Complex Numbers 2 Question 2
2.
If $a>0$ and $z=\frac{(1+i)^{2}}{a-i}$, has magnitude $\sqrt{\frac{2}{5}}$, then $\bar{z}$ is equal to
(2019 Main, 10 April I)
(a) $\frac{1}{5}-\frac{3}{5} i$
(b) $-\frac{1}{5}-\frac{3}{5} i$
(c) $-\frac{1}{5}+\frac{3}{5} i$
(d) $-\frac{3}{5}-\frac{1}{5} i$
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Answer:
Correct Answer: 2. (b)
Solution:
- The given complex number $z=\frac{(1+i)^{2}}{a-i}$
$ \begin{aligned} & =\frac{(1-1+2 i)(a+i)}{a^{2}+1} \quad\left[\because i^{2}=-1\right] \\ & =\frac{2 i(a+i)}{a^{2}+1}=\frac{-2+2 a i}{a^{2}+1} \\ & \because \quad|z|=\sqrt{2 / 5} \\ & \Rightarrow \quad \sqrt{\frac{4+4 a^{2}}{\left(a^{2}+1\right)^{2}}}=\sqrt{\frac{2}{5}} \Rightarrow \frac{2}{\sqrt{1+a^{2}}}=\sqrt{\frac{2}{5}} \\ & \Rightarrow \quad \frac{4}{1+a^{2}}=\frac{2}{5} \Rightarrow a^{2}+1=10 \\ & \Rightarrow \quad a^{2}=9 \Rightarrow a=3 \\ & \therefore \quad z=\frac{-2+6 i}{10} \end{aligned} $
So, $\bar z=(\frac{-2+6 i}{10}) = (\frac{-1}{5}+\frac{3}{5}i)$
$\Rightarrow \bar z= \frac{-1}{5}-\frac{3}{5}i$
