The Identity z + 1/z = 2 cos θ

A number of length 1 and its reciprocal

Take z with |z| = 1 and argument $\theta$θ, so $z=\cos \theta +i\sin \theta$z = cos θ + i sin θ. By De Moivre’s theorem with n = −1, $\frac{1}{z}=z^{-1}=\cos \left(-\theta \right)+i\sin \left(-\theta \right)$1/z = z⁻¹ = cos(−θ) + i sin(−θ). The cosine of $-\theta$−θ is $\cos \theta$cos θ and the sine of $-\theta$−θ is $-\sin \theta$−sin θ, so $\frac{1}{z}=\cos \theta -i\sin \theta$1/z = cos θ − i sin θ.

So when |z| = 1, the reciprocal $\frac{1}{z}$1/z is the conjugate of z. Check by multiplying: $\left(\cos \theta +i\sin \theta \right)\left(\cos \theta -i\sin \theta \right)={\cos }^2\theta -i^2{\sin }^2\theta ={\cos }^2\theta +{\sin }^2\theta =1$(cos θ + i sin θ)(cos θ − i sin θ) = cos²θ − i²sin²θ = cos²θ + sin²θ = 1.

Add them, subtract them

With |z| = 1, add z and $\frac{1}{z}$1/z: $\left(\cos \theta +i\sin \theta \right)+\left(\cos \theta -i\sin \theta \right)=2\cos \theta$(cos θ + i sin θ) + (cos θ − i sin θ) = 2cos θ. The imaginary parts cancel. Subtract instead: $\left(\cos \theta +i\sin \theta \right)-\left(\cos \theta -i\sin \theta \right)=2i\sin \theta$(cos θ + i sin θ) − (cos θ − i sin θ) = 2i sin θ. The real parts cancel.

So $z+\frac{1}{z}=2\cos \theta$z + 1/z = 2cos θ and $z-\frac{1}{z}=2i\sin \theta$z − 1/z = 2i sin θ, whenever |z| = 1.

For example, z = 0.6 + 0.8i has $|z|=\sqrt{0.36+0.64}=1$|z| = √(0.36 + 0.64) = 1, with $\cos \theta =0.6$cos θ = 0.6 and $\sin \theta =0.8$sin θ = 0.8. Then $\frac{1}{z}=0.6-0.8i$1/z = 0.6 − 0.8i, $z+\frac{1}{z}=1.2=2\times 0.6$z + 1/z = 1.2 = 2 × 0.6, and $z-\frac{1}{z}=1.6i=2i\times 0.8$z − 1/z = 1.6i = 2i × 0.8.

z = 0.6 + 0.8i has length 1, and $\frac{1}{z}=0.6-0.8i$1/z = 0.6 − 0.8i, dashed, is its mirror image in the real axis. Laid tip to tail, the 0.8 up and the 0.8 down cancel, and the sum lands on the real axis at $1.2=2\cos \theta$1.2 = 2cos θ.

Every power

With |z| = 1, De Moivre gives $z^n=\cos n\theta +i\sin n\theta$zⁿ = cos nθ + i sin nθ and $z^{-n}=\cos n\theta -i\sin n\theta$z⁻ⁿ = cos nθ − i sin nθ. Adding and subtracting as before: $z^n+\frac{1}{z^n}=2\cos n\theta$zⁿ + 1/zⁿ = 2cos nθ and $z^n-\frac{1}{z^n}=2i\sin n\theta$zⁿ − 1/zⁿ = 2i sin nθ, for every whole number n.

For z = 0.6 + 0.8i: $z^2=0.36-0.64+0.96i=-0.28+0.96i$z² = 0.36 − 0.64 + 0.96i = −0.28 + 0.96i and $\frac{1}{z^2}=-0.28-0.96i$1/z² = −0.28 − 0.96i, so $z^2+\frac{1}{z^2}=-0.56$z² + 1/z² = −0.56. And $2\cos 2\theta =2\left({\cos }^2\theta -{\sin }^2\theta \right)=2\left(0.36-0.64\right)=-0.56$2cos 2θ = 2(cos²θ − sin²θ) = 2(0.36 − 0.64) = −0.56.

A power of $\cos \theta$cos θ as multiple angles

With |z| = 1, $2\cos \theta =z+\frac{1}{z}$2cos θ = z + 1/z, so ${\left(2\cos \theta \right)}^3={\left(z+\frac{1}{z}\right)}^3$(2cos θ)³ = (z + 1/z)³. By the binomial theorem, ${\left(z+\frac{1}{z}\right)}^3=z^3+3z^2\left(\frac{1}{z}\right)+3z\left(\frac{1}{z^2}\right)+\frac{1}{z^3}=z^3+3z+\frac{3}{z}+\frac{1}{z^3}$(z + 1/z)³ = z³ + 3z²(1/z) + 3z(1/z²) + 1/z³ = z³ + 3z + 3/z + 1/z³.

Pair each power with its reciprocal: $\left(z^3+\frac{1}{z^3}\right)+3\left(z+\frac{1}{z}\right)=2\cos 3\theta +6\cos \theta$(z³ + 1/z³) + 3(z + 1/z) = 2cos 3θ + 6cos θ. The left side is $8{\cos }^3\theta$8cos³θ, so $8{\cos }^3\theta =2\cos 3\theta +6\cos \theta$8cos³θ = 2cos 3θ + 6cos θ, and ${\cos }^3\theta =\frac{\cos 3\theta +3\cos \theta }{4}$cos³θ = (cos 3θ + 3cos θ)/4.

Check at $\theta =60^{\circ }$θ = 60°: ${\cos }^{3}60^{\circ }={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$cos³60° = (1/2)³ = 1/8, and $\frac{\cos 180^{\circ }+3\cos 60^{\circ }}{4}=\frac{-1+\frac{3}{2}}{4}=\frac{1}{8}$(cos 180° + 3cos 60°)/4 = (−1 + 3/2)/4 = 1/8.

This turns a power of $\cos \theta$cos θ into cosines of single angles, which is the form an integral of ${\cos }^3\theta$cos³θ needs.

Sines the same way

With |z| = 1, $2i\sin \theta =z-\frac{1}{z}$2i sin θ = z − 1/z. Cube it: ${\left(z-\frac{1}{z}\right)}^3=z^3-3z+\frac{3}{z}-\frac{1}{z^3}=\left(z^3-\frac{1}{z^3}\right)-3\left(z-\frac{1}{z}\right)=2i\sin 3\theta -6i\sin \theta$(z − 1/z)³ = z³ − 3z + 3/z − 1/z³ = (z³ − 1/z³) − 3(z − 1/z) = 2i sin 3θ − 6i sin θ.

The left side is ${\left(2i\sin \theta \right)}^3=8i^3{\sin }^3\theta =-8i{\sin }^3\theta$(2i sin θ)³ = 8i³sin³θ = −8i sin³θ. Dividing both sides by −8i: ${\sin }^3\theta =\frac{3\sin \theta -\sin 3\theta }{4}$sin³θ = (3sin θ − sin 3θ)/4. Check at $\theta =30^{\circ }$θ = 30°: ${\sin }^{3}30^{\circ }=\frac{1}{8}$sin³30° = 1/8, and $\frac{\frac{3}{2}-1}{4}=\frac{1}{8}$(3/2 − 1)/4 = 1/8.

Only when |z| = 1

Off the unit circle the identities fail. $z=2\left(\cos 60^{\circ }+i\sin 60^{\circ }\right)=1+\sqrt{3}i$z = 2(cos 60° + i sin 60°) = 1 + √3 i has |z| = 2. Its reciprocal is $\frac{1}{2}$1/2 at −60°, which is $\frac{1}{4}-\left(\frac{\sqrt{3}}{4}\right)i$1/4 − (√3/4)i, so $z+\frac{1}{z}=\frac{5}{4}+\left(\frac{3\sqrt{3}}{4}\right)i$z + 1/z = 5/4 + (3√3/4)i, about 1.25 + 1.299i. That is not real, and not 2cos 60° = 1.

The reciprocal is the mirror image of z only when |z| = 1. Otherwise its length is 1/|z|, and the imaginary parts no longer cancel.

The usual mistakes

Changing both signs for $\frac{1}{z}$1/z. $-\cos \theta -i\sin \theta$−cos θ − i sin θ is −z, half a turn away; with |z| = 1, $\frac{1}{z}$1/z turns the angle back to $-\theta$−θ, which changes only the sign of the imaginary part.

Giving $2i\sin \theta$2i sin θ for $z+\frac{1}{z}$z + 1/z. That is the difference; adding cancels the imaginary parts and leaves $2\cos \theta$2cos θ.

Cubing the cosine instead of the angle. $z^3+\frac{1}{z^3}=2\cos 3\theta$z³ + 1/z³ = 2cos 3θ, not $2{\cos }^3\theta$2cos³θ: De Moivre multiplies the angle.

Forgetting the 2. It is $2\cos \theta$2cos θ that equals $z+\frac{1}{z}$z + 1/z, so ${\left(z+\frac{1}{z}\right)}^3$(z + 1/z)³ is $8{\cos }^3\theta$8cos³θ, not ${\cos }^3\theta$cos³θ.

A middle coefficient of 1. The middle terms of ${\left(z+\frac{1}{z}\right)}^3$(z + 1/z)³ are 3z and $\frac{3}{z}$3/z, so ${\cos }^3\theta =\frac{\cos 3\theta +3\cos \theta }{4}$cos³θ = (cos 3θ + 3cos θ)/4, not $\frac{\cos 3\theta +\cos \theta }{4}$(cos 3θ + cos θ)/4.

How much darker the corner of a photo is

The application below takes $z=\cos \theta +i\sin \theta$z = cos θ + i sin θ, so |z| = 1, and expands the fourth power with coefficients 1, 4, 6, 4, 1: ${\left(z+\frac{1}{z}\right)}^4=\left(z^4+\frac{1}{z^4}\right)+4\left(z^2+\frac{1}{z^2}\right)+6=2\cos 4\theta +8\cos 2\theta +6$(z + 1/z)⁴ = (z⁴ + 1/z⁴) + 4(z² + 1/z²) + 6 = 2cos 4θ + 8cos 2θ + 6, which is $16{\cos }^4\theta$16cos⁴θ. Its angle is in radians: $\frac{\pi }{6}$π/6 is 30°.