Question

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

• A. $\left|z\right|>1$
• B. $\left|z\right|<1$
• C. $\left|z\right|=1$
• D. None of these

Answer 1

The correct option is C $\left|z\right|=1$

Let $\frac{z-1}{z+1}=iy$, where $y$ is real

$\Rightarrow \frac{z+1}{z-1}=\frac{1}{iy}$

Applying componendo and dividdendo

$\frac{2z}{2}=\frac{1+iy}{1-iy}\Rightarrow z=\frac{1+iy}{1-iy}\Rightarrow \left|z\right|=\frac{\sqrt{1+y^2}}{\sqrt{1+y^2}}=1$