Question

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

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

The correct option is B

1

$\frac{z-i}{z+i}$ = $\frac{x+i(y-1)}{x+i(y+1)}.\frac{x-i(y+1)}{x-i(y+1)}$

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

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.