Write z^2 + 1z^2 in terms of a cosine, where z = cosθ + isinθ. [1 mark]

Using the expansion of (cosθ + isinθ)^2, find cos 2θ in terms of cosθ. [2 marks]

SEAB Original-style practice: Use de Moivre's theorem to express cos 3θ in terms of cosθ.

By de Moivre, (cosθ + isinθ)^3 = cos 3θ + isin 3θ. Expand the left side with the binomial theorem (writing c = cosθ, s = sinθ): $c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3.$ Group real and imaginary parts: $= (c^3 - 3cs^2) + i(3c^2 s - s^3).$ Equate real parts: cos 3θ = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4c^3 - 3c. So cos 3θ = 4cos^3θ - 3cosθ. Markers reward applying de Moivre, the binomial expansion, equating real parts, and using s^2 = 1 - c^2 to reach the final identity.

SEAB Original-style practice: By considering z + 1z where z = cosθ + isinθ, show that cos^4θ = 18(cos 4θ + 4cos 2θ + 3).

With z = cosθ + isinθ, de Moivre gives z^n = cos nθ + isin nθ and z^-n = cos nθ - isin nθ, so $z^n + z^-n = 2cos nθ, in particular z + 1z = 2cosθ.$ Therefore (2cosθ)^4 = (z + 1z)^4. Expand by the binomial theorem: $(z + 1z)^4 = z^4 + 4z^2 + 6 + 4z^-2 + z^-4.$ Pair the terms: (z^4 + z^-4) + 4(z^2 + z^-2) + 6 = 2cos 4θ + 4(2cos 2θ) + 6 = 2cos 4θ + 8cos 2θ + 6. So 16cos^4θ = 2cos 4θ + 8cos 2θ + 6, hence cos^4θ = 18(cos 4θ + 4cos 2θ + 3). Markers reward the result z^n + z^-n = 2cos nθ, the binomial expansion, pairing terms into cosines, and dividing through to the stated identity.

SingaporeFurther MathsSyllabus dot point

How does de Moivre's theorem let us take powers of complex numbers and derive trigonometric identities?

State and apply de Moivre's theorem to find powers of complex numbers and to derive multiple-angle and power-reduction trigonometric identities

A focused answer to the H2 Further Mathematics outcome on de Moivre's theorem. The statement for integer powers, using it to expand multiple angles, deriving cos and sin of n-theta, and the z plus one over z method for power-reduction identities.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking

SEAB wants you to state de Moivre's theorem and use it for two purposes: computing integer powers of complex numbers in polar form, and deriving trigonometric identities. Two techniques follow from it: expanding $(\cos\theta + i\sin\theta)^n$ to get multiple-angle formulae, and the $z + \tfrac{1}{z}$ method to express powers of $\cos\theta$ or $\sin\theta$ as sums of cosines or sines of multiple angles.

The answer

Statement of de Moivre's theorem

For any integer $n$ ,

(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta,

equivalently $\left(r\mathrm{e}^{i\theta}\right)^n = r^n\mathrm{e}^{in\theta}$ . Raising to a power multiplies the argument by $n$ and raises the modulus to the $n$ th power.

Multiple-angle identities

To express $\cos n\theta$ or $\sin n\theta$ in powers of $\cos\theta$ and $\sin\theta$ , expand $(\cos\theta + i\sin\theta)^n$ with the binomial theorem and equate real and imaginary parts. The real part gives $\cos n\theta$ , the imaginary part gives $\sin n\theta$ . Use $\sin^2\theta = 1 - \cos^2\theta$ to tidy the result.

Power-reduction with the z + 1/z method

Let $z = \cos\theta + i\sin\theta$ , so that

z^n + \frac{1}{z^n} = 2\cos n\theta, \qquad z^n - \frac{1}{z^n} = 2i\sin n\theta.

To express, say, $\cos^4\theta$ as a sum of multiple-angle cosines, expand $\left(z + \tfrac{1}{z}\right)^4 = (2\cos\theta)^4$ by the binomial theorem, then pair terms $z^k + z^{-k}$ into $2\cos k\theta$ . This is the standard route for integrating powers of sine and cosine.

Choosing the right technique

Use the expansion method to write a single multiple angle in powers of the basic ratio. Use the $z + \tfrac{1}{z}$ method to go the other way, writing a power of the basic ratio as a linear combination of multiple angles.

Worked example

Use de Moivre's theorem to express $\sin 3\theta$ in terms of $\sin\theta$ .

Step 1: Apply de Moivre

(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta.

Step 2: Expand the left side

With $c = \cos\theta$ , $s = \sin\theta$ :

(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3.

Step 3: Take the imaginary part

The imaginary part is $\sin 3\theta = 3c^2 s - s^3$ .

Step 4: Eliminate cosine using the Pythagorean identity

Replace $c^2 = 1 - s^2$ :

\sin 3\theta = 3(1 - s^2)s - s^3 = 3s - 3s^3 - s^3 = 3s - 4s^3.

Step 5: State the identity

\sin 3\theta = 3\sin\theta - 4\sin^3\theta.

Common traps

Mishandling powers of $i$: $(is)^2 = -s^2$ and $(is)^3 = -is^3$ ; sign errors here corrupt the whole expansion.
Equating the wrong part: $\cos n\theta$ is the real part and $\sin n\theta$ the imaginary part; swapping them gives the wrong identity.
Forgetting to substitute the Pythagorean identity: A multiple-angle identity in a single ratio needs $\sin^2 = 1 - \cos^2$ (or vice versa) to remove the other ratio.
Dropping the binomial coefficients: The expansion of $\left(z + \tfrac{1}{z}\right)^4$ has coefficients $1, 4, 6, 4, 1$ ; omitting them breaks the pairing into cosines.
Pairing terms incorrectly: Only $z^k + z^{-k}$ pairs to $2\cos k\theta$ ; track the matching positive and negative powers carefully.

Examples in context

Example 1. Integrating trigonometric powers. To integrate $\cos^4\theta$ , the $z + \tfrac{1}{z}$ method first rewrites it as $\tfrac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)$ , which integrates term by term. This is why power-reduction is taught alongside further integration.

Example 2. Signal harmonics. Expressing a power of a sinusoid as a sum of multiple-angle terms reveals the harmonic content of a distorted signal: $\cos^3\theta$ contains a $\cos 3\theta$ harmonic, exactly the frequency-tripling seen in nonlinear electronics.

Try this

Q1. State de Moivre's theorem for an integer power $n$ . [1 mark]

Cue. $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ .

Q2. Write $z^2 + \dfrac{1}{z^2}$ in terms of a cosine, where $z = \cos\theta + i\sin\theta$ . [1 mark]

Cue. $z^2 + z^{-2} = 2\cos 2\theta$ .

Q3. Using the expansion of $(\cos\theta + i\sin\theta)^2$ , find $\cos 2\theta$ in terms of $\cos\theta$ . [2 marks]

Cue. Real part of $(c + is)^2 = c^2 - s^2$ , so $\cos 2\theta = c^2 - (1 - c^2) = 2\cos^2\theta - 1$ .

Exam-style practice questions

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

Original6 marksUse de Moivre's theorem to express

\cos 3\theta

in terms of

\cos\theta

Show worked answer →

By de Moivre, $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$ .

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

c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3.

Group real and imaginary parts:

= (c^3 - 3cs^2) + i(3c^2 s - s^3).

Equate real parts:

\cos 3\theta = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4c^3 - 3c

So $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ .

Markers reward applying de Moivre, the binomial expansion, equating real parts, and using $s^2 = 1 - c^2$ to reach the final identity.

Original7 marksBy considering

z + \dfrac{1}{z}

where

z = \cos\theta + i\sin\theta

, show that

\cos^4\theta = \dfrac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)

Show worked answer →

With $z = \cos\theta + i\sin\theta$ , de Moivre gives $z^n = \cos n\theta + i\sin n\theta$ and $z^{-n} = \cos n\theta - i\sin n\theta$ , so

z^n + z^{-n} = 2\cos n\theta, \qquad \text{in particular } z + \tfrac{1}{z} = 2\cos\theta.

Therefore

(2\cos\theta)^4 = \left(z + \tfrac{1}{z}\right)^4

. Expand by the binomial theorem:

\left(z + \tfrac{1}{z}\right)^4 = z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4}.

Pair the terms:

(z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6 = 2\cos 4\theta + 4(2\cos 2\theta) + 6 = 2\cos 4\theta + 8\cos 2\theta + 6

So $16\cos^4\theta = 2\cos 4\theta + 8\cos 2\theta + 6$ , hence $\cos^4\theta = \tfrac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)$ .

Markers reward the result $z^n + z^{-n} = 2\cos n\theta$ , the binomial expansion, pairing terms into cosines, and dividing through to the stated identity.