For \(z\in\CC\smts{-1}\), show that \[ \frac{z-1}{z+1}\in i\RR \iff \abs z=1. \]
Hint: \(z\) is real iff \(\bar{z} = z\) and purely imaginary iff \(\bar{z} = -z\).
Solution
Write \(w=(z-1)/(z+1)\). Then \[ w\in i\RR &\iff w=-\overline w \\ &\iff \frac{z-1}{z+1}=\frac{1-\bar z}{1+\bar z} \\ &\iff (z-1)(1+\bar z)=(1-\bar z)(1+z) \\ &\iff 2\abs z^2-2=0 \\ &\iff \abs z=1. \]