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?

I believe you have meant purely imaginary 
Setting $z=a+ib$ where  $a,b$ are real
$$\frac{z-1}{z+1}=\frac{a+ib-1}{a+ib+1}=\frac{(a-1+ib)(a+1-ib)}{(a+1)^2+b^2}$$
$$=\frac{(a)^2-(1-ib)^2}{(a+1)^2+b^2}=\frac{a^2+b^2-1+2ib}{(a+1)^2+b^2}$$
