NESA HSC 2021-style practice: Using de Moivre's theorem and the binomial expansion of (cosθ + isinθ)^5, or otherwise, show that cos 5θ = 16cos^5θ - 20cos^3θ + 5cosθ.

By de Moivre's theorem, (cosθ + isinθ)^5 = cos 5θ + isin 5θ. Expand the left side with the binomial theorem (writing c = cosθ, s = sinθ): (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 5θ = 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 5θ = c^5 - 10 c^3 (1 - c^2) + 5 c (1 - 2c^2 + c^4) = 16 c^5 - 20 c^3 + 5 c. Hence cos 5θ = 16cos^5θ - 20cos^3θ + 5cosθ. 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^2θ to reach the required cosine-only form.

NESA HSC 2022-style practice: Consider the equation z^5 + 1 = 0. It is known that if z is a solution with z ≠ -1, then u = z + (1)/(z) is a solution of u^2 - u - 1 = 0. Hence find the exact value of cos(3π)/(5).

The non-real solutions of z^5 + 1 = 0 lie on the unit circle at z = e^iθ. For such z, (1)/(z) = e^-iθ, so u = z + (1)/(z) = e^iθ + e^-iθ = 2cosθ. The fifth roots of -1 (excluding z = -1) occur at θ = (π)/(5), (3π)/(5), (7π)/(5), (9π)/(5), which pair into conjugates giving the two distinct real values u = 2cos(π)/(5) and u = 2cos(3π)/(5). Solving u^2 - u - 1 = 0 by the quadratic formula: u = (1 sqrt(1 + 4))/(2) = (1 sqrt(5))/(2). Since (3π)/(5) is between (π)/(2) and π, cos(3π)/(5) is negative, so we take the negative root: 2cos(3π)/(5) = (1 - sqrt(5))/(2), hence cos(3π)/(5) = (1 - sqrt(5))/(4). Mark notes: 1 mark for recognising u = 2cosθ for roots on the unit circle, 1 mark for solving the quadratic, 1 mark for selecting the negative root to give cos(3π)/(5) = (1 - sqrt(5))/(4).

§-Syllabus dot point

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 built up stage by stage as a regular polygon on the unit circle, and their sum, with rigorous verified worked examples.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-20

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
De Moivre's theorem
Finding nth roots
Roots of unity
Why the roots of unity sum to zero
Multiple-angle formulae from de Moivre
Roots of a general complex number
Reducing arguments
How exam questions ask about de Moivre and 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). The geometry, $n$ points evenly spread round a circle forming a regular polygon, is the picture that makes every roots question quick, so this page builds it up carefully.

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.

Computing $(1+i)^8$ by repeated Cartesian multiplication is laborious and error-prone; in polar form it is two short lines. That contrast is the whole reason de Moivre's theorem is examined.

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 $1$ .

The figure builds the sixth roots of unity ( $z^6 = 1$ ) stage by stage, so you can see the regular polygon emerge.

Stage 1, start on the unit circle. Every root of $z^6 = 1$ has modulus $1$ , so all six lie on the unit circle. The simplest root is $k = 0$ : $\omega_0 = \cos 0 + i\sin 0 = 1$ , sitting on the positive real axis.

Stage 2, step round by $\tfrac{2\pi}{6} = 60^\circ$ . Each successive root advances the argument by $\dfrac{2\pi}{n} = 60^\circ$ . So $\omega_1 = \operatorname{cis}60^\circ$ and $\omega_2 = \operatorname{cis}120^\circ$ . Writing $\omega = \omega_1$ , the roots are the powers $1, \omega, \omega^2, \ldots$

Stage 3, all six roots. Continuing around, the six arguments are $0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ$ , equally spaced $\dfrac{360^\circ}{6} = 60^\circ$ apart. The root at $180^\circ$ is $-1$ , as expected for an even $n$ .

Stage 4, join them into a regular polygon. The six roots are the vertices of a regular hexagon inscribed in the unit circle. By symmetry the vertices balance about the origin, which is the geometric reason the roots of unity sum to zero.

Why the roots of unity sum to zero

A key property: the roots of unity sum to zero for $n \ge 2$ . There are two ways to see it. Algebraically, $z^n - 1 = (z - 1)(z^{n-1} + \dots + z + 1)$ , and the sum of all roots equals the negative of the coefficient of $z^{n-1}$ in $z^n - 1$ , which is $0$ . Geometrically, the roots are the vertices of a regular polygon centred at the origin, so the position vectors cancel in symmetric pairs (or triples), leaving a resultant of zero, exactly the balance you can see in the hexagon above.

Multiple-angle formulae from de Moivre

One of the most powerful uses of de Moivre's theorem is deriving identities for $\cos n\theta$ and $\sin n\theta$ . Expand $(\cos\theta + i\sin\theta)^n$ with the binomial theorem, then equate the real part to $\cos n\theta$ and the imaginary part to $\sin n\theta$ . Replacing $\sin^2\theta = 1 - \cos^2\theta$ converts the cosine identity to a polynomial in $\cos\theta$ alone, as in the $\cos 5\theta$ question above. This technique turns a trigonometric identity that would be hard to prove directly into a routine expansion.

Roots of a general complex number

The same method finds the roots of any complex number, not just $1$ . To solve $z^n = w$ , first write $w$ in modulus-argument form $w = R\operatorname{cis}\phi$ . The $n$ roots then sit on a circle of radius $R^{1/n}$ , equally spaced by $\frac{2\pi}{n}$ , with the first root at argument $\frac{\phi}{n}$ . Geometrically the roots of unity are the special case $w = 1$ ; multiplying every root of unity by one particular root of $z^n = w$ rotates and scales the regular polygon onto the roots of the general equation. This is why drawing the roots of unity well prepares you for every roots question.

Reducing arguments

After raising to a high power the argument often exceeds $2\pi$ or falls below $-2\pi$ . Always reduce it into a single revolution by adding or subtracting multiples of $2\pi$ before reading off the Cartesian value, since $\cos$ and $\sin$ are periodic with period $2\pi$ . Forgetting this step leaves the answer correct in principle but in an awkward, unmarked form.

How exam questions ask about de Moivre and roots of unity

"Find $z^n$ " or "evaluate $(\,\ldots\,)^n$ in the form $x + iy$ ": convert to polar form, apply de Moivre, reduce the argument, then convert back.
"Solve $z^n = w$ " or "find the $n$ th roots of ...": use the $n$ th-roots formula for $k = 0, \dots, n-1$ ; expect to plot them.
"Show that the roots ... form a regular polygon" or "sketch on an Argand diagram": state the common modulus $R^{1/n}$ and the equal spacing $\tfrac{2\pi}{n}$ .
"Show that the roots sum to zero" or " $1 + \omega + \omega^2 + \dots = 0$ ": use the factorisation of $z^n - 1$ or the symmetry of the polygon.
"Show that $\cos n\theta = \ldots$ ": expand $(\cos\theta + i\sin\theta)^n$ by the binomial theorem and take the real part.
"It is known that $u = z + \tfrac{1}{z}$ ...": recognise $z + \tfrac1z = 2\cos\theta$ for a root on the unit circle, then solve the resulting real equation.

Worked example

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

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

The cube roots of unity are the three solutions of $z^3 = 1$ . We use the general formula for $n$ th roots, then check the sum using the key property that the roots of unity always sum to zero.

Apply the $n$ th roots formula: Since $1 = \cos 0 + i\sin 0$ , we have $R = 1$ , $\phi = 0$ , and $n = 3$ . The general formula gives three roots, one for each value of $k$ from $0$ to $n - 1$ :

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

The three roots are equally spaced around the unit circle, $120^\circ$ apart.

Evaluate each root explicitly: Substitute $k = 0$ , $1$ , $2$ in turn, reading off the cosine and sine values at each angle:

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

Notice that $z_1$ and $z_2$ are complex conjugates, as expected for a real polynomial.

Verify that the three roots sum to zero: Add the three roots, grouping real and imaginary parts separately:

1 + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 1 - 1 + 0i = 0.

The imaginary parts cancel and the real parts sum to zero, confirming the result.

Check that $z_1$ is genuinely a cube root: Since $|z_1| = 1$ and $\arg z_1 = \frac{2\pi}{3}$ , de Moivre's theorem gives $z_1^3$ modulus $1^3 = 1$ and argument $3 \times \frac{2\pi}{3} = 2\pi$ , which equals the argument of $1$ . So $z_1^3 = 1$ . Correct.

Final answer: the three cube roots of unity are $1$ , $-\frac{1}{2} + \frac{\sqrt{3}}{2}i$ , and $-\frac{1}{2} - \frac{\sqrt{3}}{2}i$ , and their sum is $0$ .

Worked example: the fourth roots of a complex number

Solve $z^4 = -8 + 8\sqrt{3}\,i$ for the fourth roots

Solve $z^4 = -8 + 8\sqrt{3}\,i$ , giving the roots in modulus-argument form.

Write the right-hand side in polar form: Compute the modulus and argument of $w = -8 + 8\sqrt{3}\,i$ . The modulus is $R = \sqrt{(-8)^2 + (8\sqrt3)^2} = \sqrt{64 + 192} = \sqrt{256} = 16$ . The point is in the second quadrant with reference angle $\tan^{-1}\frac{8\sqrt3}{8} = \tan^{-1}\sqrt3 = \frac{\pi}{3}$ , so $\phi = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . Hence $w = 16\operatorname{cis}\frac{2\pi}{3}$ .

Apply the nth-roots formula: With $n = 4$ , the modulus of each root is $R^{1/4} = 16^{1/4} = 2$ , and the arguments are $\dfrac{\phi + 2\pi k}{4}$ for $k = 0, 1, 2, 3$ :

\theta_k = \frac{\frac{2\pi}{3} + 2\pi k}{4} = \frac{\pi}{6} + \frac{\pi k}{2}.

List the four roots: Substituting $k = 0, 1, 2, 3$ and keeping each argument in $(-\pi, \pi]$ :

$k = 0$ : $2\operatorname{cis}\dfrac{\pi}{6}$ .
$k = 1$ : $2\operatorname{cis}\dfrac{2\pi}{3}$ .
$k = 2$ : $2\operatorname{cis}\dfrac{7\pi}{6} = 2\operatorname{cis}\!\left(-\dfrac{5\pi}{6}\right)$ .
$k = 3$ : $2\operatorname{cis}\dfrac{5\pi}{3} = 2\operatorname{cis}\!\left(-\dfrac{\pi}{3}\right)$ .

Sense-check the geometry: All four roots have modulus $2$ and are spaced $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$ apart, so they are the vertices of a square centred at the origin. That even spacing confirms the arguments are correct.

Final answer: the four roots are $2\operatorname{cis}\dfrac{\pi}{6}$ , $2\operatorname{cis}\dfrac{2\pi}{3}$ , $2\operatorname{cis}\!\left(-\dfrac{5\pi}{6}\right)$ and $2\operatorname{cis}\!\left(-\dfrac{\pi}{3}\right)$ .

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.

HSC 20213 marksUsing de Moivre's theorem and the binomial expansion of

(\cos\theta + i\sin\theta)^5

, or otherwise, show that

\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta

Show worked answer →

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

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

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

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: $\cos 5\theta = c^5 - 10 c^3 s^2 + 5 c s^4$ .

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

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

Hence $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ .

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

HSC 20222 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 $\arctan\frac{1}{\sqrt{3}} = \frac{\pi}{6}$ , so $\arg = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ .

Exponential form: $-\sqrt{3} + i = 2 e^{i 5\pi/6}$ .

Now apply de Moivre to the tenth power:

$(-\sqrt{3} + i)^{10} = 2^{10} e^{i 50\pi/6} = 1024\, e^{i 25\pi/3}$ .

Reduce the argument modulo $2\pi$ : $\frac{25\pi}{3} - 8\pi = \frac{25\pi - 24\pi}{3} = \frac{\pi}{3}$ .

So the value is $1024\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 1024\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 512 + 512\sqrt{3}\, i$ .

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

HSC 20223 marksConsider the equation

z^5 + 1 = 0

. It is known that if

z

is a solution with

z \neq -1

, then

u = z + \frac{1}{z}

is a solution of

u^2 - u - 1 = 0

. Hence find the exact value of

\cos\frac{3\pi}{5}

Show worked answer →

The non-real solutions of $z^5 + 1 = 0$ lie on the unit circle at $z = e^{i\theta}$ . For such $z$ , $\frac{1}{z} = e^{-i\theta}$ , so

$u = z + \frac{1}{z} = e^{i\theta} + e^{-i\theta} = 2\cos\theta$ .

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

Solving $u^2 - u - 1 = 0$ by the quadratic formula: $u = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$ .

Since $\frac{3\pi}{5}$ is between $\frac{\pi}{2}$ and $\pi$ , $\cos\frac{3\pi}{5}$ is negative, so we take the negative root:

$2\cos\frac{3\pi}{5} = \frac{1 - \sqrt{5}}{2}$ , hence $\cos\frac{3\pi}{5} = \frac{1 - \sqrt{5}}{4}$ .

Mark notes: 1 mark for recognising $u = 2\cos\theta$ for roots on the unit circle, 1 mark for solving the quadratic, 1 mark for selecting the negative root to give $\cos\frac{3\pi}{5} = \frac{1 - \sqrt{5}}{4}$ .