Question

If $|z|=1,(z\ne -1)$ and $z=x+iy$, then $\left(\frac{z-1}{z+1}\right)$ is

• A. purely real
• B. purely imaginary
• C. zero
• D. undefined

Answer 1

Answer: (B)

$z=x+iy$
$\Rightarrow |z{|}^2=x^2+y^2=1\dots (i)$
Now, $\left(\frac{z-1}{z+1}\right)=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$
$=\frac{\left(x^2+y^2-1\right)+2iy}{(x+1{)}^2+y^2}=\frac{2iy}{(x+1{)}^2+y^2}$
[By equation (i)]
Hence, $\left(\frac{z-1}{z+1}\right)$ is purely imaginary