Question

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

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

Answer 1

Answer: (B)

$\frac{\pi }{4}$

Since z lies on a circle $|z-2i|=2\sqrt{2}$

Let $z=x+iy\Rightarrow {|z-2i|}^2={\left(2\sqrt{2}\right)}^2$

$\therefore x^2+{(y-2)}^2=8$

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

$\Rightarrow \frac{4y}{x^2+y^2-4}=1$

$\Rightarrow \frac{y(x+2)-y(x-2)}{x^2+y^2-4}=1$

Dividing both sides by $x^2-4$

$\Rightarrow \frac{\frac{y(x+2)-y(x-2)}{x^2-4}}{\frac{x^2+y^2-4}{x^2-4}}=1$

$\Rightarrow \frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y^2}{x^2-4}}=1$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{x-2}\right)-{\tan }^{-1}\left(\frac{y}{x+2}\right)={\tan }^{-1}\left(1\right)$

$\therefore ang(z-2)-ang(z+2)=\frac{z}{4}$