Question

If $\frac{z+2i}{z-2i}$ is purely imaginary then $\left|z\right|$ is

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $1$

$\frac{z+2i}{z-2i}$

$=\frac{z+2i}{z-2i}\times \frac{z+2i}{z+2i}$

$=\frac{z^2+4z-4}{z^2-4}$

Let $z=x+iy$

$=\frac{(x+iy{)}^2+4(x+iy)-4}{(x+iy{)}^2-4}$

$=\frac{x^2-y^2+2ixy-4+4x+4iy}{(x+iy{)}^2-4}$

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

z is purely imaginary. So, Re(z)=0

$\frac{x^2-y^2+4x-4}{(x+iy{)}^2-4}=0$

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

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

The above equation represents the hyperbola on x-axis, with $(1,0)$

$\therefore z=1$