Question

If $\frac{z-1}{z+1}$ is purely imaginary, then

• A. $|z|=\frac{1}{2}$
• B. $|z|=1$
• C. $|z|=2$
• D. $|z|=3$

Answer 1

Answer: (B)

$|z|=1$

Let $\omega =\frac{z-1}{z+1}$
Then, using componendo and dividendo, we get

$z=\frac{1+\omega }{1-\omega }$
$\Rightarrow |z|=\left|\frac{\omega +1}{\omega -1}\right|$

Put $|z|=1$
$\Rightarrow |\omega +1|=|\omega -1|$ ...... (i)

Let $\omega =u+iv$
Then, $|\omega +1|=|u+iv+1|$
$=|(u+1)+iv|=\sqrt{(u+1{)}^2+v^2}$
and

$|\omega -1|=\sqrt{(u-1{)}^2+v^2}$

$\therefore$ From Eq. (i)
$\sqrt{(u+1{)}^2+v^2}=\sqrt{(u-1{)}^2+v^2}$
$\Rightarrow (u+1{)}^2+v^2=(v-1{)}^2+v^2$
$\Rightarrow u=0$
$\therefore \omega =\frac{z-1}{z+1}$ is a pure imaginary number.