Show that $\frac1x+\frac1y+\frac1z=0$ 
We know that $x=\operatorname{cis}\alpha$, $y=\operatorname{cis}\beta$, $z=\operatorname{cis}\gamma$ and that $x+y+z=0$. Prove that $\frac1x+\frac1y+\frac1z=0$.

The only thing I could get out of this is that $\frac1x+\frac1y+\frac1z=\frac{x+y+z}{xyz}$, and the numerator equals 0 but to be honest I'd really like to know how to prove this properly.
 A: Hint: $x=\operatorname{cis}\alpha\iff|x|=1\iff\bar x=\frac1x$, and $x+y+z=0\iff\bar x+\bar y+\bar z=0$.
A: $x+y+z=0 \to \operatorname{cis}\alpha+\operatorname{cis}\beta+\operatorname{cis}\gamma=0\to (\cos\alpha+\cos\beta+\cos\gamma)+i(\sin\alpha+\sin\beta+\sin\gamma)=0$
Now as both real and imaginary parts are $0$ we can say if $a+bi=0+0i=0$ then also $a-bi=0-0i=0$, so we conclude that:
$(\cos\alpha+\cos\beta+\cos\gamma)-i(\sin\alpha+\sin\beta+\sin\gamma)=0$
$ \to (\cos(-\alpha)+\cos(-\beta)+\cos(-\gamma))+i(\sin(-\alpha)+\sin(-\beta)+\sin(-\gamma))=0$
$\to \operatorname{cis}(-\alpha)+\operatorname{cis}(-\beta)+\operatorname{cis}(-\gamma)=0\to \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$
A: Note that for $x=cisα,$  we get $ 1/x=cis(-α ).$
Thus the real part of $x+y+z$ is the same as the real part of $( 1/x + 1/y +1/z)$ and the imaginary part of $x+y+z$ is the opposite of the imaginary part of $( 1/x+1/y +1/z)$.
From $x+y+z=0$ we find out that its real and imaginary parts are both zero.
Therefore the real and imaginary part of    $( 1/x + 1/y +1/z)$ are also zero. 
