JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(S=\left\{z \in C :\left|\frac{z-6 i}{z-2 i}\right|=1\right.\) and \(\left.\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5}\right\}\). Then \(\sum_{z \in s}|z|^2\) is equal to
- A 398
- B 413
- C 423
- D 385
Answer & Solution
Correct Answer
(D) 385
Step-by-step Solution
Detailed explanation
Solving \(\left|\frac{z-6 i}{z-2 i}\right|=1 \Rightarrow y=4\)\(\quad\) ...(1) \((\)where \(z=x+i y)\) Now solving \(\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5}\) \(\Rightarrow x^2+y^2-25 x+4 y+104=0\)\(\quad\) ...(2) Solving (1) & (2) \(\Rightarrow z =17+4 i\ \&\ 8+4 i\)…
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