If a>0 and z=(1+i)2a−i, has magnitude 25, then z- is equal to :

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

1. A

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

2. B

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

3. C

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

4. D

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

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### Solution:

We have, $z=\frac{\left(1+i{\right)}^{2}}{a-i}$

$=\frac{2i}{a-i}=\frac{2i\left(a+i\right)}{{a}^{2}+1}=\frac{-2}{{a}^{2}+1}+\frac{2ai}{{a}^{2}+1}$

Since  $|z|=\sqrt{2/5}$ [Given]

$\therefore \sqrt{\frac{4}{{\left({a}^{2}+1\right)}^{2}}+\frac{4{a}^{2}}{{\left({a}^{2}+1\right)}^{2}}}=\sqrt{\frac{2}{5}}⇒\frac{2}{\sqrt{1+{a}^{2}}}=\sqrt{\frac{2}{5}}$
$⇒\frac{4}{1+{a}^{2}}=\frac{2}{5}$                                                           [Squaring both sides]

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