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JEE Mains · Maths · STD 11 - 4.1 complex nubers

\(z=\alpha+i \beta\) માટે જો \(|z+2|=z+4(1+i)\) હોય, તો \(\alpha+\beta\) અને \(\alpha \beta\) એ \(.........\) સમીકરણ ના બીજ છે.

  1. A \(x^2+7 x+12=0\)
  2. B \(x^2+3 x-4=0\)
  3. C \(x^2+2 x-3=0\)
  4. D \(x ^2+ x -12=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x^2+7 x+12=0\)

Step-by-step Solution

Detailed explanation

\(|z+2|=|\alpha+i \beta+2|\) \(=\alpha+i \beta+4+4 i\) \(\sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+ i (\beta+4)\) \(\beta+4=0\) \((\alpha+2)^2+16=(\alpha+4)^2\) \(\beta=-4\) \(\alpha^2+4+4 \alpha+16=\alpha^2+16+8 \alpha\) \(4=4 \alpha\) \(\alpha=1\) \(\alpha=1, \beta=-4\)…
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