The Identity z + 1/z = 2 cos θ

A number of length 1 and its reciprocal

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

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

Add them, subtract them

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

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

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

z = 0.6 + 0.8i has length 1, and $\frac{1}{z} = 0.6 - 0.8i$, 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 $1.2 = 2\cos \theta$.

Every power

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

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

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

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

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

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

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

Sines the same way

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

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

Only when |z| = 1

Off the unit circle the identities fail. $z = 2\left(\cos 60^\circ + i \sin 60^\circ\right) = 1 + \sqrt{3} i$ has |z| = 2. Its reciprocal is $\frac{1}{2}$ at −60°, which is $\frac{1}{4} - \left(\frac{\sqrt{3}}{4}\right)i$, so $z + \frac{1}{z} = \frac{5}{4} + \left(\frac{3\sqrt{3}}{4}\right)i$, 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 $\frac{1}{z}$. $-\cos \theta - i \sin \theta$ is −z, half a turn away; with |z| = 1, $\frac{1}{z}$ turns the angle back to $-\theta$, which changes only the sign of the imaginary part.

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

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

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

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

How much darker the corner of a photo is

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