Question

If $\frac{z-i}{z+i}(z\ne -i)$ is a purely imaginary number, then $z.\overline{z}$ is equal to

• A. \N
• B. 1
• C. 2
• D. None of these

Answer 1

The correct option is B 1

Here $\frac{z-i}{z+i}=\frac{x+i(y-1}{x+i(y+1)}.\frac{x-i(y+1)}{x-i(y+1)}$
$=\frac{(x^2+y^2-1)+i(-2x)}{x^2+(y+1{)}^2}$
As $\frac{z-i}{z+i}$ is purely imaginary, we get
$x^2+y^2-1=0\Rightarrow x^2+y^2=1\Rightarrow z\overline{z}=1.$