De Moivre's Theorem

De Moivre's theorem is the engine room of the complex-numbers strand. In a single formula it says that raising $\cos\theta + i\sin\theta$cosθ+isinθ to a power simply multiplies the angle: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$(cosθ+isinθ)n=cosnθ+isinnθ. That one line lets you compute brutal powers in seconds, derive every multiple-angle trigonometric identity by algebra rather than memory, rewrite powers of $\cos\theta$cosθ and $\sin\theta$sinθ in a form ready to integrate, and — in the next lesson — extract $n$nth roots. It is the place where the geometry of "add the arguments" becomes a precision tool, and it deserves to be at your fingertips.

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Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2, drawing on polar multiplication (the previous lesson) and the proof-by-induction technique from the proof strand. It tests AO1 (apply the theorem to compute powers), AO2 (prove the theorem by induction and justify the equating-of-parts step when deriving identities — the command word is usually "Show that" or "Prove") and AO3 (assemble De Moivre, the binomial theorem and the $z + 1/z$z+1/z substitution into a multi-step derivation). It feeds directly into $n$nth roots of unity (next lesson) and into the integration of powers of trigonometric functions in the further-calculus strand.

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Core theory: statement, proof, and the two main uses

Statement

For any integer $n$n (positive, negative or zero) and any real $\theta$θ,

$\boxed{\,(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.\,}$(cosθ+isinθ)n=cosnθ+isinnθ.​

Combined with the modulus, if $z = r(\cos\theta + i\sin\theta)$z=r(cosθ+isinθ) then $z^n = r^n(\cos n\theta + i\sin n\theta)$zn=rn(cosnθ+isinnθ) — the modulus is raised to the power and the argument multiplied by $n$n. A common shorthand is $\operatorname{cis}\theta := \cos\theta + i\sin\theta$cisθ:=cosθ+isinθ, so the theorem reads $(\operatorname{cis}\theta)^n = \operatorname{cis}(n\theta)$(cisθ)n=cis(nθ).

Proof by induction (positive integers)

Base case $n = 1$n=1: $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta = \cos(1\cdot\theta) + i\sin(1\cdot\theta)$(cosθ+isinθ)1=cosθ+isinθ=cos(1⋅θ)+isin(1⋅θ) ✓.

Inductive step. Assume the result for $n = k$n=k: $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$(cosθ+isinθ)k=coskθ+isinkθ. Then

$\begin{aligned} (\cos\theta + i\sin\theta)^{k+1} &= (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta) \\ &= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) \quad (\text{inductive hypothesis}) \\ &= \cos(k\theta + \theta) + i\sin(k\theta + \theta) \quad (\text{polar multiplication rule}) \\ &= \cos((k+1)\theta) + i\sin((k+1)\theta). \end{aligned}$(cosθ+isinθ)k+1​=(cosθ+isinθ)k(cosθ+isinθ)=(coskθ+isinkθ)(cosθ+isinθ)(inductive hypothesis)=cos(kθ+θ)+isin(kθ+θ)(polar multiplication rule)=cos((k+1)θ)+isin((k+1)θ).​

So the result holds for $n = k+1$n=k+1, and by induction for all positive integers $n$n.

Extending to $n \le 0$n≤0. For $n = 0$n=0, $(\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0$(cosθ+isinθ)0=1=cos0+isin0 ✓. For a negative integer, write $n = -m$n=−m with $m > 0$m>0:

$(\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos m\theta + i\sin m\theta} = \cos m\theta - i\sin m\theta = \cos(-m\theta) + i\sin(-m\theta),$(cosθ+isinθ)−m=(cosθ+isinθ)m1​=cosmθ+isinmθ1​=cosmθ−isinmθ=cos(−mθ)+isin(−mθ),

where the third equality realises the denominator (its modulus is $1$1, so its reciprocal is its conjugate). Thus the theorem holds for all integers. (For rational exponents it becomes the multi-valued $n$nth-root statement of the next lesson.)

Use 1 — computing powers

Convert the base to polar form, raise the modulus to the power, multiply the argument by $n$n, then reduce the argument and convert back. This replaces a length-$n$n binomial expansion with one line. The saving grows with $n$n: computing $(1 + i)^{20}$(1+i)20 by binomial expansion means twenty-one terms with alternating signs, whereas in polar form it is $(\sqrt 2)^{20}\,\operatorname{cis}(20\cdot\frac{\pi}{4}) = 2^{10}\,\operatorname{cis}(5\pi) = 1024\,\operatorname{cis}\pi = -1024$(2​)20cis(20⋅4π​)=210cis(5π)=1024cisπ=−1024 in a single line. Whenever you see a power of a complex number above the second or third, reach for polar form first.

Use 2 — multiple-angle identities

Because $(\cos\theta + i\sin\theta)^n$(cosθ+isinθ)n equals both $\cos n\theta + i\sin n\theta$cosnθ+isinnθ (De Moivre) and a binomial expansion in $\cos\theta, \sin\theta$cosθ,sinθ, equating real and imaginary parts delivers $\cos n\theta$cosnθ and $\sin n\theta$sinnθ as polynomials in $\cos\theta$cosθ and $\sin\theta$sinθ. The real part always gives $\cos n\theta$cosnθ (a polynomial in $\cos\theta$cosθ once $\sin^2\theta = 1 - \cos^2\theta$sin2θ=1−cos2θ is used to clear even powers of $\sin\theta$sinθ), and the imaginary part gives $\sin n\theta$sinnθ. This is the only systematic way to generate these identities for general $n$n; there is no point memorising $\cos 5\theta$cos5θ when you can derive it in four lines.

The $z + 1/z$z+1/z substitution — powers of cos and sin

Put $z = \cos\theta + i\sin\theta$z=cosθ+isinθ, so $|z| = 1$∣z∣=1 and $z^{-1} = \cos\theta - i\sin\theta$z−1=cosθ−isinθ. Adding and subtracting,

$z + \frac{1}{z} = 2\cos\theta, \qquad z - \frac{1}{z} = 2i\sin\theta,$z+z1​=2cosθ,z−z1​=2isinθ,

and, applying De Moivre to $z^n$zn,

$z^n + \frac{1}{z^n} = 2\cos n\theta, \qquad z^n - \frac{1}{z^n} = 2i\sin n\theta.$zn+zn1​=2cosnθ,zn−zn1​=2isinnθ.

Expanding $\left(z + \tfrac1z\right)^n$(z+z1​)n or $\left(z - \tfrac1z\right)^n$(z−z1​)n and pairing $z^k$zk with $z^{-k}$z−k then rewrites $\cos^n\theta$cosnθ or $\sin^n\theta$sinnθ as a sum of multiple-angle cosines/sines — exactly the form needed for integration.

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Worked examples (with mark scheme)

Example 1 — a power

Find $(1 + i)^{10}$(1+i)10.

Convert: $|1 + i| = \sqrt 2$∣1+i∣=2​, $\arg(1+i) = \frac{\pi}{4}$arg(1+i)=4π​, so $1 + i = \sqrt 2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$1+i=2​(cos4π​+isin4π​). (M1 polar form of base.) By De Moivre,

$(1 + i)^{10} = (\sqrt 2)^{10}\left(\cos\frac{10\pi}{4} + i\sin\frac{10\pi}{4}\right) = 32\left(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right). \quad (\text{M1 apply theorem})$(1+i)10=(2​)10(cos410π​+isin410π​)=32(cos25π​+isin25π​).(M1 apply theorem)

Reduce $\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$25π​=2π+2π​:

$(1 + i)^{10} = 32\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 32(0 + i) = 32i. \quad (\text{A1})$(1+i)10=32(cos2π​+isin2π​)=32(0+i)=32i.(A1)

(M1 base in polar form; M1 raising modulus and multiplying argument; A1 for $32i$32i. The half-power $(\sqrt 2)^{10} = 2^5 = 32$(2​)10=25=32 is a frequent slip.)

Example 2 — a power giving a real answer

Find $(\sqrt 3 - i)^6$(3​−i)6.

$|\sqrt 3 - i| = \sqrt{3 + 1} = 2$∣3​−i∣=3+1​=2; the point is in the fourth quadrant, so $\arg = -\frac{\pi}{6}$arg=−6π​. (M1 modulus and argument.)

$(\sqrt 3 - i)^6 = 2^6\left(\cos\left(6\cdot-\frac{\pi}{6}\right) + i\sin\left(6\cdot-\frac{\pi}{6}\right)\right) = 64(\cos(-\pi) + i\sin(-\pi)) = 64(-1 + 0) = -64. \quad (\text{A1})$(3​−i)6=26(cos(6⋅−6π​)+isin(6⋅−6π​))=64(cos(−π)+isin(−π))=64(−1+0)=−64.(A1)

(M1 for the polar base with the correct (negative) argument; A1 for $-64$−64. Here $6\theta = -\pi$6θ=−π lands on the negative real axis, giving a real answer; getting the quadrant of the base right is essential in general, because for many powers the sign of $\theta$θ changes the imaginary part of the result.)

Example 3 — multiple-angle identity for $\cos 3\theta$cos3θ and $\sin 3\theta$sin3θ

Express $\cos 3\theta$cos3θ and $\sin 3\theta$sin3θ in terms of $\cos\theta$cosθ and $\sin\theta$sinθ.

By De Moivre, $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$(cosθ+isinθ)3=cos3θ+isin3θ. Expand the left side by the binomial theorem (write $c = \cos\theta$c=cosθ, $s = \sin\theta$s=sinθ):

$(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3. \quad (\text{M1 binomial expand; M1 use } i^2=-1, i^3=-i)$(c+is)3=c3+3c2(is)+3c(is)2+(is)3=c3+3ic2s−3cs2−is3.(M1 binomial expand; M1 use i2=−1,i3=−i)

Group real and imaginary parts:

$(c + is)^3 = (c^3 - 3cs^2) + i(3c^2 s - s^3).$(c+is)3=(c3−3cs2)+i(3c2s−s3).

Equate with $\cos 3\theta + i\sin 3\theta$cos3θ+isin3θ:

$\cos 3\theta = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4\cos^3\theta - 3\cos\theta, \quad (\text{A1})$cos3θ=c3−3cs2=c3−3c(1−c2)=4cos3θ−3cosθ,(A1) $\sin 3\theta = 3c^2 s - s^3 = 3(1 - s^2)s - s^3 = 3\sin\theta - 4\sin^3\theta. \quad (\text{A1})$sin3θ=3c2s−s3=3(1−s2)s−s3=3sinθ−4sin3θ.(A1)

(M1 expansion; M1 simplifying powers of $i$i; A1 each identity after using $s^2 = 1 - c^2$s2=1−c2 / $c^2 = 1 - s^2$c2=1−s2.)

Example 4 — power of cosine via $z + 1/z$z+1/z

Express $\cos^4\theta$cos4θ as a sum of cosines of multiple angles.

Let $z = \cos\theta + i\sin\theta$z=cosθ+isinθ, so $z + \tfrac1z = 2\cos\theta$z+z1​=2cosθ. Then $(2\cos\theta)^4 = \left(z + \tfrac1z\right)^4$(2cosθ)4=(z+z1​)4, and by the binomial theorem

$\left(z + \frac1z\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4} = \left(z^4 + \frac{1}{z^4}\right) + 4\left(z^2 + \frac{1}{z^2}\right) + 6. \quad (\text{M1 expand and pair})$(z+z1​)4=z4+4z2+6+z24​+z41​=(z4+z41​)+4(z2+z21​)+6.(M1 expand and pair)

Using $z^k + z^{-k} = 2\cos k\theta$zk+z−k=2coskθ,

$16\cos^4\theta = 2\cos 4\theta + 8\cos 2\theta + 6 \;\Longrightarrow\; \cos^4\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}. \quad (\text{A1})$16cos4θ=2cos4θ+8cos2θ+6⟹cos4θ=81​cos4θ+21​cos2θ+83​.(A1)

(M1 for the binomial expansion and pairing $z^k$zk with $z^{-k}$z−k; A1 for the final form. This is the standard route to integrate $\cos^4\theta$cos4θ: the right-hand side integrates term-by-term.)

Example 5 — a negative power

Find $(1 + i)^{-4}$(1+i)−4, giving your answer in the form $a + bi$a+bi.

The theorem holds for negative integers, so convert and apply it directly. With $1 + i = \sqrt 2\,\operatorname{cis}\frac{\pi}{4}$1+i=2​cis4π​,

$(1 + i)^{-4} = (\sqrt 2)^{-4}\left(\cos\left(-4\cdot\frac{\pi}{4}\right) + i\sin\left(-4\cdot\frac{\pi}{4}\right)\right) = \frac{1}{4}\big(\cos(-\pi) + i\sin(-\pi)\big). \quad (\text{M1 negative power; M1 evaluate})$(1+i)−4=(2​)−4(cos(−4⋅4π​)+isin(−4⋅4π​))=41​(cos(−π)+isin(−π)).(M1 negative power; M1 evaluate)

Since $\cos(-\pi) = -1$cos(−π)=−1 and $\sin(-\pi) = 0$sin(−π)=0,

$(1 + i)^{-4} = \frac{1}{4}(-1 + 0) = -\frac{1}{4}. \quad (\text{A1})$(1+i)−4=41​(−1+0)=−41​.(A1)

(M1 applying De Moivre with $n = -4$n=−4; M1 for $(\sqrt 2)^{-4} = \tfrac14$(2​)−4=41​ and $-4\cdot\tfrac\pi4 = -\pi$−4⋅4π​=−π; A1 for $-\tfrac14$−41​. Check via positive powers: $(1+i)^4 = ((1+i)^2)^2 = (2i)^2 = -4$(1+i)4=((1+i)2)2=(2i)2=−4, so $(1+i)^{-4} = \tfrac{1}{-4} = -\tfrac14$(1+i)−4=−41​=−41​ ✓ — a reassuring confirmation that the negative-power case behaves exactly as expected.)

Example 6 — multiple-angle identity for $\sin 4\theta$sin4θ

Express $\sin 4\theta$sin4θ in terms of $\sin\theta$sinθ and $\cos\theta$cosθ.

By De Moivre, $(\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta$(cosθ+isinθ)4=cos4θ+isin4θ. Expand with $c = \cos\theta, s = \sin\theta$c=cosθ,s=sinθ:

$(c + is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4 = c^4 + 4ic^3 s - 6c^2 s^2 - 4ics^3 + s^4. \quad (\text{M1 binomial; M1 powers of } i)$(c+is)4=c4+4c3(is)+6c2(is)2+4c(is)3+(is)4=c4+4ic3s−6c2s2−4ics3+s4.(M1 binomial; M1 powers of i)

The imaginary part gives

$\sin 4\theta = 4c^3 s - 4cs^3 = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta = 4\cos\theta\sin\theta(\cos^2\theta - \sin^2\theta). \quad (\text{A1})$sin4θ=4c3s−4cs3=4cos3θsinθ−4cosθsin3θ=4cosθsinθ(cos2θ−sin2θ).(A1)

Using $2\cos\theta\sin\theta = \sin 2\theta$2cosθsinθ=sin2θ and $\cos^2\theta - \sin^2\theta = \cos 2\theta$cos2θ−sin2θ=cos2θ, this is $\sin 4\theta = 2\sin 2\theta\cos 2\theta$sin4θ=2sin2θcos2θ — the double-angle formula applied twice, recovered as a consistency check. (M1 expansion; M1 powers of $i$i; A1 the imaginary part, optionally factorised.)

Example 7 — a power with a non-trivial Cartesian answer

Find $(1 + i\sqrt 3)^5$(1+i3​)5, giving your answer in the form $a + bi$a+bi.

Convert: $|1 + i\sqrt 3| = \sqrt{1 + 3} = 2$∣1+i3​∣=1+3​=2, and the point is in the first quadrant with $\arg = \arctan\sqrt 3 = \frac{\pi}{3}$arg=arctan3​=3π​, so $1 + i\sqrt 3 = 2\,\operatorname{cis}\frac{\pi}{3}$1+i3​=2cis3π​. (M1 polar base.) By De Moivre,

$(1 + i\sqrt 3)^5 = 2^5\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right) = 32\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right). \quad (\text{M1 apply theorem})$(1+i3​)5=25(cos35π​+isin35π​)=32(cos35π​+isin35π​).(M1 apply theorem)

Now $\frac{5\pi}{3}$35π​ is in the fourth quadrant, with $\cos\frac{5\pi}{3} = \frac12$cos35π​=21​ and $\sin\frac{5\pi}{3} = -\frac{\sqrt 3}{2}$sin35π​=−23​​, so

$(1 + i\sqrt 3)^5 = 32\left(\frac12 - \frac{\sqrt 3}{2}i\right) = 16 - 16\sqrt 3\, i. \quad (\text{A1})$(1+i3​)5=32(21​−23​​i)=16−163​i.(A1)

(M1 polar base with the correct first-quadrant argument; M1 raising and multiplying the argument; A1 for $16 - 16\sqrt 3\, i$16−163​i. Note the answer is not a pure real or pure imaginary — most powers are genuinely complex, so do not expect the angle always to land on an axis.)

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Specimen-style exam question

(specimen-style — not from any past paper)

(a) Use De Moivre's theorem to show that $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$cos5θ=16cos5θ−20cos3θ+5cosθ. (b) Hence solve $16x^5 - 20x^3 + 5x = 0$16x5−20x3+5x=0 for $x \in [-1, 1]$x∈[−1,1].

(a) By De Moivre, $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$(cosθ+isinθ)5=cos5θ+isin5θ. Expand by the binomial theorem with $c = \cos\theta, s = \sin\theta$c=cosθ,s=sinθ:

$(c + is)^5 = c^5 + 5ic^4 s - 10c^3 s^2 - 10ic^2 s^3 + 5cs^4 + is^5.$(c+is)5=c5+5ic4s−10c3s2−10ic2s3+5cs4+is5.

The real part gives $\cos 5\theta = c^5 - 10c^3 s^2 + 5cs^4$cos5θ=c5−10c3s2+5cs4. Substitute $s^2 = 1 - c^2$s2=1−c2:

$\begin{aligned} \cos 5\theta &= c^5 - 10c^3(1 - c^2) + 5c(1 - c^2)^2 \\ &= c^5 - 10c^3 + 10c^5 + 5c(1 - 2c^2 + c^4) \\ &= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5 \\ &= 16c^5 - 20c^3 + 5c, \end{aligned}$cos5θ​=c5−10c3(1−c2)+5c(1−c2)2=c5−10c3+10c5+5c(1−2c2+c4)=c5−10c3+10c5+5c−10c3+5c5=16c5−20c3+5c,​

i.e. $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$cos5θ=16cos5θ−20cos3θ+5cosθ, as required.