Question

The complex number $z$ is such that $\left|z\right|=1$,$z\ne -1$ and $\omega =\left(\frac{z-1}{z+1}\right)$. Then the real part of $\omega$ is

Answer 1

Let $z=x+iy$

Given,

$\left|z\right|=1$

$\Rightarrow \sqrt{x^2+y^2}=1$

$\Rightarrow x^2+y^2=1$..............(squaring both sides)

$\omega =\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$

Rationalising the denominator, we get

$\omega =\frac{x-1+iy}{x+1+iy}\times \frac{x+1-iy}{x+1-iy}$

$\omega =\frac{(x-1+iy)(x+1-iy)}{(x+1{)}^2-(i^2{y}^2)}$

$\omega =\frac{(x-1)(x+1)+(x-1)(-iy)+iy(x+1)-\left(i^2{y}^2\right)}{(x+1{)}^2-(i^2{y}^2)}$

$\omega =\frac{(x-1)(x+1)+i(-y(x-1)+y(x+1\left)\right)-(-y^2)}{(x+1{)}^2-(-y^2)}$...................................$(\because i^2=-1)$

$=\frac{(x^2-1+y^2)+i(xy+y-xy+y)}{x^2+2x+1+y^2}$

Substituting $x^2+y^2=1$ we get

$\omega =\frac{1-1+i\left(2y\right)}{1+1+2x}$

$\omega =\frac{2yi}{2(1+x)}=0+\frac{y}{1+x}i$

$\Rightarrow$ Real part of $\omega =0$ ..............(by comparing RHS and LHS)