Question

If $z_1$ is a complex number other than $-1$ such that $|z_1|=1$ and $z_2=\frac{z_1-1}{z_1+1}$ then show that $z_2$ is purely imaginary.

Answer 1

Let $z_1=x_1+i{y}_1$, Then, $|z_1|=1\Rightarrow |z_1{|}^2=1\Rightarrow x_1^2+y_1^2=1$

$z_2=\frac{z_1-1}{z_1+1}=\frac{(x_1+i{y}_1)-1}{(x_1+i{y}_1)+1}=\frac{(x_1-1)+i{y}_1}{(x_1+1)+i{y}_1}\times \frac{(x_1+1)-i{y}_1}{(x_1+1)-i{y}_1}$

$=\frac{(x_1^2+y_1^2-1)+2i{y}_1}{(x_1+1{)}^2+y_1^2}=\frac{2i{y}_1}{(x_1+1{)}^2+y_1^2}$,

which is purely imaginary. $[\because |z_1{|}^2=1]$