Question

If $\frac{z-2}{z+2}(z\ne -2)$ is purely imaginary then $|z|$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Let $z=x+iy$

Then, $\frac{z-2}{z+2}=\frac{x+iy-2}{x+iy+2}=\frac{(x-2)+iy}{(x+2)+iy}$

$=\frac{[(x-2)+iy][(x+2)+iy]}{{(x+2)}^2+y^2}$

$=\frac{(x^2+y^2-4)+i\left(4y\right)}{{(x+2)}^2+y^2}$

Since $\frac{z-2}{z+2}$ is purely imaginary,

$\therefore x^2+y^2-4=0$

$\Rightarrow x^2+y^2=4\Rightarrow {\left|z\right|}^2=4\Rightarrow \left|z\right|=2.$