If a 0 and z = frac{(1+i)^2}{a-i},where i = | vedclass

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(A) Given $z = \frac{(1+i)^2}{a-i} = \frac{1 + 2i + i^2}{a-i} = \frac{2i}{a-i}$.Taking the modulus on both sides: $|z| = \left| \frac{2i}{a-i} \right|

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$-\frac{2}{5} - \frac{4}{5}i$

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(A) Given $z = \frac{(1+i)^2}{a-i} = \frac{1 + 2i + i^2}{a-i} = \frac{2i}{a-i}$.Taking the modulus on both sides: $|z| = \left| \frac{2i}{a-i} \right| = \frac{|2i|}{|a-i|} = \frac{2}{\sqrt{a^2 + 1}}$.Given $|z| = \frac{2}{\sqrt{5}}$,we have $\frac{2}{\sqrt{a^2 + 1}} = \frac{2}{\sqrt{5}}$.Thus,$a^2 + 1 = 5$,which implies $a^2 = 4$. Since $a > 0$,we get $a = 2$.Substituting $a = 2$ into $z$: $z = \frac{2i}{2-i} = \frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i + 2i^2}{4+1} = \frac{-2 + 4i}{5} = -\frac{2}{5} + \frac{4}{5}i$.Therefore,the conjugate $\bar{z} = -\frac{2}{5} - \frac{4}{5}i$.

If $a > 0$ and $z = \frac{(1+i)^2}{a-i}$,where $i = \sqrt{-1}$,has a magnitude of $\frac{2}{\sqrt{5}}$,then $\bar{z}$ is

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