Multiple values of $v=(\frac{z+1}{z-1})^{1/3}$ (if for example $z=i\sqrt{3}$ ) Considering $v=(\frac{z+1}{z-1})^{1/3}$.
How can I recognize its multivaluedness.
If for example $z=i\sqrt{3}$ then what are the possible values?
 A: For every $w\ne0$ in $\mathbb C$, there are three different complex numbers $u_1$, $u_2$ and $u_3$ such that $u_1^3=u_2^3=u_3^3=w$. This is a reason why none of them should be denoted $w^{1/3}$, and why in fact the notation $w^{1/3}$ should be avoided altogether on the complex plane $\mathbb C$.
Another reason is that there exists no continuous function $\varphi:\mathbb C\to\mathbb C$ such that $\varphi(w)^3=w$ for every $w$ in $\mathbb C$. The most one can do is to exhibit such a continuous function $\varphi:\mathbb C\setminus S\to\mathbb C$, where $S$ is a slit, for example $S=\mathbb R_-^*$, but not on the whole of $\mathbb C$ (even on the unit circle this cannot be done with continuity).
The determination of $(u_1,u_2,u_3)$, on the other hand, is simple. Since $w\ne0$, $w=r\mathrm e^{\mathrm it}$ with $r$ in $\mathbb R_+^*$ and $t$ in $\mathbb R$, then the numbers $u_k$ are the three complex numbers $\sqrt[3]{r}\,\mathrm e^{\mathrm it/3+\mathrm 2in\pi/3}$ for $n$ in $\{0,1,2\}$, say. Note that $\sqrt[3]{\ }$ here denotes the (perfectly well defined) third-root function, from $\mathbb R_+$ to $\mathbb R_+$.
If $z=\mathrm i\sqrt3$, then $z+1=2\mathrm e^{\mathrm i\pi/3}$ and  $z-1=2\mathrm e^{2\mathrm i\pi/3}$ hence $w=(z+1)/(z-1)=\mathrm e^{-\mathrm i\pi/3}$ and the numbers $u_k$ are the three complex numbers $\mathrm e^{-\mathrm i\pi/9+\mathrm 2in\pi/3}$, that is, $\cos(-\pi/9+2n\pi/3)+\mathrm i\sin(-\pi/9+2n\pi/3)$, for $n$ in $\{0,1,2\}$.
A: Why not use $z=0$ , so that $v=-1$. We can then write :
Let $z=re^{i\theta}$ . Then $z^3=r^3e^{i3\theta}=(-1)=1e^{i(2k+1)\pi}$ , so that, by
equality, $r=1$ and $3\theta=(2k+1)\pi$ , so $\theta=\frac{(2k+1)\pi}{3}; k=0,1,2$ ( you can check that 
the values of $z$ repeat after $k=3$. Then check that for each $k$ , you have a root for $z=-1$
