Complexes question $(z-1) \over (z+1)$ 
Given $z \ne -1$. Prove that $(z-1) \over (z+1)$ is an Imaginary number if and only if $|z| = 1$.

I tried computing $(z-1) \over (z+1)$ by multiplying like that: $(z-1) \over (z+1)$$(z-1) \over (z-1)$ and then getting: $(z-1)^2$  
but if $z = 1$ then it's zero, and it's not an imaginary number. Any suggestions?

Multiply numerator and denominator by $z^*+1$:
$$\frac{z-1}{z+1}=\frac{(z-1)(z^*+1)}{|z+1|^2}=\frac{zz^*+z-z^*-1}{|z+1|^2}=
\frac{|z|^2-1+2iIm\{z\}}{|z+1|^2}$$
Obviously, the denominator is real, and so this number is purely imaginary if $|z|^2-1=0$, i.e. if $|z|=1$.
