NSWMaths Extension 2Syllabus dot point

How does de Moivre's theorem let us compute powers and roots of complex numbers, and what is the geometry of the nth roots of unity?

Use de Moivre's theorem to find powers and nth roots of complex numbers and to derive the roots of unity and their geometric arrangement

A focused answer to the HSC Maths Extension 2 dot point on de Moivre's theorem. Powers in polar form, finding all nth roots, the roots of unity arranged on a circle, and their sum, with rigorous verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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What this dot point is asking
De Moivre's theorem
Finding nth roots
Roots of unity

What this dot point is asking

NESA wants you to use de Moivre's theorem to raise complex numbers to powers and to find all the $n$ th roots of a complex number. You must derive the $n$ th roots of unity, plot them as equally spaced points on the unit circle, and use their structure (for example, that they sum to zero).

De Moivre's theorem

If $z = r(\cos\theta + i\sin\theta)$ then for every integer $n$ ,

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

This follows from the fact that multiplication in polar form multiplies moduli and adds arguments; raising to the $n$ th power multiplies the modulus by itself $n$ times and adds the argument $n$ times. The theorem can be proved formally by induction for positive integers, then extended to negative integers via reciprocals.

A typical use is computing high powers. To find $(1 + i)^8$ , write $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$ , so

(1 + i)^8 = (\sqrt{2})^8\left(\cos 2\pi + i\sin 2\pi\right) = 16(1 + 0i) = 16.

Finding nth roots

To solve $z^n = w$ where $w = R(\cos\phi + i\sin\phi)$ , write the unknown as $z = r(\cos\theta + i\sin\theta)$ . Then $r^n = R$ gives $r = R^{1/n}$ (the positive real root), and $n\theta = \phi + 2\pi k$ for integer $k$ . The distinct solutions are

z_k = R^{1/n}\left(\cos\frac{\phi + 2\pi k}{n} + i\sin\frac{\phi + 2\pi k}{n}\right), \qquad k = 0, 1, \dots, n-1.

Values of $k$ outside this range repeat earlier roots. Geometrically the $n$ roots lie on a circle of radius $R^{1/n}$ , equally spaced by angle $\dfrac{2\pi}{n}$ .

Roots of unity

The $n$ th roots of unity solve $z^n = 1$ . Since $1 = \cos 0 + i\sin 0$ , they are

\omega_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \qquad k = 0, 1, \dots, n-1.

Writing $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$ , every root is a power $\omega^k$ . They sit at the vertices of a regular $n$ -gon inscribed in the unit circle, with one vertex at $1$ .

A key property: the roots of unity sum to zero for $n \ge 2$ . Because $z^n - 1 = (z - 1)(z^{n-1} + \dots + z + 1)$ , the sum of all roots equals the negative of the coefficient of $z^{n-1}$ in $z^n - 1$ , which is $0$ .

Worked example

Find the three cube roots of unity and verify they sum to zero

Solve $z^3 = 1$ . Here $R = 1$ , $\phi = 0$ , $n = 3$ , so

z_k = \cos\frac{2\pi k}{3} + i\sin\frac{2\pi k}{3}, \qquad k = 0, 1, 2.

$k = 0$ : $z_0 = \cos 0 + i\sin 0 = 1$ .
$k = 1$ : $z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ .
$k = 2$ : $z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$ .

Sum: $1 + \left(-\frac{1}{2} + \frac{\sqrt3}{2}i\right) + \left(-\frac{1}{2} - \frac{\sqrt3}{2}i\right) = 1 - 1 + 0i = 0$ .

Check $z_1^3 = 1$ : $|z_1| = 1$ and $\arg z_1 = \frac{2\pi}{3}$ , so $z_1^3$ has modulus $1$ and argument $2\pi$ , which is $1$ . Correct.

Exam-style practice questions

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

2021 HSC3 marksUsing de Moivre's theorem and the binomial expansion of (cos q + i sin q)^5, or otherwise, show that cos 5q = 16 cos^5 q - 20 cos^3 q + 5 cos q.

Show worked answer →

By de Moivre's theorem, (cos q + i sin q)^5 = cos 5q + i sin 5q.

Expand the left side with the binomial theorem (writing c = cos q, s = sin q):

(c + is)^5 = c^5 + 5c^4(is) + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5.

Using i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, the real part is

c^5 - 10c^3 s^2 + 5c s^4.

Equating real parts: cos 5q = c^5 - 10 c^3 s^2 + 5 c s^4.

Now replace s^2 = 1 - c^2 and s^4 = (1 - c^2)^2 = 1 - 2c^2 + c^4:

cos 5q = c^5 - 10 c^3 (1 - c^2) + 5 c (1 - 2c^2 + c^4)
= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5
= 16 c^5 - 20 c^3 + 5 c.

Hence cos 5q = 16 cos^5 q - 20 cos^3 q + 5 cos q.

Mark notes: 1 mark for applying de Moivre and the binomial expansion, 1 mark for taking the real part correctly, 1 mark for eliminating sin^2 q to reach the required cosine-only form.

2022 HSC2 marksWrite the complex number -sqrt(3) + i in exponential form. Hence, find the exact value of (-sqrt(3) + i)^10, giving your answer in the form x + iy.

Show worked answer →

Modulus: |-sqrt(3) + i| = sqrt(3 + 1) = 2.

Argument: the point is in the second quadrant, with reference angle arctan(1/sqrt(3)) = pi/6, so arg = pi - pi/6 = 5pi/6.

Exponential form: -sqrt(3) + i = 2 e^(i 5pi/6).

Now apply de Moivre to the tenth power:

(-sqrt(3) + i)^10 = 2^10 e^(i 50pi/6) = 1024 e^(i 25pi/3).

Reduce the argument modulo 2pi: 25pi/3 - 8pi = 25pi/3 - 24pi/3 = pi/3.

So the value is 1024(cos pi/3 + i sin pi/3) = 1024(1/2 + i sqrt(3)/2) = 512 + 512 sqrt(3) i.

Mark notes: 1 mark for the exponential form 2 e^(i 5pi/6), 1 mark for raising to the power and reducing the argument to give 512 + 512 sqrt(3) i.

2022 HSC3 marksConsider the equation z^5 + 1 = 0. It is known that if z is a solution with z not equal to -1, then u = z + 1/z is a solution of u^2 - u - 1 = 0. Hence find the exact value of cos(3pi/5).

Show worked answer →

The non-real solutions of z^5 + 1 = 0 lie on the unit circle at z = e^(i q). For such z, the conjugate 1/z = e^(-i q), so

u = z + 1/z = e^(i q) + e^(-i q) = 2 cos q.

The fifth roots of -1 (excluding z = -1) occur at q = pi/5, 3pi/5, 7pi/5, 9pi/5, which pair into conjugates giving the two distinct real values u = 2 cos(pi/5) and u = 2 cos(3pi/5).

Solving u^2 - u - 1 = 0 by the quadratic formula:

u = (1 +/- sqrt(1 + 4))/2 = (1 +/- sqrt(5))/2.

Since 3pi/5 is between pi/2 and pi, cos(3pi/5) is negative, so we take the negative root:

2 cos(3pi/5) = (1 - sqrt(5))/2, hence cos(3pi/5) = (1 - sqrt(5))/4.

Mark notes: 1 mark for recognising u = 2 cos q for roots on the unit circle, 1 mark for solving the quadratic, 1 mark for selecting the negative root to give cos(3pi/5) = (1 - sqrt(5))/4.