JEE Mains · Maths · STD 11 - 4.1 complex nubers
A point \(\mathrm{z}\) moves in the complex plane such that \(\arg \left(\frac{\mathrm{z}-2}{\mathrm{z}+2}\right)=\frac{\pi}{4}\), then the minimum value of \(|z-9 \sqrt{2}-2 i|^{2}\) is equal to ..... .
- A \(89\)
- B \(108\)
- C \(98\)
- D \(72\)
Answer & Solution
Correct Answer
(C) \(98\)
Step-by-step Solution
Detailed explanation
Let \(z=x+i y\) \(\arg \left(\frac{x-2+i y}{x+2+i y}\right)=\frac{\pi}{4}\) \(\arg (x-2+i y)-\arg (x+2+i y)=\frac{\pi}{4}\) \(\tan ^{-1}\left(\frac{y}{x-2}\right)-\tan ^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi}{4}\)…
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