Complex Numbers Part 2

Concept:The first condition describes points equidistant from two fixed points, yielding a line. The second condition gives a circle-like locus.Explanation:From $\left|\frac{z-6i}{z-2i}\right| = 1$, we get $|z-6i| = |z-2i|$.This means $z$ is equidistant from $(0,2)$ and $(0,6)$.Hence $z$ lies on the perpendicular bisector of these points, which is the horizontal line $y=4$.So $z = x + 4i$, with $x \in \mathbb{R}$.Substitute $z = x+4i$ into $\left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}$.This gives $\frac{|x+4i-8+2i|}{|x+4i+2i|} = \frac{3}{5}$$\Rightarrow \frac{|x-8+6i|}{|x+6i|} = \frac{3}{5}$.Cross-multiplying: $5|x-8+6i| = 3|x+6i|$.Write magnitudes: $5\sqrt{(x-8)^2+6^2} = 3\sqrt{x^2+6^2}$.Square both sides: $25[(x-8)^2+36] = 9(x^2+36)$.Expand: $25(x^2 -16x +64+36) = 9x^2+324$$\Rightarrow 25x^2 -400x +2500 = 9x^2+324$$\Rightarrow 16x^2 -400x +2176 = 0$Divide by 16: $x^2 -25x +136 = 0$Factor: $(x-17)(x-8)=0$Thus $x = 8$ or $x = 17$.So the set $S = \{8+4i,\; 17+4i\}$.Now compute sum of squares of moduli:$|8+4i|^2 = 8^2+4^2 = 64+16 = 80$$|17+4i|^2 = 17^2+4^2 = 289+16 = 305$Sum = $80+305 = 385$.Answer:$\sum_{z \in S} |z|^2 = 385$, which corresponds to option C.