# If

Question:

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

1. (1) $-\frac{1}{5}-\frac{3}{5} i$

2. (2) $-\frac{3}{5}-\frac{1}{5} i$

3. (3) $\frac{1}{5}-\frac{3}{5} i$

4. (4) $-\frac{1}{5}+\frac{3}{5} i$

Correct Option: 1

Solution:

$z=\frac{(1+i)^{2}}{a-i} \times \frac{a+i}{a+i}$

$z=\frac{(1-1+2 i)(a+i)}{a^{2}+1}=\frac{2 a i-2}{a^{2}+1}$

$|z|=\sqrt{\left(\frac{-2}{a^{2}+1}\right)^{2}+\left(\frac{2 a}{a^{2}+1}\right)^{2}}=\sqrt{\frac{4+4 a^{2}}{\left(a^{2}+1\right)^{2}}}$

$\Rightarrow|z|=\sqrt{\frac{4\left(1+a^{2}\right)}{\left(1+a^{2}\right)^{2}}}=\frac{2}{\sqrt{1+a^{2}}}$......(1)

Since, it is given that $|z|=\sqrt{\frac{2}{5}}$

Then, from equation (1),

$\sqrt{\frac{2}{5}}=\frac{2}{\sqrt{1+a^{2}}}$

Now, square on both side; we get

$\Rightarrow \frac{2}{5}=\frac{4}{1+a^{2}} \Rightarrow 1+a^{2}=10 \Rightarrow a=\pm 3$

Since, it is given that $\mathrm{a}>0 \Rightarrow \mathrm{a}=3$

Then, $z=\frac{(1+i)^{2}}{a-i}=\frac{1+i^{2}+2 i}{3-i}=\frac{2 i}{3-i}$

$=\frac{2 i(3+i)}{10}=\frac{-1+3 i}{5}$

Hence, $\bar{z}=\frac{-1}{5}-\frac{3}{5} i$