Question

If $z$ lies on the circle $|z-2i|=2\sqrt{2}$, then the value of $\arg \left[\frac{(z-2)}{(z+2)}\right]$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{4}$

Given: $z$ lies on the circel $|z-2i|=2\sqrt{2}$

Let $z=x+iy$

$|z-2i|=|x+i(y-2)|=2\sqrt{2}$

$\sqrt{x^2+(y-2{)}^2}=2\sqrt{2}$

$x^2+(y-2{)}^2=8...(i)$

To find point of intersection of circle with real axis, put $y=0$ in $\left(i\right)$
$\Rightarrow x^2+4=8\Rightarrow x^2=4\Rightarrow x=\pm 2$


In $△OBC,OB=OC=2$

$\therefore \angle OCB=\angle OCA={45}^{\circ }$

$\therefore \angle ACB={90}^{\circ }=\frac{\pi }{2}$

Therefore, $\angle APB={45}^{\circ }$
By property of circle