De Moivre's theorem | Year 12

01 / LEARN

The idea, step by step

Repeated multiplication becomes a multiplied angle

For any integer $n$ , $(cis θ)^{n} = cis (n θ)$ . For positive integers this follows by repeated use of the angle-addition multiplication rule. The $n = 0$ case gives 1; negative powers follow by reciprocals.

Hence ${(r cis θ)}^{n} = r^{n} cis (n θ)$ . Require $r > 0$ for negative $n$ . For a non-negative integer power of zero use ordinary arithmetic; $0^{0}$ and negative powers of zero are not defined here.

{(cos θ + i sin θ)}^{n} = cos (n θ) + i sin (n θ), n \in ℤ

Compare real and imaginary parts

Expanding ${(cos θ + i sin θ)}^{n}$ algebraically and comparing with de Moivre produces identities. Both real and imaginary components must match. This gives a proof for all real $θ$ , rather than a check at a few angles.

Integer powers do not require a choice of argument branch. Replacing $θ$ by $θ + 2 k π$ changes $n θ$ by an integer number of full turns and leaves the answer unchanged.

A common trap

Raising the angle itself to the nth power.

Instead: De Moivre multiplies the angle by n; only the modulus is raised to the nth power.

02 / WORKED EXAMPLES

See the thinking

Example 1

A high exact power

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

$1 + i = \sqrt{2} cis (π / 4)$ .
The eighth power is ${(\sqrt{2})}^{8} cis (2 π) = 16$ .
The imaginary part is zero because a full turn gives direction zero.

16

Example 2

Derive the triple-angle sine identity

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

Put $c = cos θ$ , $s = sin θ$ and expand ${(c + i s)}^{3} = (c^{3} - 3 c s^{2}) + i (3 c^{2} s - s^{3})$ .
Compare imaginary parts with $cis 3 θ$ : $sin 3 θ = 3 c^{2} s - s^{3}$ .
Use $c^{2} = 1 - s^{2}$ to obtain $3 s - 4 s^{3}$ .

sin 3 θ = 3 sin θ - 4 {sin}^{3} θ

Check your understanding

Find ${(2 cis (π / 6))}^{3}$ .

Show answer

$8 i$ .

$2^{3} cis (π / 2) = 8 i$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

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

Read the worked solutions

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

Calculate ${(\sqrt{3} + i)}^{3}$ .
2 marks
$\sqrt{3} + i = 2 cis (π / 6)$ , so its cube is $8 cis (π / 2) = 8 i$ .
$8 i$ .
Calculate ${(1 - i)}^{6}$ .
2 marks
$1 - i = \sqrt{2} cis (- π / 4)$ . The sixth power is $8 cis (- 3 π / 2) = 8 i$ .
$8 i$ .
Write ${(2 cis (π / 3))}^{- 2}$ in polar form.
2 marks
Raise the nonzero modulus to $- 2$ and multiply the angle by $- 2$ .
$\frac{1}{4} cis (- 2 π / 3)$ .
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 ${(cis (π / 5))}^{10} = 1$ .
2 marks
De Moivre gives $cis (10 π / 5) = cis (2 π) = 1$ .
The tenth power is 1.
Find the least positive integer $n$ for which ${(cis (2 π / 7))}^{n} = 1$ .
3 marks
We require $2 n π / 7 = 2 k π$ , 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 ↗

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