Question

If $z$ is a complex number such that arg $\left(\frac{z-2}{z+2}\right)=\frac{\pi }{4}$, then the value of $|z-2i|$ is

• A. $2\sqrt{2}$
• B. $3\sqrt{2}$
• C. $4\sqrt{2}$
• D. None of these

Answer 1

Answer: (A)

$2\sqrt{2}$

Let $z=x+iy$ and $arg(z-2)={\theta }_1$ and $arg(z+2)={\theta }_2$
$arg(\frac{z-2}{z+2})=arg(z-2)-arg(z+2)=\frac{\pi }{4}$
${tan}^{-1}{\theta }_1-{tan}^{-1}{\theta }_2=1$
Hence,
$\pi /4={tan}^{-1}\frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y^2}{x^2-4}}$
On simplifying,
$\Rightarrow x^2+y^2-4y=4$
$(x-0{)}^2+(y-2{)}^2=8$
$|x+i(y-2){|}^2=8$
$|z-2i|=2\sqrt{2}$