# The $n$ complex $n$th roots of a complex number $z$

Suppose $z$ is a nonzero complex number, so $z=re^{i\theta}$. Show that there are only $n$ distinct complex $n$-th roots, given by $r^{1/n}e^{i(2\pi k+\theta)/n}$ for $0\leq k\leq n-1$.

My proof: Let $z=re^{i\theta}\in\mathbb{C}$ and $a\in\mathbb{C}$ s.t. $a^{n}=z$. Let $a=\rho e^{i\varphi}$. Then, $a^{n}=(\rho e^{i\varphi})^{n}$. Setting $\rho^{n}e^{i\varphi n}=re^{i\theta}$ and rewriting in polar form, we get $$\rho^{n}(\cos(n\varphi)+i\sin(n\varphi))=r(\cos(θ)+i\sin(\theta)).$$ Thus $\rho^{n}=r$ $\rightarrow$ $\rho=r^{1/n}$ and $n\varphi=\theta\rightarrow\varphi=\theta/n$. Thus the roots of $a\in\mathbb{C}$ have the form $$r^{1/n}(\cos(\tfrac{\theta}{n})+i\sin(\tfrac{\theta}{n}))=r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))$$ for some $j\in\mathbb{Z}$ due to the periodicity of trig functions. Then, $$r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))=r^{1/n}e^{i\left(2\pi j+\tfrac{\theta}{n}\right)}=r^{1/n}e^{i(\theta+2\pi k)/n}$$ for $0\leq k\leq n-1 ,j=nk\in\mathbb{Z}$.

I feel like there could be a more concise way to put this.

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This way seems concise and careful to me. +1 for all your work. –  JavaMan Jan 23 '12 at 3:41

Could you please elaborate? How does writing the equation that an $n^\text{th}$ root $x$ of $z$ must satisfy give the result? –  Jonas Meyer Jan 23 '12 at 4:48
@JonasMeyer: Because $x^n=z$ is a degree $n$ equation in $x$, it cannot have more than $n$ solutions. If you know that fact and accept the easy computation giving that the solutions mentioned in the question are indeed solutions, this remark does settle the issue. –  Marc van Leeuwen Jan 23 '12 at 13:31
@JavaMan: There is no use of the FTA. A degree $n$ polynomial over a field (or integral domain) can have at most $n$ zeros. –  Jonas Meyer Jan 23 '12 at 17:38
@JonasMeyer: Doesn't the OP ask to show that there are $n$ distinct roots? –  JavaMan Jan 23 '12 at 17:45