De Moivre's Theorem

De Moivre's theorem. For every integer $n$, $$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.$$ The argument for $n \geq 0$ is a one-line induction; the cases $n < 0$ and $n \in \mathbb{Q}$ then follow.

Proof for $n \in \mathbb{Z}^+$ by induction.

1. Base case $n = 1$: $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta = \cos(1\cdot\theta) + i\sin(1\cdot\theta)$. ✓
2. Inductive step: Assume $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$ for some $k \geq 1$. Then $$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta)$$ $$= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$$ $$= \cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta)$$ $$= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos(k+1)\theta + i\sin(k+1)\theta.$$
3. By induction, the theorem holds for all $n \in \mathbb{Z}^+$.

Extension to $n = 0$: $(\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0$. ✓

Extension to negative integers. For $n < 0$, write $n = -m$ with $m > 0$. Then $(\cos\theta + i\sin\theta)^{-m} = 1/(\cos\theta + i\sin\theta)^m = 1/(\cos m\theta + i\sin m\theta)$. Multiplying top and bottom by the conjugate $\cos m\theta - i\sin m\theta$ gives $(\cos m\theta - i\sin m\theta)/1 = \cos(-m\theta) + i\sin(-m\theta) = \cos n\theta + i\sin n\theta$. ✓

Extension to rationals. For $n = p/q$ with $q \neq 0$, the statement $(\cos\theta + i\sin\theta)^{p/q} = \cos(p\theta/q) + i\sin(p\theta/q)$ is one of the values; there are $q$ distinct $q$th roots of $\cos p\theta + i\sin p\theta$ in total.

Theorem: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for $n \in \mathbb{Z}$ · Proof: induction on $n \geq 0$; base $n=1$ trivial; step uses $\cos$ and $\sin$ addition formulae · Negative integers: follow from $z^{-m} = 1/z^m$ plus conjugate trick · General form: $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)$

Pause — copy De Moivre's theorem (integer $n$), the induction proof outline, and the general form $[r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta)$ into your book.

We just saw De Moivre's theorem $[r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta)$ proved by induction, extended to negative integers via the conjugate trick. That raises a question: how do we actually use this to compute high powers and derive trig identities? This card answers it → convert $z$ to polar, apply De Moivre, then take Re or Im parts to get $\cos n\theta$ or $\sin n\theta$.

Two main classes of question lean on De Moivre:

• Computing $z^n$. Convert $z$ to polar form, apply De Moivre, convert back to Cartesian if needed.
• Deriving multiple-angle identities. Expand $(\cos\theta + i\sin\theta)^n$ using the binomial theorem, then equate real and imaginary parts of the result $\cos n\theta + i\sin n\theta$.

For the second class, the binomial expansion gives $$(\cos\theta + i\sin\theta)^n = \sum_{k=0}^n \binom{n}{k} \cos^{n-k}\theta \, (i\sin\theta)^k.$$ Powers of $i$ cycle through $1, i, -1, -i$, so the real part collects even-$k$ terms (with alternating signs) and the imaginary part collects odd-$k$ terms.

To compute $z^n$: convert $z$ to polar, apply De Moivre, convert back if asked · $\cos n\theta = \mathrm{Re}[(\cos\theta + i\sin\theta)^n]$; $\sin n\theta = \mathrm{Im}[\,]$ · Powers of $i$ cycle: $i^0=1,\, i^1=i,\, i^2=-1,\, i^3=-i$ · $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$; $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$

Pause — copy the application procedure (convert to polar, apply De Moivre, take Re/Im), the results $\cos 3\theta = 4\cos^3\theta-3\cos\theta$ and $\sin 3\theta = 3\sin\theta-4\sin^3\theta$, into your book.