A point z moves in the complex plane such

`
Question:

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

Solution:

Let $z=x+i y$

$\arg \left(\frac{x-2+i y}{x+2+i y}\right)=\frac{\pi}{4}$

$\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{4}$

$\tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}-2}\right)-\tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}+2}\right)=\frac{\pi}{4}$

$\frac{\frac{\mathrm{y}}{\mathrm{x}-2}-\frac{\mathrm{y}}{\mathrm{x}+2}}{1+\left(\frac{\mathrm{y}}{\mathrm{x}-2}\right) \cdot\left(\frac{\mathrm{y}}{\mathrm{x}+2}\right)}=\tan \frac{\pi}{4}=1$

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

$4 y=x^{2}-4+y^{2}$

$x^{2}+y^{2}-4 y-4=0$

locus is a circle with center $(0,2) \&$ radius $=2 \sqrt{2}$

min. value $=(\mathrm{AP})^{2}=(\mathrm{OP}-\mathrm{OA})^{2}$

$=(9 \sqrt{2}-2 \sqrt{2})^{2}$

$=(7 \sqrt{2})^{2}=98$

Leave a comment