If $z_{1}$ is a complex number other than $-1$ such that $\left|z_{1}\right|=1$ and $z_{2}={ }^{\frac{z_{1}-1}{z_{1}+1}}$ then show that $z 2$ is purely imaginary.
Let $z_{1}=a+i b$ such that $\left.\left|z_{1}\right|=\sqrt{(} a^{2}+b^{2}\right)=1$
Now, $z_{2}=\frac{z_{1}-1}{z_{1}+1}=\frac{a+i b-1}{a+i b+1}=\frac{(a-1)+i b}{(a+1)+i b}$
$\Rightarrow \frac{(a-1)+i b}{(a+1)+i b} \times \frac{(a+1)-i b}{(a+1)-i b}$
$=\frac{a^{2}+a-i a b-a-1+i b+i a b+i b-i^{2} b^{2}}{(a+1)^{2}+b^{2}}$
$=\frac{a^{2}+-1+i b+i b+b^{2}}{(a+1)^{2}+b^{2}}=\frac{a^{2}+b^{2}-1+2 i b}{(a+1)^{2}+b^{2}}$
$=\frac{\left(a^{2}+b^{2}\right)-1+2 i b}{(a+1)^{2}+b^{2}}=\frac{1-1+2 i b}{(a+1)^{2}+b^{2}}\left[\because a^{2}+b^{2}=1\right]$
$=0+\frac{2 i b}{(a+1)^{2}+b^{2}}$
Thus, the real part of $z_{2}$ is 0 and $z_{2}$ is purely imaginary.