Question

A point $z$ moves in the complex plane such that $\arg \left(\frac{ z -2}{ z +2}\right)=\frac{\pi}{4}$, then the minimum value of $| z -9 \sqrt{2}-2 i|^{2}$ is equal to ____

Answer 1

Let $z = x + iy$
$\arg \left(\frac{x-2+i y}{x+2+i y}\right)=\frac{\pi}{4}$
$\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{y}{x-2}\right)-\tan ^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi}{4}$
$\frac{\frac{ y }{ x -2}-\frac{ y }{ x +2}}{1+\left(\frac{ y }{ x -2}\right) \cdot\left(\frac{ y }{ x +2}\right)}=\tan \frac{\pi}{4}=1$
$\frac{x y+2 y-x y+2 y}{x^{2}-4+y^{2}}=1$
$4 y=x^{2}-4+y^{2}$
$x^{2}+y^{2}-4 y-4=0$
locus is a circle with center $(0,2)$ & radius $=2 \sqrt{2}$

min. value $=( AP )^{2}=( OP - OA )^{2}$
$=(9 \sqrt{2}-2 \sqrt{2})^{2}$
$=(7 \sqrt{2})^{2}=98$