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Module 12 · L09 of 16 ~40 min ⚡ +90 XP available

Powers of Complex Numbers — Introduction

Squaring a complex number is just $(a+bi)(a+bi)$. But to raise $z$ to the 5th or 10th power that way is brutal — the algebra explodes. The fix comes from polar form. Writing $z = r(\cos\theta + i\sin\theta)$ converts multiplication into multiplying moduli and adding arguments, so $z^n$ becomes a one-line calculation: $r^n(\cos n\theta + i\sin n\theta)$. This lesson sets up the pattern that will become de Moivre's theorem in Module 13.

Today's hook — Without expanding, predict the modulus and argument of $(1+i)^8$. Hint: write $1+i$ in polar form first — what are its modulus and argument? Then think about what happens to those two numbers when you take an 8th power. Compare your guess after card 05.

0/5QUESTS

Recall — your gut answer first

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Take $z = 1 + i$. Without using polar form, compute $z^2$ and $z^3$ by direct multiplication. What do you notice about how the modulus changes from $z$ to $z^2$ to $z^3$? Sketch your work below.

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The two moves for computing $z^n$

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Every $z^n$ question rewards two habits: convert to polar form first (find $r$ and $\theta$), then raise the modulus and multiply the argument. Mixing these up — using Cartesian form for high powers — is the single biggest cause of wasted time in Module 12.

The convert-raise-multiply reading: (1) write $z = r(\cos\theta + i\sin\theta)$, (2) raise the modulus to $r^n$, (3) multiply the argument by $n$ to get $n\theta$.

Multiplication rule: $z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$ · Power pattern: $z^n = r^n(\cos n\theta + i\sin n\theta)$

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

Polar form is the shortcut

For $n \geq 3$ direct expansion is slow and error-prone. Convert to polar first; the algebra reduces to two scalar operations on $r$ and $\theta$.

Multiplication adds arguments

$z_1 z_2$ has modulus $r_1 r_2$ and argument $\theta_1 + \theta_2$. Repeated multiplication by $z$ rotates by $\theta$ each time and scales by $r$.

Convert back to $a + bi$ at the end

Most HSC answers want $a + bi$ form. After raising to the power, evaluate $\cos n\theta$ and $\sin n\theta$ — these are usually exact values from the unit circle.

What you'll master

Know

Key facts

Polar form: $z = r(\cos\theta + i\sin\theta)$ where $r = |z|$ and $\theta = \arg z$
Multiplication: $z_1 z_2 = r_1 r_2[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)]$
Power pattern: $z^n = r^n(\cos n\theta + i\sin n\theta)$ (informal, for positive integer $n$)
$|z^n| = |z|^n$ and $\arg(z^n) = n \arg z$ (modulo $2\pi$)

Understand

Concepts

Why polar form turns multiplication into addition of arguments
Why each successive power scales by $r$ and rotates by $\theta$ on the Argand diagram
Why the pattern $z^n = r^n(\cos n\theta + i\sin n\theta)$ generalises to de Moivre's theorem (Module 13)

Can do

Skills

Convert $a + bi$ to polar form $r(\cos\theta + i\sin\theta)$
Compute small integer powers of $z$ using both Cartesian and polar methods
Convert a polar-form answer back to $a + bi$

Key terms

Modulus $|z|$For $z = a + bi$, $|z| = \sqrt{a^2 + b^2}$. The distance from the origin to $z$ on the Argand diagram. Always non-negative.

Argument $\arg z$The angle $\theta$ measured from the positive real axis to the vector $z$. Principal argument lies in $(-\pi, \pi]$.

Polar form$z = r(\cos\theta + i\sin\theta)$ where $r = |z|$ and $\theta = \arg z$. Also written $r\,\text{cis}\,\theta$.

Cartesian form$z = a + bi$ where $a = \text{Re}(z)$ and $b = \text{Im}(z)$. Best for addition and subtraction; awkward for high powers.

Power $z^n$For positive integer $n$, $z^n = z \cdot z \cdots z$ ($n$ factors). In polar form, $z^n = r^n(\cos n\theta + i\sin n\theta)$.

Argand diagramA 2D plane with real axis (horizontal) and imaginary axis (vertical). Multiplication by $z$ acts as a rotation by $\arg z$ and a scaling by $|z|$.

MEX-N1NESA outcome (Complex Numbers I): uses the relationship between polar form and powers of complex numbers, leading to de Moivre's theorem.

From multiplication to powers

core concept

Start with two complex numbers in polar form: $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$. Expanding the product and using the compound-angle identities:

$$z_1 z_2 = r_1 r_2[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)]$$

So multiplication multiplies moduli and adds arguments. Now set $z_1 = z_2 = z = r(\cos\theta + i\sin\theta)$:

$z^2 = r \cdot r [\cos(\theta + \theta) + i\sin(\theta + \theta)] = r^2(\cos 2\theta + i\sin 2\theta)$
$z^3 = z^2 \cdot z = r^2 \cdot r [\cos(2\theta + \theta) + i\sin(2\theta + \theta)] = r^3(\cos 3\theta + i\sin 3\theta)$
Continuing this pattern: $z^n = r^n(\cos n\theta + i\sin n\theta)$ for every positive integer $n$.

Worked through the hook: $z = 1 + i$ has $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $\theta = \tfrac{\pi}{4}$. So $z^8 = (\sqrt{2})^8 \big(\cos 8 \cdot \tfrac{\pi}{4} + i\sin 8 \cdot \tfrac{\pi}{4}\big) = 16(\cos 2\pi + i\sin 2\pi) = 16(1 + 0) = 16$. Direct expansion of $(1+i)^8$ would take far longer.

Connecting to de Moivre. The pattern $z^n = r^n(\cos n\theta + i\sin n\theta)$ is the informal motivation. In Module 13 you will prove it formally by induction — that proof is de Moivre's theorem, and it extends to negative and rational $n$. For now, treat the formula as a powerful computational shortcut you have observed by repeated multiplication.

Multiplication rule: $r_1 r_2$ moduli, $\theta_1 + \theta_2$ arguments · $z^n = r^n(\cos n\theta + i\sin n\theta)$ for positive integers $n$ — derived by repeated multiplication · For $z = 1 + i$: $r = \sqrt{2}$, $\theta = \pi/4$; so $z^8 = 16$, $z^4 = -4$ · Always convert to polar form before raising to a high power

Pause — copy the power formula $z^n = r^n(\cos n\theta + i\sin n\theta)$, the law $|z^n| = |z|^n$, $\arg(z^n) = n\arg z$, and the worked example $(1+i)^8 = 16$ into your book.

Quick check: If $z = 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6})$, what is $z^3$ in polar form?

Computing small integer powers

core concept

We just saw that $z^n = r^n(\cos n\theta + i\sin n\theta)$ follows from repeated application of the polar product rule, with the polar route preferred for large $n$. That raises a question: for small powers like $n = 2, 3$, when is the Cartesian route faster? This card answers it → for $n = 2, 3$ Cartesian is often quicker; for large $n$ always convert to polar first; unit-circle numbers stay on the unit circle.

Two routes exist for every $z^n$ problem. Choose by the size of $n$:

Cartesian (direct expansion). Works for $n = 2$ or $n = 3$. Just compute $(a+bi)^2$ or $(a+bi)^3$ using FOIL / the binomial expansion. Remember $i^2 = -1$.
Polar (modulus + argument). Best for $n \geq 4$. Find $r$ and $\theta$, then write $z^n = r^n(\cos n\theta + i\sin n\theta)$, and convert back if needed.

Geometric intuition. Each multiplication by $z$ on the Argand plane rotates the previous power by $\theta$ and scales by $r$. So the sequence $z, z^2, z^3, \ldots$ traces a spiral (outwards if $r > 1$, inwards if $r < 1$). If $r = 1$, the powers stay on the unit circle and just rotate.

$$|z^n| = |z|^n \;,\;\; \arg(z^n) = n\arg(z) \pmod{2\pi}$$

Common mistake. Forgetting to raise the modulus. Students often write $z^n = \cos n\theta + i\sin n\theta$, dropping the $r^n$. This is only correct if $|z| = 1$. Always carry the modulus through.

Cartesian route: good for $n = 2, 3$ · Polar route: convert, multiply $\theta$ by $n$, raise $r$ to the $n$th power · $|z^n| = |z|^n$ and $\arg(z^n) = n\arg z$ (mod $2\pi$) · If $|z| = 1$, powers stay on the unit circle · If $|z| > 1$, powers spiral outward; if $|z| < 1$, inward

Pause — copy the Cartesian-vs-polar choice rule (Cartesian for $n=2,3$; polar for large $n$), the behaviour on the unit circle ($|z|=1$ stays there), and the outward/inward spiral criterion into your book.

Did you get this? True or false: if $|z| = 1$, then $|z^n| = 1$ for every positive integer $n$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · DIRECT EXPANSION

Compute $(1 + i)^3$ by direct multiplication. Give your answer in Cartesian form $a + bi$.

First find $(1+i)^2 = (1+i)(1+i) = 1 + i + i + i^2 = 1 + 2i - 1 = 2i$.

For small powers, direct expansion is fine. Always use $i^2 = -1$ to collapse the result back to $a + bi$ form.

PROBLEM 2 · USE POLAR FORM FOR A HIGH POWER

Find $(\sqrt{3} + i)^6$ in Cartesian form. Use polar form.

Convert to polar: $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$, $\theta = \tan^{-1}\!\big(\tfrac{1}{\sqrt{3}}\big) = \tfrac{\pi}{6}$. So $\sqrt{3} + i = 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6})$.

For powers $\geq 4$, polar form is far faster than direct expansion. Always confirm the angle quadrant first.

PROBLEM 3 · MODULUS AND ARGUMENT OF A POWER

Given $z = 1 - i$, find $|z^{10}|$ and $\arg(z^{10})$.

Compute $|z|$ and $\arg z$. $|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. $1 - i$ lies in the fourth quadrant, so $\arg(1 - i) = -\tfrac{\pi}{4}$.

Use the diagram to fix the quadrant — sign errors on $\arg$ are the most common mistake. $(1, -1)$ is below the real axis.

Fill the gap: If $z = r(\cos\theta + i\sin\theta)$, then $z^n = $ $(\cos$ $+ i\sin n\theta)$ for positive integer $n$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to raise the modulus

Writing $z^n = \cos n\theta + i\sin n\theta$ drops the $r^n$ factor. This is only correct when $|z| = 1$. For every other $z$, the modulus must be raised to the $n$th power — the answer is $r^n(\cos n\theta + i\sin n\theta)$.

Trap 02

Wrong quadrant for $\arg z$

$\tan\theta = b/a$ has two solutions in $(-\pi, \pi]$. Always sketch $z$ on the Argand diagram first to fix the quadrant. For $1 - i$, $\arg = -\pi/4$ (Q4), not $+\pi/4$ (Q1).

Trap 03

Direct expansion of high powers

Computing $(a+bi)^7$ by FOIL/binomial wastes time and invites arithmetic errors. Convert to polar first. Polar form turns the problem into one multiplication for the modulus and one for the argument.

Did you get this? True or false: $(2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}))^4 = \cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Compute $(2 + i)^2$ in Cartesian form by direct expansion.

Convert $z = -1 + \sqrt{3}\,i$ to polar form $r(\cos\theta + i\sin\theta)$. Then write down $z^3$ using the power pattern.

Find $(1 + i)^4$ using polar form, then verify by computing $((1+i)^2)^2$ in Cartesian form.

Let $z = \cos\tfrac{\pi}{5} + i\sin\tfrac{\pi}{5}$. Compute $z^{10}$. (Note $|z| = 1$.)

For $z = 2(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4})$, compute the modulus and argument of $z^5$. Express $\arg(z^5)$ as a principal argument in $(-\pi, \pi]$.

Odd one out: Three of these are equal to $(1 + i)^4$. Which one is NOT?

Revisit your thinking

Earlier you computed $z^2$ and $z^3$ for $z = 1 + i$ by direct multiplication and noticed how the modulus grows.

You should have found $z^2 = 2i$ and $z^3 = -2 + 2i$, with moduli $\sqrt{2}, 2, 2\sqrt{2}$ — each step multiplies the modulus by $\sqrt{2}$. That is exactly the polar pattern: $|z^n| = |z|^n$. On the Argand diagram the points spiral outward, each rotated by $\tfrac{\pi}{4}$ from the last. This geometric picture is the key to seeing why $z^n = r^n(\cos n\theta + i\sin n\theta)$ — and it is the foundation for de Moivre's theorem in Module 13.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Compute $(1 + i)^4$ using polar form. Give your answer in Cartesian form. (2 marks)

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ApplyBand 43 marks

Q2. Express $z = -\sqrt{3} - i$ in polar form, then find $z^4$ in Cartesian form. (3 marks)

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AnalyseBand 53 marks

Q3. Let $z = r(\cos\theta + i\sin\theta)$ with $r > 0$. Using the multiplication rule $z_1 z_2 = r_1 r_2[\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)]$, show that $z^3 = r^3(\cos 3\theta + i\sin 3\theta)$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $(2+i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$.

2. $|z| = \sqrt{1 + 3} = 2$; $z$ lies in Q2, so $\arg z = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$. $z = 2(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3})$. Then $z^3 = 8(\cos 2\pi + i\sin 2\pi) = 8(1 + 0i) = 8$.

3. $r = \sqrt{2}, \theta = \tfrac{\pi}{4}$, so $(1+i)^4 = (\sqrt{2})^4(\cos\pi + i\sin\pi) = 4(-1+0i) = -4$. Check: $(1+i)^2 = 2i$, $(2i)^2 = 4i^2 = -4$. ✓

4. $|z| = 1$, $\arg z = \tfrac{\pi}{5}$, so $z^{10} = 1 \cdot (\cos 2\pi + i\sin 2\pi) = 1$.

5. $|z^5| = 2^5 = 32$; $\arg(z^5) = 5 \cdot \tfrac{\pi}{4} = \tfrac{5\pi}{4}$; principal: $\tfrac{5\pi}{4} - 2\pi = -\tfrac{3\pi}{4}$.

Q1 (2 marks): $1 + i = \sqrt{2}(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4})$ [1]. $(1+i)^4 = (\sqrt{2})^4(\cos\pi + i\sin\pi) = 4 \cdot (-1) = -4$ [1].

Q2 (3 marks): $|z| = \sqrt{3 + 1} = 2$ [1]. $z$ in Q3, so $\arg z = -\pi + \tfrac{\pi}{6} = -\tfrac{5\pi}{6}$; $z = 2(\cos(-\tfrac{5\pi}{6}) + i\sin(-\tfrac{5\pi}{6}))$ [1]. $z^4 = 16(\cos(-\tfrac{10\pi}{3}) + i\sin(-\tfrac{10\pi}{3})) = 16(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}) = 16(-\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}) = -8 + 8\sqrt{3}\,i$ [1].

Q3 (3 marks): $z^2 = z \cdot z = r \cdot r[\cos(\theta+\theta) + i\sin(\theta+\theta)] = r^2(\cos 2\theta + i\sin 2\theta)$ [1]. $z^3 = z^2 \cdot z = r^2 \cdot r[\cos(2\theta+\theta) + i\sin(2\theta+\theta)]$ [1] $= r^3(\cos 3\theta + i\sin 3\theta)$ [1].

Boss battle · The Power Spiral

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Five timed questions on polar form, multiplication, and computing $z^n$. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick complex-number questions. Lighter alternative to the boss.

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