B8 - De Moivre’s theorem | AQA A-Level Further Mathematics

Raising a complex number to a power can be easier than multiplying out its brackets. You will use de Moivre’s theorem, explain it through repeated angle addition, include zero and negative integer powers where they are defined, and derive multiple-angle identities by comparing real and imaginary parts.

Positive integer powers

Can you cube $\sqrt3+\mathrm{i}$ without expanding a cubic? Its modulus is $2$ and its argument is $\pi/6$ . Multiplying three copies multiplies the moduli to give $8$ and adds three equal arguments to give $\pi/2$ .

De Moivre’s theorem

For real $\theta$ and a positive integer $n$ ,

[r(\cos\theta+\mathrm{i}\sin\theta)]^n =r^n(\cos n\theta+\mathrm{i}\sin n\theta)

Here $r>0$ for the modulus-argument form of a nonzero complex number. Equivalently, $(r\mathrm{e}^{\mathrm{i}\theta})^n=r^n\mathrm{e}^{\mathrm{i}n\theta}$ . The trigonometric form is in the formula booklet.

The angle is multiplied by $n$ ; the sine and cosine values are not each raised to $n$ . First convert the whole number to modulus-argument form.

Worked example 1: cubing a first-quadrant number

Question. Express $(\sqrt3+\mathrm{i})^3$ in Cartesian form.

Plan. Write the base as $2\mathrm{e}^{\mathrm{i}\pi/6}$ , cube the modulus and triple the argument.

\begin{aligned} \sqrt3+\mathrm{i}&=2\mathrm{e}^{\mathrm{i}\pi/6} && \text{modulus }2,\ \text{argument }\pi/6\\ (\sqrt3+\mathrm{i})^3&=8\mathrm{e}^{\mathrm{i}\pi/2} && \text{de Moivre}\\ &=8\mathrm{i} && \cos(\pi/2)=0,\ \sin(\pi/2)=1 \end{aligned}

Answer. $8\mathrm{i}$ .

Check. Direct algebra gives $(\sqrt3+\mathrm{i})^2=2+2\sqrt3\mathrm{i}$ , and multiplying this by $\sqrt3+\mathrm{i}$ gives $8\mathrm{i}$ .

Change only the sign of the imaginary part. The modulus stays fixed, while the argument changes sign.

Worked example 2: cubing the reflected number

Question. Express $(\sqrt3-\mathrm{i})^3$ in Cartesian form.

Plan. Use the fourth-quadrant argument $-\pi/6$ , then apply the same power rule.

\begin{aligned} \sqrt3-\mathrm{i}&=2\mathrm{e}^{-\mathrm{i}\pi/6} && \text{same modulus, opposite angle}\\ (\sqrt3-\mathrm{i})^3&=8\mathrm{e}^{-\mathrm{i}\pi/2} && \text{triple the argument}\\ &=-8\mathrm{i} \end{aligned}

Answer. $-8\mathrm{i}$ .

Check. Multiplication commutes with conjugation: the conjugate of $(\sqrt3+\mathrm{i})^3=8\mathrm{i}$ is $(\sqrt3-\mathrm{i})^3=-8\mathrm{i}$ . Also, its modulus is $8=2^3$ .

The cube has become purely imaginary even though the original number has both parts. Powering changes direction as well as size.

Watch out

Wrong: $(\cos\theta+\mathrm{i}\sin\theta)^n=\cos^n\theta+\mathrm{i}\sin^n\theta$ . Cross terms are missing. The correct result is $\cos(n\theta)+\mathrm{i}\sin(n\theta)$ .

Try a new base and power.

Repeated angle addition

The power rule comes from multiplying unit-circle numbers, so it can be understood directly through angle addition. Write $u(\theta)=\cos\theta+\mathrm{i}\sin\theta$ for this explanation.

Expanding a product gives

\begin{aligned} u(\alpha)u(\beta) &=(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\\ &\quad+\mathrm{i}(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\\ &=\cos(\alpha+\beta)+\mathrm{i}\sin(\alpha+\beta)\\ &=u(\alpha+\beta) \end{aligned}

The second equality uses the cosine and sine addition identities, which are supplied in the formula booklet. Setting the two angles equal gives $u(\theta)^2=u(2\theta)$ . Each further multiplication adds another $\theta$ , so $n$ copies give $u(n\theta)$ . The real factor $r$ simply appears $n$ times, giving $r^n$ .

Worked example 3: building a cube from angle addition

Question. Starting from $u(\alpha)u(\beta)=u(\alpha+\beta)$ , show that $u(\theta)^3=u(3\theta)$ .

Plan. Combine the first two factors, then add the final angle.

\begin{aligned} u(\theta)^3&=[u(\theta)u(\theta)]u(\theta) && \text{group three factors}\\ &=u(2\theta)u(\theta) && \text{add the first two angles}\\ &=u(3\theta) && \text{add the remaining angle} \end{aligned}

Answer. $(\cos\theta+\mathrm{i}\sin\theta)^3=\cos3\theta+\mathrm{i}\sin3\theta$ .

Check. At $\theta=\pi/2$ , the left side is $\mathrm{i}^3=-\mathrm{i}$ , and the right side is $\cos(3\pi/2)+\mathrm{i}\sin(3\pi/2)=-\mathrm{i}$ . This checks a case; the angle-addition argument proves the identity.

Increase the exponent by one. The same reasoning adds one more angle, rather than multiplying trigonometric values.

Worked example 4: adding a fourth factor

Question. Use the cube result and angle addition to show that $u(\theta)^4=u(4\theta)$ .

Plan. Separate one factor and apply the known product identity.

\begin{aligned} u(\theta)^4&=u(\theta)^3u(\theta) && \text{separate the last factor}\\ &=u(3\theta)u(\theta) && \text{use the cube result}\\ &=u(4\theta) && \text{add }3\theta\text{ and }\theta \end{aligned}

Answer. $(\cos\theta+\mathrm{i}\sin\theta)^4=\cos4\theta+\mathrm{i}\sin4\theta$ .

Check. At $\theta=\pi/4$ , $u(\theta)=(1+\mathrm{i})/\sqrt2$ . Its square is $\mathrm{i}$ , so its fourth power is $-1=u(\pi)$ .

Repeated multiplication establishes the pattern for positive integer powers. It does not justify replacing $n$ by an arbitrary fraction: a root can have several values, which you will study in later lessons.

Watch out

Knowing an identity at a few numerical angles is a useful check, but it is not a proof for every angle. The angle-addition identity supplies the general reason.

Use a different grouping of the same factors.

Zero and negative integer powers

For a nonzero complex number, index laws also define the zero power and negative integer powers. The nonzero condition matters because negative powers involve division.

If $z=r\mathrm{e}^{\mathrm{i}\theta}$ with $r>0$ , then

z^0=1=r^0\mathrm{e}^{0}, \qquad z^{-m}=\frac1{z^m}=r^{-m}\mathrm{e}^{-\mathrm{i}m\theta} \quad(m\text{ a positive integer})

This extends de Moivre’s theorem to every integer exponent for $z\ne0$ . At zero, positive integer powers are zero, but negative powers are undefined; $0^0$ is not defined here. The booklet’s de Moivre formula supplies the power pattern, while these domain conditions must be understood.

Worked example 5: a reciprocal

Question. Express $(2\mathrm{e}^{\mathrm{i}\pi/3})^{-1}$ in Cartesian form.

Plan. Take the reciprocal radius and reverse the argument.

\begin{aligned} (2\mathrm{e}^{\mathrm{i}\pi/3})^{-1} &=\frac12\mathrm{e}^{-\mathrm{i}\pi/3} && \text{reciprocal}\\ &=\frac12\left(\frac12-\frac{\sqrt3}{2}\mathrm{i}\right) && \text{exact values}\\ &=\frac14-\frac{\sqrt3}{4}\mathrm{i} \end{aligned}

Answer. $1/4-\sqrt3\mathrm{i}/4$ .

Check. The original number is $1+\sqrt3\mathrm{i}$ . Multiplying it by the answer gives $(1+3)/4=1$ , with cancelling imaginary terms.

Keep the base fixed and change the exponent from $-1$ to $-2$ . Both the reciprocal radius and the reversed angle must now be squared or doubled appropriately.

Worked example 6: a reciprocal square

Question. Express $(2\mathrm{e}^{\mathrm{i}\pi/3})^{-2}$ in Cartesian form.

Plan. Use radius $2^{-2}$ and angle $-2\pi/3$ .

\begin{aligned} (2\mathrm{e}^{\mathrm{i}\pi/3})^{-2} &=\frac14\mathrm{e}^{-2\pi\mathrm{i}/3} && \text{multiply the angle by }-2\\ &=\frac14\left(-\frac12-\frac{\sqrt3}{2}\mathrm{i}\right)\\ &=-\frac18-\frac{\sqrt3}{8}\mathrm{i} \end{aligned}

Answer. $-1/8-\sqrt3\mathrm{i}/8$ .

Check. Square the previous reciprocal directly: $(1/4-\sqrt3\mathrm{i}/4)^2=(1-3)/16-\sqrt3\mathrm{i}/8$ , the same result.

For a unit-circle number $u=\cos\theta+\mathrm{i}\sin\theta$ , the reciprocal has a particularly useful form:

u^{-m}=\cos(-m\theta)+\mathrm{i}\sin(-m\theta) =\cos(m\theta)-\mathrm{i}\sin(m\theta)

Consequently, $u^m+u^{-m}=2\cos(m\theta)$ and $u^m-u^{-m}=2\mathrm{i}\sin(m\theta)$ . These identities connect complex powers to trigonometric sums. The reciprocal equals the conjugate here because $|u|=1$ ; this is not true for every nonzero complex number.

Watch out

Wrong: $z^{-2}=-z^2$ . A negative exponent means a reciprocal: $z^{-2}=1/z^2$ , requiring $z\ne0$ .

Change the base angle for your calculation.

Multiple-angle formulae from real and imaginary parts

The same complex power can be written with a multiplied angle or expanded algebraically. Equating its real parts gives a cosine identity; equating its imaginary parts gives a sine identity.

For example, squaring and using de Moivre gives

\cos2\theta+\mathrm{i}\sin2\theta =\cos^2\theta-\sin^2\theta+2\mathrm{i}\sin\theta\cos\theta

Thus $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$ , using $\sin^2\theta+\cos^2\theta=1$ . Also $\sin2\theta=2\sin\theta\cos\theta$ . Dividing gives $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$ where $\cos\theta\ne0$ and $\cos2\theta\ne0$ ; the sine and cosine identities have no such exclusions.

Dividing $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ gives $1+\tan^2\theta=\sec^2\theta$ when $\cos\theta\ne0$ . Dividing instead by $\sin^2\theta$ gives $1+\cot^2\theta=\operatorname{cosec}^2\theta$ when $\sin\theta\ne0$ . These squared and double-angle identities are not separately listed in the formula booklet: learn them or derive them from its addition identities and the unit-circle identity.

Worked example 7: derive a triple-angle cosine identity

Question. Use de Moivre’s theorem to show that $\cos3\theta=4\cos^3\theta-3\cos\theta$ .

Plan. Expand the cube, equate real parts, then remove $\sin^2\theta$ .

Put $c=\cos\theta$ and $s=\sin\theta$ to keep the expansion readable.

\begin{aligned} \cos3\theta+\mathrm{i}\sin3\theta &=(c+\mathrm{i}s)^3 && \text{de Moivre}\\ &=c^3+3\mathrm{i}c^2s-3cs^2-\mathrm{i}s^3 && \mathrm{i}^2=-1,\ \mathrm{i}^3=-\mathrm{i}\\ \cos3\theta&=c^3-3cs^2 && \text{equate real parts}\\ &=c^3-3c(1-c^2) && s^2=1-c^2\\ &=4c^3-3c \end{aligned}

Answer. $\cos3\theta=4\cos^3\theta-3\cos\theta$ , as required.

Check. At $\theta=\pi/3$ , the left side is $-1$ ; the right side is $4(1/2)^3-3(1/2)=-1$ .

Keep the same cube and change only which component you compare. This gives a second identity without a second expansion.

Worked example 8: derive the matching sine identity

Question. Use de Moivre’s theorem to express $\sin3\theta$ in terms of $\sin\theta$ .

Plan. Take imaginary parts of the cube, then remove $\cos^2\theta$ .

\begin{aligned} \cos3\theta+\mathrm{i}\sin3\theta &=c^3-3cs^2+\mathrm{i}(3c^2s-s^3) && \text{the same expansion}\\ \sin3\theta&=3c^2s-s^3 && \text{equate coefficients of }\mathrm{i}\\ &=3(1-s^2)s-s^3 && c^2=1-s^2\\ &=3s-4s^3 \end{aligned}

Answer. $\sin3\theta=3\sin\theta-4\sin^3\theta$ .

Check. At $\theta=\pi/6$ , the left side is $1$ ; the right side is $3/2-4(1/8)=1$ .

The opening cube is now checked another way: with $r=2$ and $\theta=\pi/6$ , these identities give $\cos3\theta=0$ , $\sin3\theta=1$ , hence $(\sqrt3+\mathrm{i})^3=8\mathrm{i}$ .

Watch out

$\operatorname{Im}((c+\mathrm{i}s)^3)=3c^2s-s^3$ , not $\mathrm{i}(3c^2s-s^3)$ . Equate the real coefficients of $\mathrm{i}$ . For a proof using de Moivre, show the theorem’s connection to the expansion and state the final identity.

Increase the power by one and compare imaginary parts again.

Quick Recap

A positive integer power raises the modulus and multiplies an argument.
Repeated multiplication adds the same angle repeatedly, explaining de Moivre’s theorem.
For a nonzero base, zero and negative integer powers follow the same rule; negative powers take reciprocals.
Expand a power and compare real or imaginary parts to derive multiple-angle identities.

Yes: $\sqrt3+\mathrm{i}=2\mathrm{e}^{\mathrm{i}\pi/6}$ , so its cube is $8\mathrm{e}^{\mathrm{i}\pi/2}=8\mathrm{i}$ , without a cubic expansion.

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B8 - De Moivre’s theorem

Positive integer powers

Repeated angle addition

Zero and negative integer powers

Multiple-angle formulae from real and imaginary parts

Quick Recap

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