Complex NumbersHard
Question
Let $S = \left\{ z \in \mathbb{C}:\left| \frac{z - 6i}{z - 2i} \right| = 1 \right.\ $ and $\left. \ \left| \frac{z - 8 + 2i}{z + 2i} \right| = \frac{3}{5} \right\}$.
Then $\sum_{z \in s}\mspace{2mu}|z|^{2}$ is equal to
Options
A.398
B.413
C.423
D.385
Solution
Solving $\left| \frac{z - 6i}{z - 2i} \right| = 1 \Rightarrow y = 4\ldots.$.
(where $z = x + iy$ )
Now solving $\left| \frac{z - 8 + 2i}{z + 2i} \right| = \frac{3}{5}$
$$\begin{array}{r} \Rightarrow x^{2} + y^{2} - 25x + 4y + 104 = 0\#(2) \end{array}$$
Solving (1) & (2) ⇒ $z = 17 + 4i\& 8 + 4i$
$$\Rightarrow \Sigma|z|^{2} = (17)^{2} + (4)^{2} + (8)^{2} + (4)^{2} = 385$$
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