If $a>0$ and $z = \frac{(1+i)^2}{a - i}$ ($i = \sqrt{-1}$) has magnitude $\frac{2}{\sqrt{5}}$, then $\overline{z}$ is
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Algebra Tip:Memorize that $(1+i)^2 = 2i$ and $(1-i)^2 = -2i$. This saves valuable time in complex number calculations.Properties of modulus can also be used: $|z| = \frac{|(1+i)^2|}{|a-i|} = \frac{(\sqrt{2})^2}{\sqrt{a^2+1}} = \frac{2}{\sqrt{a^2+1}}$. This skips the rationalization step entirely!