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(C) Let $z=x+iy$.$\arg \left(\frac{z-2}{z+2}ight)=\frac{\pi}{4}$ represents an arc of a circle.Let $z-2 = r_1 e^{i	heta_1}$ and $z+2 = r_2 e^{i	he

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$98$

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(C) Let $z=x+iy$.$\arg \left(\frac{z-2}{z+2}ight)=\frac{\pi}{4}$ represents an arc of a circle.Let $z-2 = r_1 e^{i	heta_1}$ and $z+2 = r_2 e^{i	heta_2}$. Then $	heta_1 - 	heta_2 = \frac{\pi}{4}$.This is the locus of points $z$ such that the angle subtended by the segment joining $(-2, 0)$ and $(2, 0)$ at $z$ is $\frac{\pi}{4}$.The equation is $	an^{-1}\left(\frac{y}{x-2}ight) - 	an^{-1}\left(\frac{y}{x+2}ight) = \frac{\pi}{4}$.Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get $\frac{\frac{y}{x-2} - \frac{y}{x+2}}{1 + \frac{y^2}{x^2-4}} = 1$.$\frac{4y}{x^2+y^2-4} = 1 \implies x^2+y^2-4y-4=0$.This is a circle with center $O(0, 2)$ and radius $R = \sqrt{0^2 + 2^2 - (-4)} = \sqrt{8} = 2\sqrt{2}$.We want to minimize $|z - (9\sqrt{2} + 2i)|^2$. Let $P = 9\sqrt{2} + 2i$,which is the point $(9\sqrt{2}, 2)$.The distance $OP = \sqrt{(9\sqrt{2}-0)^2 + (2-2)^2} = 9\sqrt{2}$.The minimum distance from $P$ to the circle is $OP - R = 9\sqrt{2} - 2\sqrt{2} = 7\sqrt{2}$.The minimum value of $|z-P|^2$ is $(7\sqrt{2})^2 = 49 	imes 2 = 98$.

$A$ 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 ..... .

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