Question

If $\frac{z-1}{z+1}$ is purely imaginary then $\left|z\right|=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$1$

$\text{Given:}$

$\frac{z-1}{z+1}\text{is purely imaginary}$

$\text{Let}z=x+iy$

$\frac{z-1}{z+1}=\frac{x+iy+(-1)}{x+iy+1}\times \frac{x-iy+1}{(x+1)-iy}$

$\frac{z-1}{z+1}=\frac{x^2-iyx+x+i{y}^{}x+y^2+iy-x+iy-1}{(x+1{)}^2+y^2}$

$\frac{z-1}{z+1}=\frac{x^2+y^2-1+2iy}{(x+1{)}^2+y^2}$

$\frac{z-1}{z+1}=\frac{x^2+y^2-1}{x^2+1+2x+y^2}+\frac{2iy}{(x+1{)}^2+y^2}$

For purely imaginary,

$\frac{x^2+y^2-1}{x^2+1+2x+y^2}=0$

$x^2+y^2=1$

$|z{|}^2=1$

$|z|=1$ option $A$ is correct