If a > 0 and z = ( 1 + i ) ^ 2 a - i , has magnitude (2)/(5) , then...

$Z = \dfrac{{\left(1 + i\right)}^{2}}{a - i} = \dfrac{2i}{a - i}$
\left|z\right| = \sqrt{\dfrac{2}{5}} ⇒ \left|\dfrac{2i}{a - i}\right| = \dfrac{2}{\sqrt{{a}^{2} + 1}} = \sqrt{\dfrac{2}{5}}$<br />$⇒ a = 3\left(∵ a > 0\right)$<br />So,$\overset{-}{z} = \overset{-}{\left(\dfrac{2i}{3 - i}\right)} = \dfrac{- 2i}{3 + i} = \dfrac{- 2i}{3 + i} \times \dfrac{3 - i}{3 - i}$<br />$= \dfrac{- 6i - 2}{{3}^{2} + 1} = - \dfrac{1}{5} - \dfrac{3}{5}i