Complex Numbers — IB Math AA (HL)

1. Introduction to Complex Numbers

Complex numbers extend the real number system to include solutions of equations like $x^2 + 1 = 0$. We define the imaginary unit $i$ such that $i^2 = -1$. Every complex number has a real part and an imaginary part.

Cartesian Form

A complex number: $z = a + bi$, where $a = \operatorname{Re}(z)$ (real part) and $b = \operatorname{Im}(z)$ (imaginary part).

Modulus: $|z| = \sqrt{a^2 + b^2}$

Argument: $\arg(z) = \theta$ where $\tan\theta = \dfrac{b}{a}$, adjusted for the correct quadrant. The principal argument satisfies $-\pi < \theta \le \pi$.

Complex Conjugate: $\bar{z} = z^* = a - bi$

Key conjugate properties: $z \cdot \bar{z} = |z|^2$ and $z + \bar{z} = 2\operatorname{Re}(z)$. To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator.

Operations in Cartesian Form

Addition: $(a+bi) + (c+di) = (a+c) + (b+d)i$

Multiplication (using $i^2=-1$): $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$

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

2. Polar Form and the Argand Diagram

Every complex number $z = a+bi$ can be represented as a point $(a,b)$ in the Argand diagram. The polar form uses modulus $r=|z|$ and argument $\theta = \arg(z)$.

Polar Form

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

where $r = |z| \ge 0$ and $\theta = \arg(z)$.

From Cartesian: $r = \sqrt{a^2+b^2}$, $\theta = \arctan\!\left(\dfrac{b}{a}\right)$ (adjust for quadrant!)

To Cartesian: $a = r\cos\theta$, $b = r\sin\theta$

Key Insight — Multiplication in Polar Form: When multiplying two complex numbers in polar form, multiply the moduli and add the arguments: $$z_1 z_2 = r_1 r_2 \,[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)]$$

Worked Example 1

Convert $z = 1 + \sqrt{3}\,i$ to polar form

Find the modulus: $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$

Find the argument: $\tan\theta = \dfrac{\sqrt{3}}{1} = \sqrt{3}$. Since $a = 1 > 0$ and $b = \sqrt{3} > 0$ (Quadrant I), $\theta = \dfrac{\pi}{3}$.

Write polar form: $z = 2\!\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)$

3. De Moivre's Theorem

De Moivre's Theorem

For any integer $n$ and $z = r(\cos\theta + i\sin\theta)$:

$$z^n = r^n(\cos n\theta + i\sin n\theta)$$

De Moivre's theorem also derives multiple angle identities. With $n=2$: expanding $(\cos\theta + i\sin\theta)^2$ and equating real/imaginary parts recovers the double angle formulas.

Worked Example 2

Find $(1+i)^8$ using de Moivre's theorem

Polar form: $r = \sqrt{2}$, $\theta = \pi/4$. So $1+i = \sqrt{2}\,\operatorname{cis}\dfrac{\pi}{4}$.

Apply de Moivre: $(1+i)^8 = (\sqrt{2})^8 \cdot \operatorname{cis}\!\left(8 \cdot \dfrac{\pi}{4}\right) = 16 \cdot \operatorname{cis}(2\pi)$

Simplify: $\operatorname{cis}(2\pi) = 1$, so $(1+i)^8 = \boxed{16}$

4. Roots of Complex Numbers

To find the $n$-th roots of $w = R\,\operatorname{cis}\,\phi$, we solve $z^n = w$. There are exactly $n$ distinct complex roots, equally spaced on a circle of radius $R^{1/n}$.

The $n$-th Roots Formula

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

Cube Roots of Unity

The three cube roots of $1$ (i.e., solutions of $z^3 = 1$) are: $1,\; \omega,\; \omega^2$ where

$$\omega = \operatorname{cis}\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \qquad \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$$

HL Key Property: The cube roots of unity satisfy $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. Note also that $\omega^2 = \bar{\omega}$. These identities appear frequently in HL Paper 1.

Worked Example 3

Solve $z^3 = 8$

Polar form of 8: $8 = 8\,\operatorname{cis}\,0$ (real positive, $r=8$, $\phi=0$).

Roots formula ($n=3$): $z_k = 8^{1/3}\,\operatorname{cis}\!\left(\dfrac{2\pi k}{3}\right) = 2\,\operatorname{cis}\!\left(\dfrac{2\pi k}{3}\right)$

Three roots:

$k=0$: $z_0 = 2$
$k=1$: $z_1 = 2\,\operatorname{cis}\dfrac{2\pi}{3} = -1 + \sqrt{3}\,i$
$k=2$: $z_2 = 2\,\operatorname{cis}\dfrac{4\pi}{3} = -1 - \sqrt{3}\,i$

Practice Problems

Q1. Express $z = -2 + 2i$ in polar form. Hence find $z^6$.

Show Solution

$r = \sqrt{4+4} = 2\sqrt{2}$. Since $a < 0$, $b > 0$ (Quadrant II): $\theta = \pi - \pi/4 = 3\pi/4$. So $z = 2\sqrt{2}\,\operatorname{cis}\dfrac{3\pi}{4}$.

$z^6 = (2\sqrt{2})^6\,\operatorname{cis}(9\pi/2) = 512\,\operatorname{cis}(\pi/2) = 512i$

Q2. Find all solutions to $z^4 = -16$, expressing answers in Cartesian form.

Show Solution

$-16 = 16\,\operatorname{cis}\,\pi$. Roots: $z_k = 2\,\operatorname{cis}\!\left(\dfrac{(2k+1)\pi}{4}\right)$ for $k=0,1,2,3$.

$z_0 = \sqrt{2}+\sqrt{2}i$, $z_1 = -\sqrt{2}+\sqrt{2}i$, $z_2 = -\sqrt{2}-\sqrt{2}i$, $z_3 = \sqrt{2}-\sqrt{2}i$

Q3. If $\omega$ is a non-real cube root of unity, show that $(1+\omega)(1+\omega^2) = 1$.

Show Solution

From $1+\omega+\omega^2=0$: $1+\omega^2 = -\omega$ and $1+\omega = -\omega^2$.

$(1+\omega)(1+\omega^2) = (-\omega^2)(-\omega) = \omega^3 = 1$. $\square$

5. Exam Tips

Always verify the quadrant when computing $\arg(z)$; $\arctan(b/a)$ alone is insufficient for Q2 and Q3.
When finding $n$-th roots, generate all $n$ values using $k = 0, 1, \ldots, n-1$.
For de Moivre, convert to polar first — never expand $(a+bi)^n$ directly for large $n$.
Remember: $|z_1 z_2| = |z_1||z_2|$ and $\arg(z_1 z_2) = \arg z_1 + \arg z_2$ (mod $2\pi$).
Roots of unity are equally spaced on a circle — sketch the Argand diagram to verify spacing.