De Moivre's theorem | Year 12

01 / LEARN

Understand the idea.

Repeated multiplication becomes a multiplied angle

A power repeats the same complex multiplication. Each multiplication multiplies the modulus by $r$ and adds another $θ$ to the argument. For $r > 0$ , de Moivre's theorem gives ${(r cis θ)}^{n} = r^{n} cis (n θ)$ for every integer $n$ .

For a nonzero base, exponent zero gives $1$ and negative powers use reciprocals. Adding a full turn to the starting argument leaves an integer power unchanged. Positive powers of zero are zero; $0^{0}$ and negative powers of zero are undefined here.

\begin{array}{c} {(cos θ + i sin θ)}^{n} \\ = cos (n θ) + i sin (n θ) \\ n \in ℤ \end{array}

Compare real and imaginary parts

The same power can be calculated by bracket expansion and by de Moivre. Because equal complex numbers have equal real and imaginary parts, comparing the results proves trigonometric identities for every real angle.

Keep sine and cosine symbolic during a proof. After comparing the required component, use ${sin}^{2} θ + {cos}^{2} θ = 1$ to simplify. Substitution of particular angles can check arithmetic, but cannot establish an identity for all angles.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A high exact power

Calculate ${(1 + i)}^{8}$ .

Worked solution

1
Express the base in polar form.
$1 + i = \sqrt{2} cis \frac{π}{4}$
The point has modulus $\sqrt{2}$ and lies at angle $\frac{π}{4}$ .
2
Raise the modulus and multiply the angle.
${(1 + i)}^{8} = {(\sqrt{2})}^{8} cis (8 \cdot \frac{π}{4})$
De Moivre applies because the exponent is an integer.
3
Evaluate the full turn.
$16 cis (2 π) = 16 (1 + 0 i) = 16$
A full turn lies on the positive real axis, so the imaginary component vanishes.

Answer

$16$ .

EXAMPLE 02

Derive the triple-angle sine identity

Use de Moivre to derive a formula for $sin 3 θ$ .

Worked solution

1
Write the cubic using de Moivre.
${(cos θ + i sin θ)}^{3} = cos 3 θ + i sin 3 θ$
The third power triples the angle for every real $θ$ .
2
Expand the same cubic algebraically.
$\begin{aligned} {(c + i s)}^{3} & = c^{3} + 3 i c^{2} s - 3 c s^{2} - i s^{3} \\ = (c^{3} - 3 c s^{2}) + i (3 c^{2} s - s^{3}) \end{aligned}$
Here $c = cos θ$ and $s = sin θ$ ; binomial expansion uses $i^{2} = - 1$ and $i^{3} = - i$ .
3
Equate imaginary components.
$sin 3 θ = 3 {cos}^{2} θ sin θ - {sin}^{3} θ$
Two equal complex numbers have the same imaginary component.
4
Eliminate the squared cosine.
$\begin{aligned} sin 3 θ & = 3 (1 - {sin}^{2} θ) sin θ - {sin}^{3} θ \\ = 3 sin θ - 4 {sin}^{3} θ \end{aligned}$
The Pythagorean identity holds for every real $θ$ , completing the proof.

Answer

$sin 3 θ = 3 sin θ - 4 {sin}^{3} θ$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find ${(2 cis (\frac{π}{6}))}^{3}$ .

Show answer & reasoning

$8 i$ .

The modulus becomes $2^{3} = 8$ and the argument becomes $3 (\frac{π}{6}) = \frac{π}{2}$ . Since $cis (\frac{π}{2}) = i$ , the result is $8 i$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Calculate ${(\sqrt{3} + i)}^{3}$ .
2 marks
Calculate ${(1 - i)}^{6}$ .
2 marks
Write ${(2 cis (\frac{π}{3}))}^{- 2}$ in polar form.
2 marks
Use de Moivre and a square expansion to obtain a formula for $cos 2 θ$ .
3 marks
Show that ${(cis (\frac{π}{5}))}^{10} = 1$ .
2 marks
Find the least positive integer $n$ for which ${(cis (\frac{2 π}{7}))}^{n} = 1$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Calculate ${(\sqrt{3} + i)}^{3}$ .
2 marks
$\sqrt{3} + i = 2 cis (\frac{π}{6})$ , so its cube is $8 cis (\frac{π}{2}) = 8 i$ .
$8 i$ .
Calculate ${(1 - i)}^{6}$ .
2 marks
$1 - i = \sqrt{2} cis (- \frac{π}{4})$ . The sixth power is $8 cis (- \frac{3 π}{2}) = 8 i$ .
$8 i$ .
Write ${(2 cis (\frac{π}{3}))}^{- 2}$ in polar form.
2 marks
Raise the nonzero modulus to $- 2$ and multiply the angle by $- 2$ .
$\frac{1}{4} cis (- \frac{2 π}{3})$ .
Use de Moivre and a square expansion to obtain a formula for $cos 2 θ$ .
3 marks
The real part of ${(cos θ + i sin θ)}^{2}$ is ${cos}^{2} θ - {sin}^{2} θ$ ; de Moivre gives real part $cos 2 θ$ .
$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ$ .
Show that ${(cis (\frac{π}{5}))}^{10} = 1$ .
2 marks
De Moivre gives $cis (\frac{10 π}{5}) = cis (2 π) = 1$ .
The tenth power is 1.
Find the least positive integer $n$ for which ${(cis (\frac{2 π}{7}))}^{n} = 1$ .
3 marks
We require $\frac{2 n π}{7} = 2 k π$ , so $n = 7 k$ . The least positive integer is 7.
$n = 7$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Repeated multiplication becomes a multiplied angle

Compare real and imaginary parts

See the method in action.

A high exact power

Express the base in polar form.

Raise the modulus and multiply the angle.

Evaluate the full turn.

Derive the triple-angle sine identity

Write the cubic using de Moivre.

Expand the same cubic algebraically.

Equate imaginary components.

Eliminate the squared cosine.

Try it yourself.

Check your understanding

Keep practising