---
tag: J29BR10TM7
kind: proof
name: Proof of the existence of nth roots of complex numbers
msc: [12D99]
proves: 0YM6P6C206
uses: [0BRXFNQ5JV, 0KW2X0TFBX, 03MTBVG3VE, 0RMVS7RXZX, 0RKW2X63YR, 0E8HNVBCSP, 0FRSR10DWF]
sources:
  - url: https://en.wikipedia.org/wiki/Nth_root
    locator: Complex roots
---

## Proof

We take the $n$th root of the modulus and divide the argument by $n$.

1. Let $z$ be a complex number and $n \ge 1$ an integer. If $z = 0$, then $w = 0$ works, since $0^n = 0$.
2. Now let $z \neq 0$. Let $r = |z| > 0$ be its [[modulus|0KW2X0TFBX]] and $\theta = \operatorname{Arg} z$.
By the [[polar form|0BRXFNQ5JV]], $z = r (\cos \theta + i \sin \theta)$.
3. Let $s = \exp\bigl(\ln(r) / n\bigr)$, a real number, with $\exp$ the real [[exponential function|0RMVS7RXZX]].
Here $\ln r$ exists because $r > 0$.
4. Claim: $\exp(y)^m = \exp(m y)$ for real $y$ and integers $m \ge 0$. For $m = 0$ both sides are $1$, since all terms of the series for $\exp(0)$ after the first vanish.
If it holds for $m$, the [[exponential of a sum|0E8HNVBCSP]] gives $\exp(y)^{m+1} = \exp(m y) \exp(y) = \exp((m + 1) y)$.
5. By step 4 and [[the logarithm is the inverse of the exponential|0RKW2X63YR]], $s^n = \exp(\ln r) = r$.
6. Let $w = s \bigl(\cos(\theta / n) + i \sin(\theta / n)\bigr)$.
7. Multiplication in $\C$ is commutative and associative, since [[$\C$ is a field|0FRSR10DWF]].
So $w^n = s^n \bigl(\cos(\theta / n) + i \sin(\theta / n)\bigr)^n$.
8. By [[de Moivre's formula|03MTBVG3VE]] with $x = \theta / n$, the last power is $\cos \theta + i \sin \theta$.
9. By steps 5, 7 and 8, $w^n = r (\cos \theta + i \sin \theta) = z$ by step 2.
So $z$ has an $n$th root $w$.

## Idea

In polar form, raising to the $n$th power raises the modulus to the $n$th power and multiplies the angle by $n$.
So to undo it, take the positive real $n$th root of the modulus and divide the angle by $n$.