De Moivre's Theorem (Integer Exponent)

For any integer n, raising cis θ to the power n simply multiplies the angle. In Euler's form: (e^(iθ))^n = e^(inθ) — the result is immediate. Proof for positive integers uses induction; extends to negative integers via (cis θ)^(−1) = cis(−θ).

De Moivre's theorem says that raising $\cos\theta + i\sin\theta$ to the power $n$ simply multiplies the angle. The proof runs in three parts: positive integers by induction, then $n = 0$, then negative integers.

Part 1 — Positive Integers (Induction)

Claim: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for all $n \in \mathbb{Z}^+$.

Base case ($n = 1$): Both sides equal $\cos\theta + i\sin\theta$. ✓

Inductive step: Assume the result holds for $n = k$:

$$(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$$

Multiply both sides by $(\cos\theta + i\sin\theta)$:

$$(\cos\theta + i\sin\theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$$

Expanding the right side:

$$= \cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\cos k\theta\sin\theta + \sin k\theta\cos\theta)$$

Recognising the compound angle formulas:

$$\cos k\theta\cos\theta - \sin k\theta\sin\theta = \cos(k\theta + \theta) = \cos(k+1)\theta$$ $$\cos k\theta\sin\theta + \sin k\theta\cos\theta = \sin(k\theta + \theta) = \sin(k+1)\theta$$

Therefore:

$$(\cos\theta + i\sin\theta)^{k+1} = \cos(k+1)\theta + i\sin(k+1)\theta$$

The statement holds for $n = k+1$. By induction, it holds for all positive integers. $\blacksquare$

Part 2 — Zero

$$(\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0 \checkmark$$

Part 3 — Negative Integers

For $n \in \mathbb{Z}^-$, write $n = -m$ where $m > 0$. First observe:

$$(\cos\theta + i\sin\theta)^{-1} = \frac{1}{\cos\theta + i\sin\theta} = \frac{\cos\theta - i\sin\theta}{\cos^2\theta + \sin^2\theta} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$$

Now using Part 1 on $(\cos\theta + i\sin\theta)^{-1} = \cos(-\theta) + i\sin(-\theta)$:

$$(\cos\theta + i\sin\theta)^{-m} = \left[(\cos\theta + i\sin\theta)^{-1}\right]^m = (\cos(-\theta) + i\sin(-\theta))^m = \cos(-m\theta) + i\sin(-m\theta)$$

Which is exactly $\cos n\theta + i\sin n\theta$ with $n = -m$. $\blacksquare$

The Euler Form Perspective

If you have already established Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$, De Moivre's theorem is immediate:

$$(e^{i\theta})^n = e^{in\theta} = \cos n\theta + i\sin n\theta$$

The induction proof above is the elementary path that does not assume Euler's formula. Both are valid — they are two routes to the same result.