JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(a > 0\) and \(z = \frac{{{{\left( {1 + i} \right)}^2}}}{{a - i}}\), has magnitude \(\sqrt {\frac{2}{5}} \), then \(\bar z\) is equal to:
- A \( - \frac{3}{5} - \frac{1}{5}i\)
- B \( - \frac{1}{5} - \frac{3}{5}i\)
- C \( - \frac{1}{5} + \frac{3}{5}i\)
- D \( \frac{1}{5} - \frac{3}{5}i\)
Answer & Solution
Correct Answer
(B) \( - \frac{1}{5} - \frac{3}{5}i\)
Step-by-step Solution
Detailed explanation
\(z=\frac{(1+i)^{2}}{a-i}=\frac{2 i(a+i)}{a^{2}+1}\) \(|z|=\frac{2}{\sqrt{a^{2}+1}}=\sqrt{\frac{2}{5}} \Rightarrow a=3\) \(\therefore \bar{z}=\frac{-2 i(3-i)}{10}\) \(\Rightarrow \frac{-1-3 i}{5}\)
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