$-th roots: z_k = R^{1/n}\,\text{cis}\!\left(\frac{\phi + 2k\pi}{n}\right) ($k = 0, 1, \ldots, n-1$)

Complex number problem strategy

• Division: multiply by conjugate: Multiply top and bottom by $\bar{z}_{\text{denominator}}$ to make denominator real.
• Powers: use polar form: Convert to polar first, apply de Moivre, convert back if needed.
• Argument: adjust for quadrant: $\tan^{-1}(b/a)$ gives an angle in $(-90°, 90°)$ — check quadrant of $z$ and adjust.
• Roots: $n$ equally spaced: Start at $k=0$, add $2\pi/n$ (or $360°/n$) for each subsequent root.
• Argand loci: $|z - a| = r$ is a circle; $\arg(z - a) = \theta$ is a ray.

Quick reference

| Concept | Rule |
|---|---|
| i² | −1 |
| z = a + bi | Cartesian form |
| Conjugate | a − bi |
| Modulus | √(a² + b²) |
| Division | Multiply by conjugate |

Worked example (Easy)

Let $z_1 = 3 + 2i$ and $z_2 = 1 - i$. Find (a) $z_1 + z_2$, (b) $z_1 z_2$, (c) $z_1/z_2$.

1. (a) Addition. z_1 + z_2 = (3+1) + (2-1)i = 4 + i
2. (b) Multiplication. FOIL and use $i^2 = -1$: (3+2i)(1-i) = 3 - 3i + 2i - 2i^2 = 3 - i + 2 = 5 - i
3. (c) Division — multiply by conjugate of denominator. Conjugate of $1-i$ is $1+i$: \frac{3+2i}{1-i} = \frac{(3+2i)(1+i)}{(1-i)(1+i)} = \frac{3 + 3i + 2i + 2i^2}{1+1} = \frac{3 + 5i - 2}{2} = \frac{1+5i}{2}

Answer: (a) $4+i$; (b) $5-i$; (c) $\frac{1}{2} + \frac{5}{2}i$.

Worked example (Medium)

Express $z = (1 + i)(2 - 3i)$ in $a + bi$ form and find $|z|$.

1. Expand. Distribute. (1)(2) + (1)(-3i) + (i)(2) + (i)(-3i) = 2 - 3i + 2i + 3 = 5 - i
2. Modulus. $|a + bi| = \sqrt{a^2 + b^2}$. |z| = \sqrt{25 + 1} = \sqrt{26}

Answer: $z = 5 - i,\; |z| = \sqrt{26}$.

Worked example (Medium)

Express $z = -1 + \sqrt{3}\,i$ in polar form. Hence find $z^6$.

1. Modulus. |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2
2. Argument — $z$ is in Q2. $\tan^{-1}(\sqrt{3}/|-1|) = \tan^{-1}(\sqrt{3}) = 60°$, but Q2, so $\theta = 180° - 60° = 120° = \frac{2\pi}{3}$. \arg(z) = \frac{2\pi}{3}
3. Polar form. z = 2\,\text{cis}\!\left(\frac{2\pi}{3}\right)
4. Apply de Moivre. z^6 = 2^6\,\text{cis}\!\left(6 \cdot \frac{2\pi}{3}\right) = 64\,\text{cis}(4\pi) = 64(\cos 4\pi + i\sin 4\pi)
5. Evaluate. $\cos 4\pi = 1$, $\sin 4\pi = 0$: z^6 = 64 \cdot 1 + 64 \cdot 0 \cdot i = 64

Answer: $z = 2\,\text{cis}(2\pi/3)$; $z^6 = 64$ (a real number!).

Worked example (Hard)

Find all cube roots of $8i$. Express in Cartesian form.

1. Write $8i$ in polar form. $8i = 8(\cos 90° + i\sin 90°) = 8\,\text{cis}(\pi/2)$. R = 8, \quad \phi = \pi/2
2. Apply root formula. $n = 3$, roots are $z_k = 8^{1/3}\,\text{cis}\!\left(\frac{\pi/2 + 2k\pi}{3}\right)$ for $k = 0, 1, 2$. 8^{1/3} = 2
3. $k = 0$. z_0 = 2\,\text{cis}\!\left(\frac{\pi}{6}\right) = 2\!\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \sqrt{3} + i
4. $k = 1$. z_1 = 2\,\text{cis}\!\left(\frac{\pi/2 + 2\pi}{3}\right) = 2\,\text{cis}\!\left(\frac{5\pi}{6}\right) = 2\!\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = -\sqrt{3} + i
5. $k = 2$. z_2 = 2\,\text{cis}\!\left(\frac{\pi/2 + 4\pi}{3}\right) = 2\,\text{cis}\!\left(\frac{3\pi}{2}\right) = 2(0 - i) = -2i

Answer: Cube roots of $8i$: $z = \sqrt{3}+i$, $z = -\sqrt{3}+i$, $z = -2i$. (Equally spaced at $120°$ apart on circle of radius 2.)

Common mistakes

• Forgetting that argument is not simply $\tan^{-1}(b/a)$ $\tan^{-1}$ returns values in $(-90°, 90°)$ only. A complex number in Q2 has $a < 0$, $b > 0$, so $\tan^{-1}(b/a)$ gives a negative angle — the wrong quadrant entirely. Fix: Always locate $z = a + bi$ in the correct quadrant. Q1: $\theta = \tan^{-1}(b/a)$. Q2: $\theta = \pi - \tan^{-1}(|b|/|a|)$. Q3: $\theta = -\pi + \tan^{-1}(|b|/|a|)$. Q4: $\theta = -\tan^{-1}(|b|/|a|)$.
• Only giving one root when asked for all $n$-th roots There are exactly $n$ distinct $n$-th roots. A common error is stopping at $k=0$. Fix: State all $n$ values: $k = 0, 1, 2, \ldots, n-1$. Plot them on an Argand diagram to check they're equally spaced.

Proof of Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$

Euler's formula connects exponentials and trigonometry in a stunning way. It can be derived via Maclaurin series:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

Substitute $x = i\theta$:

$$e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \cdots$$

Use $i^2 = -1$, $i^3 = -i$, $i^4 = 1$:

$$= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots$$

Separate real and imaginary parts:

$$= \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right)}_{\cos\theta} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\sin\theta}$$

Therefore:
$$\boxed{e^{i\theta} = \cos\theta + i\sin\theta}$$

The most famous special case ($\theta = \pi$):

$$e^{i\pi} + 1 = 0$$

This Euler's identity combines $e$, $i$, $\pi$, $1$ and $0$ in a single equation — often called the most beautiful equation in mathematics.

Practice questions

1. What is $i^{23}$?

   1. $i$
   2. $-i$
   3. $1$
   4. $-1$

   (1 marks)

2. Find $\dfrac{2+3i}{1+2i}$ in the form $a + bi$.

   (2 marks)

3. Express $z = -2 - 2i$ in polar form $r\,\text{cis}\,\theta$ where $\theta \in (-\pi, \pi]$.

   (3 marks)

4. Use de Moivre's theorem to express $\cos 3\theta$ in terms of $\cos\theta$.

   (3 marks)

5. Find all fourth roots of $16$. Express in polar form and plot on the Argand diagram.

   (4 marks)

Summary

• $z = a + bi$; $i^2 = -1$; conjugate $\bar{z} = a - bi$; $|z| = \sqrt{a^2+b^2}$.
• Division: multiply numerator and denominator by conjugate of denominator.
• Polar: $z = r\,\text{cis}\,\theta = re^{i\theta}$; multiply in polar = multiply moduli + add arguments.
• De Moivre: $z^n = r^n\,\text{cis}(n\theta)$.
• $n$ roots of $Re^{i\phi}$: $R^{1/n}\,\text{cis}((\phi+2k\pi)/n)$, $k = 0,\ldots,n-1$.
• Argument adjustment: $\tan^{-1}$ only correct in Q1/Q4 — check quadrant always.