Question

If $z(\ne -1)$ is complex number such that $\frac{z-1}{z+1}$ is purely imaginary, then $|z|$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is A $1$

Given $\frac{z-1}{z+1}$ is purely imaginary

Let $z=x+iy$

Now $\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}=\frac{(x-1)+iy}{(x+1)+iy}$

$=\frac{(x-1)+iy}{(x+1)+iy}\cdot \frac{(x+1)-iy}{(x+1)-iy}$

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

$=\frac{x^2-1+y^2+i2y}{(x+1{)}^2+y^2}$

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

But $\frac{z-1}{z+1}$ is imaginary, $\Rightarrow$ Re $\left(\frac{z-1}{z+1}\right)=0$

$\therefore \frac{x^2-1+y^2}{(x+1{)}^2+y^2}=0$

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

$\Rightarrow |z|=1$.