Complexes question $(z-1) \over (z+1)$

Question

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?

Answer 1

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 $|z|=1$.

Answer 2

I believe you have meant purely imaginary

Setting $z=a+ib$ where $a,b$ are real

$$\frac{z-1}{z+1}=\frac{a+ib-1}{a+ib+1}=\frac{(a-1+ib)(a+1-ib)}{(a+1)^2+b^2}$$

$$=\frac{(a)^2-(1-ib)^2}{(a+1)^2+b^2}=\frac{a^2+b^2-1+2ib}{(a+1)^2+b^2}$$