JEE Mains · Maths · STD 11 - 4.1 complex nubers
If for \(z=\alpha+i \beta,|z+2|=z+4(1+i)\), then \(\alpha+\beta\) and \(\alpha \beta\) are the roots of the equation
- A \(x^2+7 x+12=0\)
- B \(x^2+3 x-4=0\)
- C \(x^2+2 x-3=0\)
- D \(x ^2+ x -12=0\)
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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