Question

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

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

Answer 1

Answer: (B)

$\left|z\right|=1$

Answer 2

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 $\omega=u +i v$
Then, $|\omega+1|=|u +i v+1|$
$=|(u+1)+i v|=\sqrt{(u+1)^{2}+v^{2}}$
and $|\omega-1|=\sqrt{(u-1)^{2}+v^{2}}$
$\therefore$ From E (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.