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Module 12 · L15 of 16 ~40 min ⚡ +90 XP available

Applications of Complex Numbers

Complex numbers are not just an algebraic curiosity — they are a precision tool for geometry and trigonometry. Multiplying by $e^{i\theta}$ is a rotation. The ratio $(z_3 - z_1)/(z_2 - z_1)$ encodes both relative angle and scale. Expanding $(\cos\theta + i\sin\theta)^n$ unlocks multiple-angle identities. This lesson shows how to wield those ideas in HSC-style geometric proofs and identity derivations.

Today's hook — Sketch the points $A(1)$, $B(3 + 2i)$ on the Argand plane. If $C$ is obtained by rotating $B$ about $A$ through $90^{\circ}$ anticlockwise, what complex number represents $C$? Try to do it without coordinates — just multiply by $i$. Check your answer after card 05.

0/5QUESTS

Recall — your gut answer first

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Without notes — what is the geometric effect of multiplying a complex number $z$ by $e^{i\theta}$? And by a real $r > 0$? Write a one-line answer for each before reading on.

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The two moves for geometric proof

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Every "use complex numbers to prove a geometric result" question rewards two habits: translate the picture into complex numbers (assign each point a complex coordinate, write displacements as differences), then multiply by a rotation / use modulus and argument to encode angles and lengths.

The translate-rotate-conclude workflow: (1) assign each labelled point $A, B, C, \ldots$ a complex number $z_A, z_B, z_C, \ldots$, (2) express the relationship using a ratio or product (e.g. $z_C - z_A = e^{i\theta}(z_B - z_A)$ rotates $B$ about $A$ by $\theta$), (3) take modulus for length, argument for angle.

Rotation about $z_0$: $w - z_0 = e^{i\theta}(z - z_0)$ · Ratio: $\arg\!\left(\dfrac{z_3 - z_1}{z_2 - z_1}\right)$ = angle $\angle z_2 z_1 z_3$

$w - z_0 \;=\; e^{i\theta}(z - z_0)$

Rotation has a centre

To rotate $z$ about a point $z_0$ (not the origin), translate first: $w - z_0 = e^{i\theta}(z - z_0)$, so $w = z_0 + e^{i\theta}(z - z_0)$.

Ratio = angle + scale

$\dfrac{z_3 - z_1}{z_2 - z_1}$: modulus = ratio of lengths $|AC|/|AB|$; argument = directed angle from $AB$ to $AC$.

Distance is modulus

$|z_2 - z_1|$ is the distance between the points $z_1$ and $z_2$ on the Argand plane. So $|w| = r$ is a circle of radius $r$; $|z - z_0| = r$ is a circle centred at $z_0$.

What you'll master

Know

Key facts

Multiplying by $e^{i\theta}$ is rotation by $\theta$ about the origin
Rotation of $z$ about $z_0$ by $\theta$: $w = z_0 + e^{i\theta}(z - z_0)$
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ (de Moivre — context only here)
$|z_2 - z_1|$ = distance; $\arg(z_2 - z_1)$ = angle of the segment with the positive real axis

Understand

Concepts

Why two triangles with the same complex ratio $(z_3 - z_1)/(z_2 - z_1)$ are similar
Why equating real and imaginary parts of $(\cos\theta + i\sin\theta)^n$ gives multiple-angle identities
Why a perpendicularity condition becomes "ratio is purely imaginary"

Can do

Skills

Translate a geometric configuration into complex coordinates and solve for an unknown point
Use the rotation formula to prove perpendicularity, distance or similarity
Expand $(\cos\theta + i\sin\theta)^3$ and read off the identities for $\cos 3\theta$ and $\sin 3\theta$

Key terms

Rotation factor $e^{i\theta}$A unit complex number. Multiplying $z$ by $e^{i\theta}$ leaves $|z|$ unchanged and increases $\arg z$ by $\theta$ — a pure rotation about $0$.

Rotation about a pointTo rotate $z$ about $z_0$ by angle $\theta$: subtract $z_0$, multiply by $e^{i\theta}$, add $z_0$ back. Formula: $w = z_0 + e^{i\theta}(z - z_0)$.

Complex ratio$\dfrac{z_3 - z_1}{z_2 - z_1}$ has modulus $|AC|/|AB|$ and argument equal to the directed angle $\angle z_2 z_1 z_3$. Used for similarity and angle proofs.

Perpendicularity$AB \perp CD$ iff $\dfrac{z_B - z_A}{z_D - z_C}$ is purely imaginary (argument $\pm \pi/2$).

Similar triangles (directly)$\triangle z_1 z_2 z_3$ and $\triangle w_1 w_2 w_3$ are directly similar iff $\dfrac{z_3 - z_1}{z_2 - z_1} = \dfrac{w_3 - w_1}{w_2 - w_1}$.

Multiple-angle identityAn identity expressing $\cos n\theta$ or $\sin n\theta$ in terms of powers of $\cos\theta$ and $\sin\theta$, obtained by expanding $(\cos\theta + i\sin\theta)^n$ and matching real/imaginary parts.

MEX-N1NESA outcome (Complex Numbers I): performs operations using the various representations of complex numbers and uses them to solve problems, including in geometry and trigonometry.

Rotation and the complex ratio

core concept

The single most useful idea in this lesson is: a rotation of $z$ about the point $z_0$ through angle $\theta$ is multiplication of the displacement $z - z_0$ by $e^{i\theta}$. Symbolically

$$w \;=\; z_0 + e^{i\theta}(z - z_0).$$

This encodes both angle and length in a single complex equation. From it, all the standard geometric tests fall out:

Perpendicularity: $AB \perp CD \;\Leftrightarrow\; \dfrac{z_B - z_A}{z_D - z_C} \in i\mathbb{R}$ (i.e. purely imaginary).
Equal length: $|AB| = |CD| \;\Leftrightarrow\; |z_B - z_A| = |z_D - z_C|$.
Similar triangles (directly): $\dfrac{z_3 - z_1}{z_2 - z_1} = \dfrac{w_3 - w_1}{w_2 - w_1}$.
Collinear: $\dfrac{z_3 - z_1}{z_2 - z_1} \in \mathbb{R}$ (argument is $0$ or $\pi$).

Worked through the hook. $A = 1$, $B = 3 + 2i$. To rotate $B$ about $A$ by $90^{\circ}$ anticlockwise, multiplication by $e^{i\pi/2} = i$:

$C = A + i(B - A) = 1 + i(2 + 2i) = 1 + 2i + 2i^2 = 1 + 2i - 2 = -1 + 2i.$

Why this is so powerful. Classical geometry needs you to chase angles, drop perpendiculars, invoke similar triangles. Complex coordinates collapse all of that into algebra. Once the picture is translated to $z$'s, what was a multi-step construction becomes a single line of expansion.

Rotation about $z_0$: $w = z_0 + e^{i\theta}(z - z_0)$ · Multiplication by $i$ = anticlockwise $90^{\circ}$ rotation about the origin · Perpendicular $\Leftrightarrow$ ratio is purely imaginary; collinear $\Leftrightarrow$ ratio is real · Direct similarity of $\triangle z_1 z_2 z_3$ and $\triangle w_1 w_2 w_3$: $(z_3-z_1)/(z_2-z_1) = (w_3-w_1)/(w_2-w_1)$

Pause — copy the rotation formula $w = z_0+e^{i\theta}(z-z_0)$, the perpendicularity test (ratio purely imaginary), the collinearity test (ratio real), and the direct-similarity condition into your book.

Quick check: Points $A$ and $B$ represent $2$ and $4 + i$ respectively. Which complex number represents the image of $B$ when rotated $90^{\circ}$ anticlockwise about $A$?

Multiple-angle identities via expansion

core concept · awareness

We just saw that rotation about $z_0$ by $\theta$ is $w = z_0 + e^{i\theta}(z-z_0)$, and that a ratio $z_3-z_1)/(z_2-z_1)$ is purely imaginary iff $z_1z_2 \perp z_1z_3$. That raises a question: how does De Moivre's theorem let us derive $\cos 3\theta$, $\sin 3\theta$ and higher multiples directly? This card answers it → expand $(\cos\theta+i\sin\theta)^n$ binomially and equate real and imaginary parts.

This module ends with a glimpse of de Moivre's theorem (formally developed in Complex Numbers II). The key idea: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$. Expand the left-hand side with the binomial theorem; match real and imaginary parts of both sides; out fall identities for $\cos n\theta$ and $\sin n\theta$.

Worked sketch — deriving $\sin 3\theta$ and $\cos 3\theta$. Let $c = \cos\theta$, $s = \sin\theta$. Then

$(c + is)^3 \;=\; c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 \;=\; (c^3 - 3cs^2) + i(3c^2 s - s^3).$

By the identity $(c + is)^3 = \cos 3\theta + i\sin 3\theta$, equate real and imaginary parts:

$\cos 3\theta \;=\; \cos^3\theta - 3\cos\theta\sin^2\theta \;=\; 4\cos^3\theta - 3\cos\theta$.
$\sin 3\theta \;=\; 3\cos^2\theta\sin\theta - \sin^3\theta \;=\; 3\sin\theta - 4\sin^3\theta$.

$$\cos 3\theta = 4\cos^3\theta - 3\cos\theta,\qquad \sin 3\theta = 3\sin\theta - 4\sin^3\theta.$$

Awareness only. In Module 12 you only need to recognise this method and apply it for small $n$ ($n = 2, 3$). The general theorem and applications (roots of unity, polynomial factorisation) belong to Complex Numbers II. But knowing the method now means $\cos 3\theta$ is no longer something to memorise — it is something to derive in three lines.

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ (de Moivre, used here) · Expand by binomial; match real and imaginary parts · $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$; $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ · Use $\sin^2\theta = 1 - \cos^2\theta$ to convert between forms

Pause — copy $\cos 3\theta = 4\cos^3\theta-3\cos\theta$, $\sin 3\theta = 3\sin\theta-4\sin^3\theta$, and the method (expand $(c+is)^n$, match Re and Im, use $\sin^2=1-\cos^2$) into your book.

Did you get this? True or false: expanding $(\cos\theta + i\sin\theta)^2$ and matching real parts gives the identity $\cos 2\theta = \cos^2\theta - \sin^2\theta$.

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

PROBLEM 1 · ROTATION ABOUT A POINT

Triangle $ABC$ has $A = 2 + i$, $B = 5 + 2i$. Point $C$ is obtained by rotating $B$ about $A$ through $90^{\circ}$ anticlockwise. Find $C$, then verify $|AB| = |AC|$ and $AB \perp AC$.

Use rotation formula with $\theta = \pi/2$, so $e^{i\theta} = i$: $C = A + i(B - A) = (2 + i) + i((5 + 2i) - (2 + i)) = (2 + i) + i(3 + i)$.

Always translate so the centre of rotation sits at the origin before multiplying. Here we subtract $A$, multiply by $i$, then add $A$ back.

PROBLEM 2 · SIMILAR TRIANGLES VIA COMPLEX RATIO

Let $z_1 = 1$, $z_2 = 3 + 2i$, $z_3 = 2i$, and $w_1 = 4$, $w_2 = 6 + 4i$, $w_3 = 2 + 4i$. Show that triangles $z_1 z_2 z_3$ and $w_1 w_2 w_3$ are directly similar.

Use the direct-similarity test: $\dfrac{z_3 - z_1}{z_2 - z_1} = \dfrac{w_3 - w_1}{w_2 - w_1}$. Compute each side.

Direct similarity preserves orientation. The complex-ratio test captures both the scale factor (modulus) and the angle (argument) in one equation.

PROBLEM 3 · DERIVE $\sin 3\theta$

Use the expansion of $(\cos\theta + i\sin\theta)^3$ to derive the identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.

Apply de Moivre: $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$. Set $c = \cos\theta$, $s = \sin\theta$.

De Moivre is the bridge between algebraic expansion and trigonometric identity. Stating it up-front makes the rest a binomial exercise.

Fill the gap: The rotation of $z$ about the point $z_0$ through angle $\theta$ anticlockwise is $w = z_0 + $ $\cdot \, (z - $ $)$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to translate to the centre

Rotating $z$ about a point $z_0 \neq 0$ is NOT just $e^{i\theta}z$. You must subtract $z_0$ first, multiply by $e^{i\theta}$, then add $z_0$ back. Skipping the translation rotates about the origin instead of about $z_0$.

Trap 02

Sign of the rotation angle

$e^{i\theta}$ rotates anticlockwise (positive direction). For a clockwise rotation by $\theta$, use $e^{-i\theta}$ — equivalently, replace $i$ with $-i$. A $90^{\circ}$ clockwise rotation is multiplication by $-i$, not $i$.

Trap 03

Mismatching real and imaginary parts

When expanding $(\cos\theta + i\sin\theta)^n$ to derive a multiple-angle identity, you must carefully collect terms by their $i$-content. Real parts give $\cos n\theta$; imaginary parts (the coefficients of $i$, NOT including $i$ itself) give $\sin n\theta$.

Did you get this? True or false: rotating the complex number $z = 2 + 3i$ about the point $z_0 = 1 + i$ by $90^{\circ}$ anticlockwise gives $w = -1 + 2i$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the image of $z = 4 + i$ under a $90^{\circ}$ anticlockwise rotation about $z_0 = 2 + i$.

$A = 1$, $B = i$, $C = 1 + i$. Show that $\triangle ABC$ is a right-angled isoceles triangle, using only complex-number arguments (modulus and arg).

Use the expansion of $(\cos\theta + i\sin\theta)^2$ to derive both $\cos 2\theta = \cos^2\theta - \sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$.

Points $P$ and $Q$ on the Argand plane represent $z$ and $w$ with $|z| = |w|$ and $\arg(z) - \arg(w) = \pi/3$. Find the modulus and argument of $z/w$.

Triangle $ABC$ has $A = 0$, $B = 1$. If $\triangle ABC$ is equilateral with $C$ in the upper half-plane, find $C$ using a rotation argument.

Odd one out: Three of these complex-number conditions correctly describe a geometric property. Which one is INCORRECT?

Revisit your thinking

Earlier you sketched $A = 1$, $B = 3 + 2i$, and rotated $B$ about $A$ by $90^{\circ}$ anticlockwise.

Using $C = A + i(B - A) = 1 + i(2 + 2i) = 1 + 2i - 2 = -1 + 2i$. The key move was to subtract $A$ before multiplying by $i$ (the rotation factor), then add $A$ back to undo the translation. Multiplication by $i$ alone would have rotated about the origin, not about $A$. This translate-rotate-translate-back pattern is the spine of every geometric proof in this module.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Let $A = 3 + i$ and $B = 5 + 4i$. Find the complex number representing the image of $B$ when it is rotated $90^{\circ}$ anticlockwise about $A$. (2 marks)

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ApplyBand 43 marks

Q2. Use the expansion of $(\cos\theta + i\sin\theta)^3$ and de Moivre's theorem to prove that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. (3 marks)

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AnalyseBand 53 marks

Q3. $A$, $B$, $C$ are the points $z_1 = 2$, $z_2 = 5 + 3i$, $z_3 = 2 + 4i$ on the Argand plane. By computing $\dfrac{z_3 - z_1}{z_2 - z_1}$, decide whether $\angle BAC$ is acute, right or obtuse, and find $|AB|$ and $|AC|$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

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

2. $|AB| = |i - 1| = \sqrt{2}$; $|AC| = |(1 + i) - 1| = |i| = 1$. So not isoceles as stated; the right angle is at $C$. Check: $(C - A)/(C - B) = ((1+i)-1)/((1+i)-i) = i/1 = i$ (purely imaginary), so $CA \perp CB$. The triangle is right-angled at $C$ with $|CA| = |CB| = 1$, so it IS right-angled isoceles (with the right angle at $C$, not $A$).

3. $(c + is)^2 = c^2 + 2ics - s^2 = (c^2 - s^2) + i(2cs)$. By de Moivre this equals $\cos 2\theta + i\sin 2\theta$. Real: $\cos 2\theta = c^2 - s^2$. Imaginary: $\sin 2\theta = 2cs$.

4. $z/w$ has modulus $|z|/|w| = 1$ and argument $\arg z - \arg w = \pi/3$. So $z/w = e^{i\pi/3} = \tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}$.

5. $C = A + e^{i\pi/3}(B - A) = 0 + e^{i\pi/3}(1) = \cos 60^{\circ} + i\sin 60^{\circ} = \tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}$.

Q1 (2 marks): $C = A + i(B - A) = (3 + i) + i(2 + 3i) = (3 + i) + (-3 + 2i) = 0 + 3i = 3i$. [1 mark setup, 1 mark final.]

Q2 (3 marks): By de Moivre, $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$ [1]. Expand: $(c + is)^3 = c^3 + 3ic^2s - 3cs^2 - is^3 = (c^3 - 3cs^2) + i(3c^2s - s^3)$ [1]. Equate real parts: $\cos 3\theta = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4c^3 - 3c = 4\cos^3\theta - 3\cos\theta$ [1]. $\blacksquare$

Q3 (3 marks): $\dfrac{z_3 - z_1}{z_2 - z_1} = \dfrac{4i}{3 + 3i} = \dfrac{4i(3 - 3i)}{18} = \dfrac{12 + 12i}{18} = \dfrac{2 + 2i}{3}$ [1]. Argument $= \arg(2 + 2i) = \pi/4$, so $\angle BAC = \pi/4$ — acute [1]. $|AB| = |3 + 3i| = 3\sqrt{2}$; $|AC| = |4i| = 4$ [1].

Boss battle · The Rotation Architect

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Five timed questions on rotation, similarity, perpendicularity and multiple-angle identities. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

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

Mark lesson as complete

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