Question

If $\frac{z-1}{z+1}$ is purely imaginary number ($z\ne -1$), find the value of $\left|z\right|$.

Answer 1

Let $z=x+iy$.
Then,
$$\begin{gathered} \frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1} \\ =\frac{\left(x-1\right)+iy}{\left(x+1\right)+iy}\times \frac{\left(x+1\right)-iy}{\left(x+1\right)-iy} \\ =\frac{x^2+x-ixy-x-1+iy+ixy+iy-i^2{y}^2}{{\left(x+1\right)}^2-i^2{y}^2} \\ =\frac{x^2+y^2-1+2iy}{x^2+1+2x+y^2}[\because i^2=-1] \end{gathered}$$

If $\frac{z-1}{z+1}$ is purely imaginary number, then
$$\begin{gathered} \mathrm{Re}\left(\frac{z-1}{z+1}\right)=0 \\ \Rightarrow x^2+y^2-1=0 \\ \Rightarrow x^2+y^2=1 \\ \Rightarrow {\left|z\right|}^2=1 \\ \Rightarrow \left|z\right|=1 \end{gathered}$$

Thus, the value of $\left|z\right|$ is 1.