De Moivre's Theorem

Euler's formula told us that multiplying complex numbers in polar form adds the angles. Apply that rule to a number multiplied by itself over and over, and the angle simply gets multiplied. That is De Moivre's theorem:

(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.

Raising a unit-modulus number to the nth power rotates its angle to n\theta. It is the single most useful tool for computing powers of complex numbers — and, surprisingly, for proving trigonometric identities.

Straight from Euler's formula

With e^{i\theta} = \cos\theta + i\sin\theta in hand, the proof is two lines of exponent arithmetic.

Step 1 — rewrite the bracket as an exponential. The thing being raised to the power is exactly e^{i\theta}:

(\cos\theta + i\sin\theta)^n = \left(e^{i\theta}\right)^n.

Step 2 — apply the exponent law (e^x)^n = e^{nx}:

\left(e^{i\theta}\right)^n = e^{in\theta}.

Step 3 — read it back through Euler's formula at angle n\theta:

e^{in\theta} = \cos n\theta + i\sin n\theta.

Chaining the three lines gives the theorem. And for a general modulus r, the same exponent law gives (re^{i\theta})^n = r^n e^{in\theta}: the modulus is raised to the power, the angle is multiplied.

For every integer n and real angle \theta:

De Moivre's theorem can also be proved without exponentials, using only the angle-addition identities. For n = 1 the statement is trivial. Assume it holds for some n; then

(\cos\theta + i\sin\theta)^{n+1} = (\cos n\theta + i\sin n\theta)(\cos\theta + i\sin\theta).

Expanding and using i^2 = -1, the real part is \cos n\theta\cos\theta - \sin n\theta\sin\theta = \cos(n+1)\theta and the imaginary part is \sin n\theta\cos\theta + \cos n\theta\sin\theta = \sin(n+1)\theta — precisely the angle-addition formulas. So the statement holds for n + 1, and by induction for all positive integers.

Application: powers made painless

Computing (1 + i)^8 by multiplying out eight brackets would be brutal. In polar form it is a one-liner.

Step 1 — write 1 + i in polar form. Its modulus is \sqrt{1^2 + 1^2} = \sqrt{2} and its argument is \tfrac{\pi}{4} (it points at 45°):

1 + i = \sqrt{2}\,e^{i\pi/4}.

Step 2 — apply De Moivre. Power the modulus, multiply the angle by 8:

(1 + i)^8 = \left(\sqrt{2}\right)^8 e^{i\,8\cdot\pi/4} = 16\,e^{i2\pi}.

Step 3 — simplify. A full turn e^{i2\pi} = 1, so

(1 + i)^8 = 16.

Application: triple-angle identities

De Moivre's real magic is turning trig identities into algebra. Expand (\cos\theta + i\sin\theta)^3 two different ways and match the real and imaginary parts.

Step 1 — De Moivre's side. With n = 3:

(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta.

Step 2 — expand with the binomial theorem. Writing c = \cos\theta and s = \sin\theta:

(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3.

Step 3 — use i^2 = -1 and i^3 = -i. The middle terms pick up the sign flips:

(c + is)^3 = c^3 + 3ic^2 s - 3cs^2 - is^3 = (c^3 - 3cs^2) + i(3c^2 s - s^3).

Step 4 — match real parts. Equate the real part to \cos 3\theta, then replace s^2 = 1 - c^2:

\cos 3\theta = c^3 - 3c(1 - c^2) = 4\cos^3\theta - 3\cos\theta.

Step 5 — match imaginary parts. Equate the imaginary part to \sin 3\theta, then replace c^2 = 1 - s^2:

\sin 3\theta = 3(1 - s^2)s - s^3 = 3\sin\theta - 4\sin^3\theta.

Two classic identities, derived with no trig trickery at all — just De Moivre and the binomial theorem.

Watch the angle multiply

The slider \theta sets a point \cos\theta + i\sin\theta on the unit circle; the slider n raises it to a power. By De Moivre the result \cos n\theta + i\sin n\theta sits at angle n\theta — same circle, angle scaled by n.