Question

If ${}^{\prime }{z}^{\prime }$ 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. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

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


$|z-2i|=2\sqrt{2}$
$\Rightarrow x^2+(y-2{)}^2=8$
Let , $A=2+0i,B=-2+0i$
Now $AB=4$ and $(AB{)}^2=(OA{)}^2+(OB{)}^2$
Hence $\angle BOA=\frac{\pi }{2}$

By angle at the centre of circle is twice the angle subtended at circumference property
$\angle BPA=\frac{\pi }{4}$
Hence, $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi }{4}$