MHT CET · Maths · Complex Number
If \(\mathrm{a}>0\) and \(\mathrm{z}=\frac{(1+\mathrm{i})^2}{\mathrm{a}-\mathrm{i}}, \mathrm{i}=\sqrt{-1}\), has magnitude \(\frac{2}{\sqrt{5}}\), then \(\bar{z}\) is
- A \(-\frac{2}{5}-\frac{4}{5} \mathrm{i}\)
- B \(-\frac{2}{5}+\frac{4}{5} i\)
- C \(\frac{2}{5}-\frac{4}{5} \mathrm{i}\)
- D \(\frac{2}{5}+\frac{4}{5} \mathrm{i}\)
Answer & Solution
Correct Answer
(A) \(-\frac{2}{5}-\frac{4}{5} \mathrm{i}\)
Step-by-step Solution
Detailed explanation
\(z =\frac{(1+i)^2}{a-i} \)
\( =\frac{2 i}{a-i}=\frac{2 i(a+i)}{a^2+1} \)
\( \therefore |z| =\sqrt{\frac{-2+2 a i}{a^2+1}} \)
\( \Rightarrow \frac{2}{\sqrt{5}}=\frac{2 a^2}{\sqrt{a^2+1}}\)
\(\Rightarrow \mathrm{a}=2 \ldots[\because \mathrm{a}>0]\)
\(\therefore z =\frac{-2+4 i}{5}=\frac{-2}{5}+\frac{4}{5} i \)
\( \Rightarrow \bar{z}=-\frac{2}{5}-\frac{4}{5} \mathrm{i}\)
\( =\frac{2 i}{a-i}=\frac{2 i(a+i)}{a^2+1} \)
\( \therefore |z| =\sqrt{\frac{-2+2 a i}{a^2+1}} \)
\( \Rightarrow \frac{2}{\sqrt{5}}=\frac{2 a^2}{\sqrt{a^2+1}}\)
\(\Rightarrow \mathrm{a}=2 \ldots[\because \mathrm{a}>0]\)
\(\therefore z =\frac{-2+4 i}{5}=\frac{-2}{5}+\frac{4}{5} i \)
\( \Rightarrow \bar{z}=-\frac{2}{5}-\frac{4}{5} \mathrm{i}\)
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