Pj12, Pj13 - De Moivre's theorem and multiple angles

Pj12, Pj13 - De Moivre's theorem and multiple angles

De Moivre's theorem turns powers of complex numbers into multiples of angles. Pj12 gives the theorem and its meaning; Pj13 uses it to derive multiple-angle identities and to express powers of sine and cosine in terms of multiple angles.

Powers multiply arguments

Can you find (3+i)12(\sqrt3 + i)^{12} exactly without expanding twelve brackets? The answer will be a real integer, although the number being multiplied has a non-zero imaginary part.

You already know

For non-zero complex numbers, multiplication multiplies moduli and adds arguments. In modulus-argument form, z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta), with r>0r > 0; arguments differing by 2π2\pi describe the same point.

Take u=cos⁡θ+isin⁡θu = \cos\theta + i\sin\theta. Multiplying uu by itself gives modulus 11 and argument 2θ2\theta, so u2=cos⁡2θ+isin⁡2θu^2 = \cos2\theta + i\sin2\theta. Each further multiplication adds another θ\theta. For example, a point at angle π/3\pi/3 reaches angle 2π2\pi after six multiplications: its sixth power is 11, despite the point itself being non-real.

[DIAGRAM: asset_name: Powers turn through equal angles; asset_slug: h645_b008_powers; caption: Successive powers of a unit complex number add the same angle each time; description: An Argand diagram shows the unit circle and u at pi over 3, u squared at 2 pi over 3, u cubed at pi, and u to the sixth at 1. Blue arrows from the origin indicate u and u squared, with equal angle arcs. A side panel states that moduli stay 1 and arguments are multiplied by the power.]
Successive powers of a unit complex number add the same angle each time

De Moivre's theorem

For real θ\theta and any integer nn,

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta

Thus, for r>0r > 0,

{r(cos⁡θ+isin⁡θ)}n=rn(cos⁡nθ+isin⁡nθ)\{r(\cos\theta + i\sin\theta)\}^n = r^n(\cos n\theta + i\sin n\theta)

In the formula booklet.

The multiplication rule proves the positive-integer case directly by repeated multiplication. At n=0n = 0, both sides are 11. Also u−1=cos⁡θ−isin⁡θ=cos⁡(−θ)+isin⁡(−θ)u^{-1} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta); applying the positive case to this reciprocal gives the negative-integer case. The theorem as stated is for integers: substituting n=1/2n = 1/2 does not produce both square roots.

Worked example 1: a high power

Question. Find (3+i)6(\sqrt3 + i)^6 exactly.

Plan. Write the base in modulus-argument form, then raise the modulus and multiply the argument by 66.

∣3+i∣=3+1=2Pythagorasarg⁡(3+i)=π6first quadrant3+i=2(cos⁡π6+isin⁡π6)(3+i)6=26(cos⁡π+isin⁡π)de Moivre=−64cos⁡π=−1\begin{aligned} |\sqrt3 + i| &= \sqrt{3 + 1} = 2 && \text{Pythagoras} \\ \arg(\sqrt3 + i) &= \frac{\pi}{6} && \text{first quadrant} \\ \sqrt3 + i &= 2\left(\cos\frac\pi6 + i\sin\frac\pi6\right) \\ (\sqrt3 + i)^6 &= 2^6(\cos\pi + i\sin\pi) && \text{de Moivre} \\ &= -64 && \cos\pi = -1 \end{aligned}

Why this step? The whole base has modulus 22. Raising only the angle and forgetting 262^6 would put the answer on the wrong circle.

Answer. −64-64.

Check. Squaring the base gives 2+23 i2 + 2\sqrt3\,i; cubing it gives 8i8i. Squaring 8i8i gives −64-64.

The next example changes the sign of the power. The argument now turns backwards, and the modulus becomes a reciprocal.

Worked example 2: a negative power

Question. Find (3+i)−6(\sqrt3 + i)^{-6} exactly.

Plan. Use the same modulus and argument, with exponent −6-6.

(3+i)−6=2−6(cos⁡(−π)+isin⁡(−π))de Moivre=164(−1+0i)evaluate the angle=−164\begin{aligned} (\sqrt3 + i)^{-6} &= 2^{-6}\left(\cos\left(-\pi\right) + i\sin\left(-\pi\right)\right) && \text{de Moivre} \\ &= \frac1{64}(-1 + 0i) && \text{evaluate the angle} \\ &= -\frac1{64} \end{aligned}

Why this step? A negative power changes both rnr^n and nθn\theta. It does not merely change the sign of the final number.

Answer. −1/64-1/64.

Check. Multiplying this by the sixth power −64-64 gives 11, as reciprocal powers must.

For the opening question, doubling the sixth-power angle gives 2π2\pi, while squaring its modulus gives 40964096. We will return to this after seeing what else the theorem can do.

Watch out

Wrong: {2(cos⁡θ+isin⁡θ)}n=2(cos⁡nθ+isin⁡nθ)\{2(\cos\theta + i\sin\theta)\}^n = 2(\cos n\theta + i\sin n\theta).

Right: the modulus is 2n2^n, so the expression is 2n(cos⁡nθ+isin⁡nθ)2^n(\cos n\theta + i\sin n\theta).

Write the modulus and the multiplied angle before evaluating the trigonometric functions; this also makes a negative power easy to check.

Extracting multiple-angle identities

One complex expression can yield two real identities. Expand (cos⁡θ+isin⁡θ)n(\cos\theta + i\sin\theta)^n by the binomial theorem, and also write it using de Moivre. Equating real parts gives a cosine identity; equating imaginary parts gives a sine identity.

For the next calculations, write c=cos⁡θc = \cos\theta and s=sin⁡θs = \sin\theta. These are real numbers, so terms containing ii can be separated cleanly.

Worked example 3: the cosine triple-angle identity

Question. Use de Moivre's theorem to show that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta = 4\cos^3\theta - 3\cos\theta.

Plan. Expand the cube, equate real parts, and replace s2s^2 by 1−c21 - c^2.

(c+is)3=c3+3c2is+3c(is)2+(is)3binomial expansion=(c3−3cs2)+i(3c2s−s3)i2=−1, i3=−icos⁡3θ=c3−3cs2real parts=c3−3c(1−c2)c2+s2=1=4c3−3c\begin{aligned} (c + is)^3 &= c^3 + 3c^2is + 3c(is)^2 + (is)^3 && \text{binomial expansion} \\ &= (c^3 - 3cs^2) + i(3c^2s - s^3) && i^2=-1,\ i^3=-i \\ \cos3\theta &= c^3 - 3cs^2 && \text{real parts} \\ &= c^3 - 3c(1 - c^2) && c^2+s^2=1 \\ &= 4c^3 - 3c \end{aligned}

Why this step? Replacing s2s^2 leaves a formula entirely in cos⁡θ\cos\theta, as the requested form requires.

Answer. cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta = 4\cos^3\theta - 3\cos\theta, as required.

Check. At θ=π/3\theta = \pi/3, the right side is 4(1/2)3−3/2=−1=cos⁡π4(1/2)^3 - 3/2 = -1 = \cos\pi.

Keep the same expansion, but now change the part you extract: the imaginary part gives the sine identity.

Worked example 4: the sine triple-angle identity

Question. Use de Moivre's theorem to show that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta = 3\sin\theta - 4\sin^3\theta.

Plan. Equate imaginary parts of the expanded cube, then eliminate c2c^2.

(c+is)3=(c3−3cs2)+i(3c2s−s3)expand using i2=−1sin⁡3θ=3c2s−s3imaginary parts=3(1−s2)s−s3c2=1−s2=3s−4s3\begin{aligned} (c + is)^3 &= (c^3 - 3cs^2) + i(3c^2s - s^3) && \text{expand using }i^2=-1 \\ \sin3\theta &= 3c^2s - s^3 && \text{imaginary parts} \\ &= 3(1 - s^2)s - s^3 && c^2=1-s^2 \\ &= 3s - 4s^3 \end{aligned}

Why this step? The imaginary part of a+bia + bi is bb, so there is no ii in the resulting real identity.

Answer. sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta = 3\sin\theta - 4\sin^3\theta, as required.

Check. At θ=π/6\theta = \pi/6, the right side is 3/2−4/8=1=sin⁡(π/2)3/2 - 4/8 = 1 = \sin(\pi/2).

Dividing the imaginary part by the real part gives a tangent identity. The fourth power, for example, gives

(c+is)4=(c4−6c2s2+s4)+i(4c3s−4cs3)tan⁡4θ=4c3s−4cs3c4−6c2s2+s4=4t−4t31−6t2+t4t=tan⁡θ\begin{aligned} (c+is)^4 &= (c^4-6c^2s^2+s^4)+i(4c^3s-4cs^3) \\ \tan4\theta &= \frac{4c^3s-4cs^3}{c^4-6c^2s^2+s^4} \\ &= \frac{4t-4t^3}{1-6t^2+t^4} && t=\tan\theta \end{aligned}

The last step divides every term by c4c^4, so it requires c≠0c\ne0. The denominator must also be non-zero: otherwise tan⁡4θ\tan4\theta is undefined. At θ=π/4\theta=\pi/4, the expression gives 0/(−4)=00/(-4)=0, agreeing with tan⁡π\tan\pi.

Derived multiple-angle identities

For real θ\theta,

cos⁡3θ=4cos⁡3θ−3cos⁡θ,sin⁡3θ=3sin⁡θ−4sin⁡3θ\cos3\theta=4\cos^3\theta-3\cos\theta,\qquad \sin3\theta=3\sin\theta-4\sin^3\theta

With t=tan⁡θt=\tan\theta, where both sides are defined,

tan⁡4θ=4t−4t31−6t2+t4\tan4\theta=\frac{4t-4t^3}{1-6t^2+t^4}

Not in the formula booklet: learn it.

You can reconstruct these identities from the booklet's theorem; the important skill is choosing the part or ratio that matches the requested function.

Watch out

Wrong: sin⁡3θ=i(3c2s−s3)\sin3\theta = i(3c^2s-s^3).

Right: sin⁡3θ=3c2s−s3\sin3\theta = 3c^2s-s^3, because the imaginary part is the real coefficient of ii. Keep ii during the expansion, then equate coefficients.

Apply that same ratio idea to the cube. The powers are smaller, but every term still needs to be divided by the same power of cc.

Turning powers into multiple angles

The previous part turned a multiple angle into powers. Reversing the direction is useful when an expression contains cos⁡4θ\cos^4\theta or sin⁡3θ\sin^3\theta: a sum of multiple angles is often easier to integrate later.

Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta. Since ∣z∣=1|z|=1, its reciprocal is z−1=cos⁡θ−isin⁡θz^{-1}=\cos\theta-i\sin\theta. Adding and subtracting gives the two starting points below.

Pairing reciprocal powers

For z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta and any positive integer nn,

zn+z−n=2cos⁡nθzn−z−n=2isin⁡nθ\begin{aligned} z^n+z^{-n}&=2\cos n\theta \\ z^n-z^{-n}&=2i\sin n\theta \end{aligned}

Not in the formula booklet: learn it.

Choose a sum for a cosine power and a difference for a sine power. After expanding, pair terms with opposite exponents.

Worked example 5: a cube of cosine

Question. Express cos⁡3θ\cos^3\theta as a sum of cosines of multiples of θ\theta.

Plan. Cube z+z−1=2cos⁡θz+z^{-1}=2\cos\theta, then pair reciprocal powers.

8cos⁡3θ=(z+z−1)3cube both sides=z3+3z+3z−1+z−3binomial expansion=(z3+z−3)+3(z+z−1)pair opposite powers=2cos⁡3θ+6cos⁡θde Moivre\begin{aligned} 8\cos^3\theta&=(z+z^{-1})^3 && \text{cube both sides} \\ &=z^3+3z+3z^{-1}+z^{-3} && \text{binomial expansion} \\ &=(z^3+z^{-3})+3(z+z^{-1}) && \text{pair opposite powers} \\ &=2\cos3\theta+6\cos\theta && \text{de Moivre} \end{aligned}

Why this step? Pairing a positive power with its reciprocal makes a real cosine term, so all complex numbers disappear from the answer.

Answer. cos⁡3θ=14cos⁡3θ+34cos⁡θ\cos^3\theta=\frac14\cos3\theta+\frac34\cos\theta.

Check. At θ=0\theta=0, both sides are 11; rearranging also recovers the triple-angle identity.

Increase only the exponent, from 33 to 44. Now a constant term survives because the middle binomial term has exponent zero.

Worked example 6: a fourth power of cosine

Question. Express cos⁡4θ\cos^4\theta in terms of cosines of multiple angles.

Plan. Expand (z+z−1)4(z+z^{-1})^4, pair opposite powers and retain the constant.

16cos⁡4θ=(z+z−1)4fourth powers=z4+4z2+6+4z−2+z−4expand=(z4+z−4)+4(z2+z−2)+6pair=2cos⁡4θ+8cos⁡2θ+6de Moivre\begin{aligned} 16\cos^4\theta&=(z+z^{-1})^4 && \text{fourth powers} \\ &=z^4+4z^2+6+4z^{-2}+z^{-4} && \text{expand} \\ &=(z^4+z^{-4})+4(z^2+z^{-2})+6 && \text{pair} \\ &=2\cos4\theta+8\cos2\theta+6 && \text{de Moivre} \end{aligned}

Why this step? The middle term is 6z2z−2=66z^2z^{-2}=6, not 6z6z. Tracking exponents prevents a spurious angle term.

Answer. cos⁡4θ=18cos⁡4θ+12cos⁡2θ+38\cos^4\theta=\frac18\cos4\theta+\frac12\cos2\theta+\frac38.

Check. At θ=π/2\theta=\pi/2, the right side is 1/8−1/2+3/8=01/8-1/2+3/8=0, agreeing with cos⁡4(π/2)\cos^4(\pi/2).

These power-reduction identities are not in the formula booklet: learn the method of deriving them. For sine, the factor ii must stay until the last division: (2isin⁡θ)3=−8isin⁡3θ(2i\sin\theta)^3=-8i\sin^3\theta. Losing that minus sign reverses the answer.

Exam tip

In a “Show that” identity, write the expansion, the paired powers and the conversion to sine or cosine. A correct final identity copied without the connecting algebra does not establish a given answer.

The method has also answered our opening puzzle: 3+i\sqrt3+i has modulus 22 and argument π/6\pi/6, so its twelfth power is 212(cos⁡2π+isin⁡2π)=40962^{12}(\cos2\pi+i\sin2\pi)=4096. A large expansion has become one turn around a circle.

Watch out

Wrong: (2isin⁡θ)3=8isin⁡3θ(2i\sin\theta)^3=8i\sin^3\theta. Cubing includes i3=−ii^3=-i. The correct line is (2isin⁡θ)3=−8isin⁡3θ(2i\sin\theta)^3=-8i\sin^3\theta; keep that minus sign when dividing.

Track the power of ii alongside the binomial coefficients in your proof.

Quick Recap

  • De Moivre raises the modulus to the power and multiplies the argument by the integer exponent; it includes negative integer powers.
  • Expand a power and equate real or imaginary parts to derive multiple-angle formulae. Divide those parts for tangent, checking where the expressions are defined.
  • For a unit complex number, pair zn+z−nz^n+z^{-n} for cosine and zn−z−nz^n-z^{-n} for sine.
  • Expanding (z+z−1)n(z+z^{-1})^n or (z−z−1)n(z-z^{-1})^n expresses powers of trigonometric functions in multiple angles.

The opening power is (3+i)12=4096(\sqrt3+i)^{12}=4096: its argument completes one turn and its modulus becomes 2122^{12}.