De Moivre's theorem — statement, proof, applications

WHAT is the statement?

WHY does it look this way? Because $\cos\theta+i\sin\theta = e^{i\theta}$ (Euler), so the LHS is $(e^{i\theta})^n = e^{in\theta} = \cos n\theta + i\sin n\theta$. That's the "cheat" proof — but we should earn it from scratch.

HOW to prove it (from first principles)

Step 0 — the key multiplication fact

Take $z_1 = \operatorname{cis}\alpha$ and $z_2 = \operatorname{cis}\beta$. Multiply: $z_1z_2 = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta).$ Why expand? To collect real and imaginary parts. $= (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta).$ Why does this simplify? Those two brackets are exactly the angle-addition formulas: $= \cos(\alpha+\beta) + i\sin(\alpha+\beta) = \operatorname{cis}(\alpha+\beta).$ So multiplying cis's adds the angles. This single fact powers everything.

Step 1 — positive integers by induction

Base case $n=1$: $(\operatorname{cis}\theta)^1 = \operatorname{cis}\theta = \operatorname{cis}(1\cdot\theta)$. ✔

Inductive step: assume $(\operatorname{cis}\theta)^k = \operatorname{cis}(k\theta)$. Then $(\operatorname{cis}\theta)^{k+1} = (\operatorname{cis}\theta)^k \cdot \operatorname{cis}\theta = \operatorname{cis}(k\theta)\cdot\operatorname{cis}\theta.$ Why can I use Step 0 now? Because we're multiplying two cis's — so angles add: $= \operatorname{cis}(k\theta + \theta) = \operatorname{cis}\big((k+1)\theta\big). \ \checkmark$ By induction it holds for all positive integers.

Step 2 — $n = 0$

$(\operatorname{cis}\theta)^0 = 1 = \cos 0 + i\sin 0 = \operatorname{cis}(0\cdot\theta)$. ✔

Step 3 — negative integers

Let $n=-m$ with $m>0$. Then $(\operatorname{cis}\theta)^{-m} = \dfrac{1}{(\operatorname{cis}\theta)^m} = \dfrac{1}{\operatorname{cis}(m\theta)}$. Why is the reciprocal easy? Because $|\operatorname{cis}(m\theta)| = 1$, so its inverse is its conjugate: $\frac{1}{\cos m\theta + i\sin m\theta} = \frac{\cos m\theta - i\sin m\theta}{\cos^2 m\theta + \sin^2 m\theta} = \cos m\theta - i\sin m\theta = \operatorname{cis}(-m\theta).$ So $(\operatorname{cis}\theta)^{-m} = \operatorname{cis}(-m\theta) = \operatorname{cis}(n\theta)$. ✔ Theorem holds for all integers.

Applications

A) $n$-th roots of a complex number

To solve $z^n = w$ where $w = r\operatorname{cis}\phi$: write $z = \rho\operatorname{cis}\psi$. Then $z^n = \rho^n\operatorname{cis}(n\psi) = r\operatorname{cis}\phi$. Why add $2\pi k$? Because angles $\phi$ and $\phi+2\pi k$ label the same $w$. Matching: $\rho = r^{1/n}, \qquad n\psi = \phi + 2\pi k \implies \psi = \frac{\phi + 2\pi k}{n}.$ $\boxed{\,z_k = r^{1/n}\operatorname{cis}\!\left(\frac{\phi + 2\pi k}{n}\right),\quad k = 0,1,\dots,n-1.\,}$ These are $n$ points equally spaced by $2\pi/n$ on a circle of radius $r^{1/n}$.

B) Multiple-angle identities

Expand $(\cos\theta + i\sin\theta)^n$ with the binomial theorem, then equate real/imaginary parts to $\cos n\theta$ and $\sin n\theta$.

C) Sums like $\sum\cos k\theta$ via geometric series of $\operatorname{cis}k\theta$.

Worked examples

Flashcards

Connections

• Polar form of complex numbers — the $r\operatorname{cis}\theta$ representation this relies on.
• Euler's formula — $e^{i\theta}=\operatorname{cis}\theta$ makes De Moivre a one-liner.
• Roots of unity — the cleanest application (equally spaced points).
• Binomial theorem — used to extract multiple-angle identities.
• Argand diagram — geometric picture of spin-and-stretch.
• Trigonometric identities — angle-addition formulas power the proof.

Concept Map

Hinglish (regional understanding)