De Moivre’s Theorem

A power is repeated multiplying

Multiplying two complex numbers multiplies their lengths and adds their angles. Squaring z is multiplying z by itself, so if z is r at $\theta$, then $z^{2}$ is r × r at $\theta + \theta$, which is $r^{2}$ at $2\theta$.

One more factor of z multiplies the length by r again and adds $\theta$ again: $z^{3} = r^{3}$ at $3\theta$. After n factors the length is $r^{n}$ and the angle is $n\theta$.

The powers of 1 + i

1 + i has modulus $\sqrt{2}$ and argument 45°. Each power is the one before it stretched by $\sqrt{2}$ and turned a further 45°: $\left(1 + i\right)^{2} = 2$ at 90° = 2i, $\left(1 + i\right)^{3} = 2\sqrt{2}$ at 135° = −2 + 2i, and $\left(1 + i\right)^{4} = 4$ at 180° = −4.

Check by multiplying out: $\left(1 + i\right)^{2} = 1 + 2i + i^{2} = 2i$. Then $\left(1 + i\right)^{3} = 2i\left(1 + i\right) = 2i + 2i^{2} = -2 + 2i$, and $\left(1 + i\right)^{4} = \left(2i\right)^{2} = 4i^{2} = -4$.

z = 1 + i and its next three powers. The arrows are $\sqrt{2}$, 2, $2\sqrt{2}$ and 4 long, at 45°, 90°, 135° and 180°: each is $\sqrt{2}$ times the length of the one before and 45° further round.

The theorem

De Moivre’s theorem: (r at $\theta)^{n} = r^{n}$ at $n\theta$. In the full notation, $\left[r\left(\cos \theta + i \sin \theta\right)\right]^{n} = r^{n}\left(\cos n\theta + i \sin n\theta\right)$.

For z = 2(cos 30° + i sin 30°), $z^{3} = 2^{3}\left(\cos 90^\circ + i \sin 90^\circ\right) = 8i$. In parts, $z = \sqrt{3} + i$, $z^{2} = 3 + 2\sqrt{3} i + i^{2} = 2 + 2\sqrt{3} i$, and $z^{3} = \left(2 + 2\sqrt{3} i\right)\left(\sqrt{3} + i\right) = 2\sqrt{3} + 2i + 6i + 2\sqrt{3} i^{2} = 8i$.

It holds for every positive whole number n. It holds for n = 1. If it holds for some n, then multiplying $z^{n} = r^{n}$ at $n\theta$ by z = r at $\theta$ gives $r^{n+1}$ at $\left(n + 1\right)\theta$, so it holds for n + 1 as well. By induction it holds for every n.

Large powers

$\left(1 + i\right)^{8} = \left(\sqrt{2}\right)^{8}$ at 8 × 45° = 16 at 360°. A full turn points the same way as 0°, so $\left(1 + i\right)^{8} = 16$. Multiplying out would take seven multiplications.

$1 + \sqrt{3} i$ has modulus $\sqrt{1 + 3} = 2$ and argument 60°, since $\tan \theta = \sqrt{3}$ in the first quadrant. So $\left(1 + \sqrt{3} i\right)^{5} = 2^{5}$ at 5 × 60° = 32 at 300°. 300° is past 180°, so take off a full turn: 300° − 360° = −60°. Then $\left(1 + \sqrt{3} i\right)^{5} = 32\left(\cos\left(-60^\circ\right) + i \sin\left(-60^\circ\right)\right) = 16 - 16\sqrt{3} i$, about 16 − 27.713i.

$\sqrt{3} + i$ is 2 at 30°, so $\left(\sqrt{3} + i\right)^{6} = 2^{6}$ at 180° = 64 at 180° = −64.

On the unit circle

When r = 1, every power has length 1 too, and the theorem reads $\left(\cos \theta + i \sin \theta\right)^{n} = \cos n\theta + i \sin n\theta$. So $\left(\cos 20^\circ + i \sin 20^\circ\right)^{9} = \cos 180^\circ + i \sin 180^\circ = -1$.

Raising $\cos \theta + i \sin \theta$ to a power multiplies the angle inside the cosine and the sine. Multiple Angles by De Moivre expands the left side to find formulas for $\cos n\theta$ and $\sin n\theta$.

Negative and zero powers

$z^{-1} = \frac{1}{z}$, and dividing 1 by r at $\theta$ gives $\frac{1}{r}$ at $-\theta$. So $z^{-n} = \left(\frac{1}{z}\right)^{n} = r^{-n}$ at $-n\theta$, and the theorem holds for negative whole numbers too.

$\left(1 + i\right)^{-4} = \left(\sqrt{2}\right)^{-4}$ at $-4 \times 45^\circ = \frac{1}{4}$ at −180°. −180° is outside the range, so add a full turn: $\frac{1}{4}$ at 180°, which is $- \frac{1}{4}$. Check: $\left(1 + i\right)^{4} = -4$, and $1 \div \left(-4\right) = - \frac{1}{4}$.

With n = 0 the theorem gives $r^{0}$ at 0°, which is 1, as $z^{0}$ is for any z other than 0.

The usual mistakes

Multiplying the length by n. (2 at $30^\circ)^{3}$ is 8 at 90°, not 6 at 90°: each of the three factors multiplies the length by 2, so it is $2^{3}$.

Leaving the angle as it was. Each factor adds another 30°, so the cube is at 90°, not 30°.

Raising the cosine and the sine to the power. $\left(\cos 30^\circ + i \sin 30^\circ\right)^{3}$ is cos 90° + i sin 90° = i, not $\cos^{3}30^\circ + i \sin^{3}30^\circ \approx 0.650 + 0.125i$, a number about 0.66 long rather than 1.

Leaving the argument out of range. $\left(1 + \sqrt{3} i\right)^{5}$ has argument −60°; 300° names the same direction but is outside $-180^\circ < \theta \le 180^\circ$.

A Ferris wheel that stops every twelfth of a turn

The application below measures turns in radians: $\frac{\pi}{6}$ is 30°. Each stop multiplies the car’s position by 1 at 30°, so n stops multiply it by 1 at 30n°. Five stops make 150°. Sixteen make 480°, which is 120° once a full turn is taken off.