Question:
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 the real parts of $z_{2}$ is zero.
Solution:
Let $z=x+i y$.
Then,
$z_{2}=\frac{z_{1}-1}{z_{1}+1}$
$=\frac{x+i y-1}{x+i y+1}$
$=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$
$=\frac{x^{2}+x-i x y-x-1+i y+i x y+i y-i^{2} y^{2}}{(x+1)^{2}-i^{2} y^{2}}$
$=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+1+2 x+y^{2}} \quad\left[\because i^{2}=-1\right]$
Now,
$\operatorname{Re}\left(z_{2}\right)=\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+1+2 x}$
$=0 \quad\left[\because\left|z_{1}\right|=1 \Rightarrow x^{2}+y^{2}=1\right]$
Thus, the real parts of $z_{2}$ is zero.