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.

  • 4
    $\begingroup$ This way seems concise and careful to me. +1 for all your work. $\endgroup$
    – JavaMan
    Commented Jan 23, 2012 at 3:41

1 Answer 1


If by concise, you are implying brevity which relates the same message we can get rid of your section on re-writing in polar form. When you have $$ \rho^n\exp(in\varphi) = r\exp(i\theta), $$ we can jump straight to $\rho^n = r$ and $n\varphi = \theta + 2k\pi$ for $k,n\in\mathbb{Z}$, so $\rho = r^{1/n}$ and $\varphi = \frac{\theta + 2k\pi}{n}$. Now, we can jump straight to your final thought that the $n$-th roots are $$ w_k = r^{1/n}\exp\biggl[\frac{i(\theta + 2k\pi)}{n}\biggr] $$ where $0\leq k < n-1$. If you want to be clear about the range of $k$, you can say when $k = n$, we have $\theta/n + 2\pi=\theta/n$ since the exponential is periodic $f(\theta) = f(\theta + 2\pi)$.

  • $\begingroup$ Shouldn't it be $0 \le k \le n−1$? But how do we justify this choice of limiting $k$ to this range in the first place? Is this the only choice that makes sense? $\endgroup$
    – mdcq
    Commented Jan 12, 2018 at 20:58

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