SACE Board SACE 2022-style practice: Calculator-free. Let z = cisθ be a complex number with modulus 1. Using de Moivre's theorem, show that z^n + z^-n = 2cos(nθ).

Since z = cisθ = cosθ + isinθ, de Moivre's theorem gives z^n = cos(nθ) + isin(nθ). [1 mark] The negative power: z^-n = cos(-nθ) + isin(-nθ) = cos(nθ) - isin(nθ), using that cosine is even and sine is odd. Adding: z^n + z^-n = [cos(nθ) + isin(nθ)] + [cos(nθ) - isin(nθ)] = 2cos(nθ), since the imaginary parts cancel. [1 mark]

SACE Board SACE 2023-style practice: Calculator-free. With z = cisθ, expand (z + z^-1)^3 to show that cos^3θ = 14cos(3θ) + 34cosθ.

From the previous result, z + z^-1 = 2cosθ and z^3 + z^-3 = 2cos(3θ). Expand by the binomial theorem: (z + z^-1)^3 = z^3 + 3z + 3z^-1 + z^-3. [1 mark] Group the symmetric pairs: = (z^3 + z^-3) + 3(z + z^-1) = 2cos(3θ) + 6cosθ. [1 mark] But (z + z^-1)^3 = (2cosθ)^3 = 8cos^3θ. So 8cos^3θ = 2cos(3θ) + 6cosθ; dividing by 8 gives cos^3θ = 14cos(3θ) + 34cosθ. [1 mark]

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How does writing a complex number by its modulus and angle make multiplication, division and powers easy?

Polar form expresses a complex number by its modulus and argument, and De Moivre's theorem raises it to integer powers.

Converting between Cartesian and polar form, the modulus and argument, multiplication and division in polar form, and using De Moivre's theorem to compute integer powers of complex numbers.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Modulus and argument
Multiplication and division in polar form
De Moivre's theorem
Why rotation explains multiplication
Negative powers and reciprocals
Common errors
Why it matters

What this dot point is asking

Cartesian form is convenient for adding, but clumsy for multiplying and taking powers. Polar form describes the same number by how far it is from the origin and in what direction.

Modulus and argument

For $z=a+bi$ the modulus is $r=|z|=\sqrt{a^2+b^2}$ and the argument $\theta=\arg(z)$ is the angle the vector makes with the positive real axis, measured anticlockwise. Then $a=r\cos\theta$ and $b=r\sin\theta$ , so

z=r(\cos\theta+i\sin\theta),

often abbreviated

z=r\operatorname{cis}\theta

Converting to polar form

Worked example

Write $z=-1+i\sqrt{3}$ in polar form.

Step 1: Compute the modulus

The modulus $r=|z|$ is the distance from the origin to the point $(-1,\sqrt{3})$ in the plane, found using the distance formula.

r=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.

Step 2: Find the argument by identifying the quadrant and reference angle

Because the real part is negative and the imaginary part is positive, the point $(-1,\sqrt3)$ lies in the second quadrant. The reference angle $\alpha$ satisfies $\tan\alpha=\sqrt3/1$ , giving $\alpha=\pi/3$ . In the second quadrant the principal argument is $\pi$ minus the reference angle.

\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}.

Step 3: Write in polar form

Combine the modulus and argument using the polar form $z=r\operatorname{cis}\theta$ .

z=2\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)=2\operatorname{cis}\dfrac{2\pi}{3}.

Final answer: $z=2\operatorname{cis}\dfrac{2\pi}{3}$ .

Multiplication and division in polar form

If $z_1=r_1\operatorname{cis}\theta_1$ and $z_2=r_2\operatorname{cis}\theta_2$ , then

z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2),\qquad \frac{z_1}{z_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2).

So multiplication is a scaling and rotation: the new vector is $r_2$ times as long and rotated by $\theta_2$ . This follows from the compound-angle formulae applied to $\cos$ and $\sin$ .

De Moivre's theorem

Applying the multiplication rule $n$ times gives De Moivre's theorem: for any integer $n$ ,

\big(r(\cos\theta+i\sin\theta)\big)^n=r^n(\cos n\theta+i\sin n\theta).

This makes large powers, which would be hopeless to expand in Cartesian form, immediate.

Using De Moivre's theorem

Worked example

Find $(1+i)^8$ .

Step 1: Convert to polar form

Expanding $(1+i)^8$ in Cartesian form would require seven multiplications. Instead, convert $1+i$ to polar form once. The modulus is $r=\sqrt{1^2+1^2}=\sqrt2$ , and the point $(1,1)$ is in the first quadrant with $\tan\theta=1$ , so $\theta=\dfrac{\pi}{4}$ .

1+i=\sqrt2\operatorname{cis}\dfrac{\pi}{4}.

Step 2: Apply De Moivre's theorem

De Moivre's theorem states $(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ . Raise the modulus to the power $8$ and multiply the argument by $8$ .

(1+i)^8=(\sqrt2)^8\operatorname{cis}\!\left(8\cdot\frac{\pi}{4}\right)=2^4\operatorname{cis}(2\pi).

Step 3: Evaluate and simplify

Since $\cos 2\pi=1$ and $\sin 2\pi=0$ , the result is purely real.

2^4\operatorname{cis}(2\pi)=16(\cos 2\pi+i\sin 2\pi)=16.

Final answer: $(1+i)^8=16$ .

The theorem also derives multiple-angle identities. Expanding $(\cos\theta+i\sin\theta)^3$ by the binomial theorem and equating real parts with $\cos 3\theta$ gives $\cos 3\theta=4\cos^3\theta-3\cos\theta$ .

Why rotation explains multiplication

The polar multiplication rule is the deepest idea in this dot point: multiplying by $r\operatorname{cis}\theta$ scales a vector by $r$ and rotates it anticlockwise by $\theta$ . Multiplying by $i=\operatorname{cis}\tfrac{\pi}{2}$ , for instance, is a pure $90^\circ$ rotation with no scaling, which is why applying $i$ twice (giving $i^2=-1$ ) rotates by $180^\circ$ and lands on the negative real axis. Seeing multiplication as a geometric transformation turns otherwise abstract algebra into a picture, and it is the reason de Moivre's theorem - repeated multiplication - simply repeats the rotation $n$ times.

Negative powers and reciprocals

De Moivre's theorem holds for negative integers too. The reciprocal $z^{-1}$ of $z=r\operatorname{cis}\theta$ is $\dfrac{1}{r}\operatorname{cis}(-\theta)$ : take the reciprocal of the modulus and negate the argument. This follows from the division rule with numerator $1=1\operatorname{cis}0$ . It means any negative power $z^{-n}=r^{-n}\operatorname{cis}(-n\theta)$ is found by the same theorem, a fact used directly in the $z^n+z^{-n}$ identity questions above.

Common errors

Why it matters

Polar form turns multiplication into rotation and division into the reverse, and De Moivre's theorem is the gateway to finding the roots of complex numbers. The conversion technique builds directly on the modulus and conjugate ideas from complex arithmetic.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-free. Let

z = \operatorname{cis}\theta

be a complex number with modulus

1

. Using de Moivre's theorem, show that

z^n + z^{-n} = 2\cos(n\theta)

Show worked answer →

Since $z = \operatorname{cis}\theta = \cos\theta + i\sin\theta$ , de Moivre's theorem gives $z^n = \cos(n\theta) + i\sin(n\theta)$ . [1 mark]

The negative power: $z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)$ , using that cosine is even and sine is odd.

Adding: $z^n + z^{-n} = [\cos(n\theta) + i\sin(n\theta)] + [\cos(n\theta) - i\sin(n\theta)] = 2\cos(n\theta)$ , since the imaginary parts cancel. [1 mark]

SACE 20233 marksCalculator-free. With

z = \operatorname{cis}\theta

, expand

(z + z^{-1})^3

to show that

\cos^3\theta = \tfrac{1}{4}\cos(3\theta) + \tfrac{3}{4}\cos\theta

Show worked answer →

From the previous result, $z + z^{-1} = 2\cos\theta$ and $z^3 + z^{-3} = 2\cos(3\theta)$ .

Expand by the binomial theorem: $(z + z^{-1})^3 = z^3 + 3z + 3z^{-1} + z^{-3}$ . [1 mark]

Group the symmetric pairs: $= (z^3 + z^{-3}) + 3(z + z^{-1}) = 2\cos(3\theta) + 6\cos\theta$ . [1 mark]

But $(z + z^{-1})^3 = (2\cos\theta)^3 = 8\cos^3\theta$ . So $8\cos^3\theta = 2\cos(3\theta) + 6\cos\theta$ ; dividing by $8$ gives $\cos^3\theta = \tfrac{1}{4}\cos(3\theta) + \tfrac{3}{4}\cos\theta$ . [1 mark]

SACE 20213 marksCalculator-free. Express

z = \sqrt{3} + i

in polar form, then use de Moivre's theorem to find

z^6

in Cartesian form.

Show worked answer →

Modulus: $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$ . The point $(\sqrt{3}, 1)$ is in the first quadrant with $\tan\theta = \tfrac{1}{\sqrt{3}}$ , so $\theta = \tfrac{\pi}{6}$ . Thus $z = 2\operatorname{cis}\tfrac{\pi}{6}$ .

By de Moivre, $z^6 = 2^6\operatorname{cis}\!\left(6\times\tfrac{\pi}{6}\right) = 64\operatorname{cis}\pi = 64(\cos\pi + i\sin\pi) = -64$ .

Marks: one for the modulus, one for the argument $\tfrac{\pi}{6}$ , one for applying de Moivre to reach $-64$ .