Question

If $a>0$ and $z=\frac{(1+i{)}^2}{a-i}$, has magnitude $\sqrt{\frac{2}{5}}$ , then $\overline{z}$ is equal to :

• A. $-\frac{1}{5}-\frac{3}{5}i$
• B. $-\frac{1}{5}+\frac{3}{5}i$
• C. $-\frac{3}{5}-\frac{1}{5}i$
• D. $\frac{1}{5}-\frac{3}{5}i$

Answer 1

The correct option is A $-\frac{1}{5}-\frac{3}{5}i$

$z=\frac{(1+i{)}^2}{a-i}$
On rationalising the denominator, we get
$z=\frac{2ai-2}{a^2+1}$
$\Rightarrow |z|=\frac{2}{\sqrt{a^2+1}}=\sqrt{\frac{2}{5}}\Rightarrow 20=2(a^2+1)\Rightarrow a^2+1=10\Rightarrow a=3(\because a>0)$

$\therefore z=\frac{2\left(3\right)i-2}{3^2+1}$
$z=\frac{-1+3i}{5}\Rightarrow \overline{z}=\frac{-1-3i}{5}$