Question

If $w=\frac{z+i}{z-i}$ is purely real then find arg z.

Answer 1

$Letz=x+iyw=\frac{x+i(y+1)}{x+i(y-1)}Onrationalising,=\frac{(x+i(y+1))(x-i(y-1))}{x^2-i^2{(y-1)}^2}=\frac{x^2-ix(y-1)+ix(y+1)+y^2-1}{x^2+{(y-1)}^2}=\frac{x^2+y^2-1+i(xy+x-xy+x)}{x^2+{(y-1)}^2}Im\left(w\right)=0\frac{2x}{x^2+{(y-1)}^2}=0x=0w=\frac{y+1}{y-1}{\tan }^{-1}\frac{y}{x}={\tan }^{-1}\frac{y}{0}=\frac{\pi }{2}$