Question

If $\left|z\right|=1,$ then the value of $\left(\frac{z-1}{z+1}\right)$ is

• A. $0$
• B. Purely real
• C. Purely imaginary
• D. Complex number

Answer 1

The correct option is C Purely imaginary

We have, $\left|z\right|=1,$ or $\sqrt{x^2+y^2}=1$

$\Rightarrow x^2+y^2-1=0$

Now $\left(\frac{z-1}{z+1}\right)=\left[\frac{(x-1)+iy}{(x+1)+iy}\right]$

$\Rightarrow \left(\frac{z-1}{z+1}\right)=\left[\frac{(x-1)+iy}{(x+1)+iy}\right]\times \left[\frac{(x+1)-iy}{(x+1)-iy}\right]$

$\Rightarrow \frac{(x^2+y^2-1)+2iy}{{(x+1)}^2+y^2}$

$=\frac{2iy}{{(x+1)}^2+y^2}$ $[\because x^2+y^2-1=0]$