Question

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 1

Answer: (C)

Given $a > 0$
$z= \frac{\left(1+i\right)^{2}}{a-i} = \frac{2i\left(a+i\right)}{a^{2}+1}$
Also $ \left|z\right| = \sqrt{\frac{2}{5}} \Rightarrow \frac{2}{\sqrt{a^{2}+1}} = \sqrt{\frac{2}{5}} \Rightarrow a=3 $
So $ \bar{z} = \frac{-2i\left(3-i\right)}{10} = \frac{-1-3i}{5} $