Question

If $a=\frac{z-1}{z+1}$ is purely imaginary number $z\ne 1$, then $\left|\begin{array}{c}z\end{array}\right|$is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step1. Given that

Given, $a=\frac{z-1}{z+1}$ is & $z\ne 1$

Let, $z=x+iy$

$\left|z\right|=\sqrt{x^2+y^2}$

Step2. finding the value of $\left|z\right|$

$$\begin{gathered} \Rightarrow Re\frac{z-1}{z+1}=Re\frac{x+iy-1}{x+iy+1},z\ne 1 \\ \Rightarrow Re\frac{z-1}{z+1}=Re\frac{x-1+iy}{x+1+iy} \\ \Rightarrow Re\frac{z-1}{z+1}=Re\frac{(x-1+iy)\times (x+1-iy)}{(x+1+iy)\times (x+1-iy)} \\ \Rightarrow Re\frac{z-1}{z+1}=\frac{x^2-1+y^2}{{(x+1)}^2+y^2} \end{gathered}$$

Put $\begin{array}{rcl}Re\frac{z-1}{z+1}& =& 0\end{array}$

Therefore,

$$\begin{gathered} \Rightarrow \begin{array}{l}\frac{x^2-1+y^2}{{(x+1)}^2+y^2}\end{array}=0 \\ \Rightarrow x^2+y^2=1 \end{gathered}$$

Taking root both side

$$\begin{gathered} \Rightarrow \sqrt{x^2+y^2}=1 \\ \Rightarrow \left|z\right|=1 \end{gathered}$$

Hence, the correct option is option(A).