Polar form and De Moivre's theorem (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

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

8 min readUpdated 2 Jun 2026

Quick answer

Polar form writes $z=r(\cos\theta+i\sin\theta)$ where $r=|z|$ is the modulus and $\theta=\arg(z)$ is the argument. Multiplying multiplies the moduli and adds the arguments; dividing divides moduli and subtracts arguments. De Moivre's theorem gives $z^n=r^n(\cos n\theta+i\sin n\theta)$ for integer $n$ .

Jump to a section

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
Exam-style questions

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

Principal argument

The argument is only defined up to multiples of $2\pi$ . The principal argument $\operatorname{Arg}(z)$ is the value in $(-\pi,\pi]$ . Always check which quadrant $(a,b)$ lies in before quoting $\theta$ , because $\tan^{-1}(b/a)$ alone cannot distinguish quadrants.

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 $\sin$ .

Polar multiplication and division

r_1\operatorname{cis}\theta_1\cdot r_2\operatorname{cis}\theta_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2),\qquad \frac{r_1\operatorname{cis}\theta_1}{r_2\operatorname{cis}\theta_2}=\frac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2)

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

Quoting the wrong quadrant for the argument

Using $\theta=\tan^{-1}(b/a)$ blindly puts every argument in $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$ , which is wrong for second- and third-quadrant numbers. For $z=-1-i$ the calculator gives $\tfrac{\pi}{4}$ but the true principal argument is $-\dfrac{3\pi}{4}$ . Always sketch the point first.

Forgetting to raise the modulus to the power

De Moivre's theorem raises the modulus to the power $n$ as well as multiplying the angle. Writing $(1+i)^8=\operatorname{cis}(2\pi)=1$ and omitting the $(\sqrt2)^8=16$ factor is a common loss of marks.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks

Calculator-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 2023-style3 marks

Calculator-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 2021-style3 marks

Calculator-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 $-64$ .

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board