Sum of roots of index $2n$ of a complex number is $0$ I need to show that the sum of the roots $\sqrt[2n]{z}$ is $0$. That's what I did:
$$\sqrt[2n]{z} = \sqrt[2n]{|z|}\left(\cos\left(\frac{\theta}{2n}+\frac{2k\pi}{2n}\right)+\sin\left(\frac{\theta}{2n}+\frac{2k\pi}{2n}\right)\right)=$$
$$\sqrt[2n]{|z|}\left(\cos\left(\frac{\theta}{2n}+\frac{k\pi}{n}\right)+\sin\left(\frac{\theta}{2n}+\frac{k\pi}{n}\right)\right)$$
I have to show that:
$$\sum_{k=0}^{n-1} \sqrt[2n]{z} = 0$$
But I couldn't find any relations. I tried to think about periodicity, tried to think about a sum of $\cos$ (generalizing a formula for it) but I couldn't think of any.
 A: If $z=0$, then all the $2n$th roots of $z$ are also $0$, and you're done.
If $z\neq 0$, show that for any $w$ that is a $2n$th root of $z$, then $w\neq 0$ (obviously) and that $-w$ is also a $2n$th root of $z$, and since $w\neq -w$ the $2n$th roots of $z$ can be paired off with their negatives and their sum together must be $0$.

A more calculational approach (and one that works for $k$th roots for any $k>1$) would be to observe that if $z=re^{i\theta}$, then the $k$th roots of $z$ are $w,w\zeta,\ldots,w\zeta^{k-1}$ where $w=re^{i\theta/k}$ and $\zeta=e^{2\pi i /k}$ is a primitive $k$th root of unity, so that 
$$\sum_{r=0}^{k-1}w\zeta^r=w\left(\sum_{r=0}^{k-1}\zeta^r\right)=w\left(\frac{\zeta^k-1}{\zeta-1}\right)=w\left(\frac{0}{\zeta-1}\right)=0$$
A: Given your formula:
$$\sqrt[2n]{|z|}\left(\cos\left(\frac{\theta}{2n}+\frac{k\pi}{n}\right)+\sin\left(\frac{\theta}{2n}+\frac{k\pi}{n}\right)\right) $$
for the $2n$-th roots, let
 $$e_k=\left(\cos\left(\frac{\theta}{2n}+\frac{k\pi}{n}\right)+\sin\left(\frac{\theta}{2n}+\frac{k\pi}{n}\right)\right)$$
Show that $e_{k+n}=-e_k$. Therefore, 
  $$ \sum_{k=0}^{2n-1} e_k = \sum_{k=0}^{n-1} (e_k+e_{n+k})=0.$$
(This was already implicitly contained in previous answers).
A: The $2n$-th roots are exactly the solutions to the polynomial equation $x^{2n}-z=0$.  By Fundamental Theorem of Algebra, if $z_1, z_2, \ldots, z_{2n}$ are the $2n$-th roots, then $x^{2n}-z = (x-z_1)(x-z_2)\cdots(x-z_{2n})$.  If we multiply out this last expression, we get $x^{2n} +(\sum z_i)x^{2n-1}+ (\sum z_i z_j)x^{2n-2} + \cdots + \prod z_i.$  Note that the coefficient on the $x^{2n-1}$ term is the sum of the roots and also that this must be equal to $0$, since the polynomial must multiply out to $x^{2n}-z$.
