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JEE Mains · Maths · STD 11 - 4.1 complex nubers

બિંદુ \(\mathrm{z}\) એ આર્ગંડ સમતલમાં એવી રીતે ગતિ કરે છે કે જેથી \(\arg \left(\frac{\mathrm{z}-2}{\mathrm{z}+2}\right)=\frac{\pi}{4}\) થાય છે તો \(|z-9 \sqrt{2}-2 i|^{2}\) ની ન્યૂનતમ કિમંત્ત મેળવો.

  1. A \(89\)
  2. B \(108\)
  3. C \(98\)
  4. D \(72\)
Verified Solution

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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