JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(z=\frac{1}{2}-2 i\), is such that \(|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}\) and \(\alpha, \beta \in R \quad\), then \(\alpha+\beta\) is equal to
- A \(-4\)
- B \(3\)
- C \(2\)
- D \(-1\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\( \mathrm{z}=\frac{1}{2}-2 \mathrm{i} \) \( |\mathrm{z}+1|=\alpha \mathrm{z}+\beta(1+\mathrm{i}) \) \( \left|\frac{3}{2}-2 \mathrm{i}\right|=\frac{\alpha}{2}-2 \alpha \mathrm{i}+\beta+\beta \mathrm{i} \)…
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