De Moivre's theorem raises a number to a power by multiplying its angle. Running it backward finds roots, and reveals something surprising: every nonzero complex number has exactly $n$ different $n$th roots.
Key idea
The $n$th roots of $r(\cos\theta + i\sin\theta)$ are $$\sqrt[n]{r}\left(\cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n}\right), \quad k = 0, 1, \ldots, n - 1.$$ They lie on a circle of radius $\sqrt[n]{r}$, spaced $\frac{2\pi}{n}$ apart.
Why it works
Suppose $w = s(\cos\varphi + i\sin\varphi)$ and $w^n = r(\cos\theta + i\sin\theta)$. De Moivre says $s^n = r$, so $s = \sqrt[n]{r}$, and $n\varphi$ must point the same way as $\theta$: $n\varphi = \theta + 2\pi k$ for some whole number $k$. Dividing by $n$, each $k$ from $0$ to $n - 1$ gives a different angle; at $k = n$ you are back where you started, a full turn later.
Worked examples
Worked example: Cube roots of 8
Find all cube roots of $8$.
- $8 = 8(\cos 0 + i\sin 0)$. Modulus of each root: $\sqrt[3]{8} = 2$.
- Angles: $\frac{0 + 2\pi k}{3}$ for $k = 0, 1, 2$: $0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$.
- $2(\cos 0 + i\sin 0) = 2$; $2\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -1 + i\sqrt{3}$; and its conjugate $-1 - i\sqrt{3}$.
Answer: $2$, $-1 + i\sqrt{3}$, $-1 - i\sqrt{3}$
Worked example: Fourth roots of a negative
Find the fourth roots of $-16$.
- $-16 = 16(\cos\pi + i\sin\pi)$. Modulus of each root: $\sqrt[4]{16} = 2$.
- Angles: $\frac{\pi + 2\pi k}{4} = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
- $2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = \sqrt{2} + \sqrt{2}\,i$, and the others are its rotations by quarter turns.
Answer: $\pm\sqrt{2} \pm \sqrt{2}\,i$ (all four sign choices)
Worked example: Square roots of $2i$
Find the square roots of $2i$.
- $2i = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$; root modulus $\sqrt{2}$.
- Angles $\frac{\pi}{4}$ and $\frac{\pi}{4} + \pi = \frac{5\pi}{4}$.
- Check: $(1 + i)^2 = 1 + 2i + i^2 = 2i$. ✓
Answer: $1 + i$ and $-1 - i$
Common mistake
Finding only one root, the one with $k = 0$. A cube root problem has three answers, a fourth root problem four. Add $\frac{2\pi}{n}$ to the angle again and again until you have $n$ roots evenly spaced around the circle.