# If a>0 and $z=\frac{(1+i)^{2}}{a-i}$ , has magnitude $\sqrt{\frac{2}{5}}$ ,then $\bar{z}$ is equal to : Option 1) $-\frac{1}{5}-\frac{3}{5}i$ Option 2) $-\frac{3}{5}-\frac{1}{5}i$ Option 3) $\frac{1}{5}-\frac{3}{5}i$ Option 4) $-\frac{1}{5}+\frac{3}{5}i$

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

$=\frac{1+2i-1}{a-i}=\frac{2i}{a-i}$

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

given that

$|z|=\sqrt{\frac{2}{5}}$

$|\frac{2i(a+i)}{a^{2}+1}|=\sqrt{\frac{2}{5}}$

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

$a=3$

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

Option (1) is correct.

Option 1)

$-\frac{1}{5}-\frac{3}{5}i$

Option 2)

$-\frac{3}{5}-\frac{1}{5}i$

Option 3)

$\frac{1}{5}-\frac{3}{5}i$

Option 4)

$-\frac{1}{5}+\frac{3}{5}i$

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