Question

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

• A. 398
• B. 385
• C. 423
• D. 413
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Hint:

Equations of the form $|z-a|=|z-b|$ always represent perpendicular bisectors in the complex plane.

Answer 1

The Correct Option is A

Step 1: Simplifying the first condition.
\[ \left|\frac{z-6i}{z-2i}\right|=1 \Rightarrow |z-6i|=|z-2i| \] This represents the perpendicular bisector of the line joining $6i$ and $2i$.
Hence, \[ \text{Im}(z)=4 \] Step 2: Simplifying the second condition.
\[ \left|\frac{z-(8-2i)}{z+2i}\right|=\frac{3}{5} \Rightarrow \frac{|z-(8-2i)|}{|z+2i|}=\frac{3}{5} \] Squaring both sides, \[ 25|z-(8-2i)|^2=9|z+2i|^2 \] Let $z=x+iy$. Substituting $y=4$, \[ 25[(x-8)^2+(4+2)^2]=9[x^2+(4+2)^2] \] \[ 25[(x-8)^2+36]=9(x^2+36) \] Solving, \[ x=2 \text{ or } x=14 \] Thus, \[ z_1=2+4i,\quad z_2=14+4i \] Step 3: Calculating $\sum |z|^2$.
\[ |z_1|^2=2^2+4^2=20 \] \[ |z_2|^2=14^2+4^2=212 \] \[ \sum |z|^2=20+212=398 \]