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.
+5 XP warm-up Without notes, what is the geometric effect of multiplying a complex number by ?
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.
+5 XP to read Every "use complex numbers to prove a geometric result" question rewards two habits: translate the picture into...
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 $\angle z_2 z_1 z_3$
The single most useful idea in this lesson is: a rotation of about the point through angle is multiplication of the...
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
This encodes both angle and length in a single complex equation. From it, all the standard geometric tests fall out:
Worked through the hook. $A = 1$, $B = 3 + 2i$. To rotate $B$ about $A$ by $90^{\circ}$ anticlockwise, multiplication by $e^{i\pi/2} = i$:
$C = A + i(B - A) = 1 + i(2 + 2i) = 1 + 2i + 2i^2 = 1 + 2i - 2 = -1 + 2i.$
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 $A$?
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 $z_1z_2 \perp z_1z_3$.
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:
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ (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
Worked examples · 3 in a row, reveal as you go
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 $AB \perp AC$.
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 expansion of $(\cos\theta + i\sin\theta)^3$ to derive the identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.
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
Misconceptions to fix · the 3 traps that cost marks
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 $w = -1 + 2i$.
Activities · practice with the ideas
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?
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.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
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)
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)
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].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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