Let S = {z in mathbb{C} : left|frac{z-6i}{z-2 | vedclass

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(D) Given $\left|\frac{z-6i}{z-2i}\right| = 1$,where $z = x + iy$. This implies $|z-6i| = |z-2i|$.Squaring both sides: $x^2 + (y-6)^2 = x^2 + (y-2)^2$

Vedclass · Accepted Answer

$385$

Vedclass · Answer

(D) Given $\left|\frac{z-6i}{z-2i}ight| = 1$,where $z = x + iy$. This implies $|z-6i| = |z-2i|$.Squaring both sides: $x^2 + (y-6)^2 = x^2 + (y-2)^2$.$y^2 - 12y + 36 = y^2 - 4y + 4$ $\Rightarrow 8y = 32$ $\Rightarrow y = 4$.Now,$\left|\frac{z-8+2i}{z+2i}ight| = \frac{3}{5} \Rightarrow 25|z-(8-2i)|^2 = 9|z+2i|^2$.$25((x-8)^2 + (y+2)^2) = 9(x^2 + (y+2)^2)$.Substitute $y=4$: $25((x-8)^2 + 36) = 9(x^2 + 36)$.$25(x^2 - 16x + 64 + 36) = 9(x^2 + 36)$.$25x^2 - 400x + 2500 = 9x^2 + 324$.$16x^2 - 400x + 2176 = 0 \Rightarrow x^2 - 25x + 136 = 0$.$(x-8)(x-17) = 0 \Rightarrow x = 8 	ext{ or } x = 17$.Thus,$z_1 = 8+4i$ and $z_2 = 17+4i$.$|z_1|^2 = 8^2 + 4^2 = 64 + 16 = 80$.$|z_2|^2 = 17^2 + 4^2 = 289 + 16 = 305$.$\sum_{z \in S} |z|^2 = 80 + 305 = 385$.

Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

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