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

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(B) Given $z = \frac{(1 + i)^2}{a - i} = \frac{2i}{a - i}$.Multiplying numerator and denominator by the conjugate $(a + i)$,we get $z = \frac{2i(a + i

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

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(B) Given $z = \frac{(1 + i)^2}{a - i} = \frac{2i}{a - i}$.Multiplying numerator and denominator by the conjugate $(a + i)$,we get $z = \frac{2i(a + i)}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}$.The magnitude is $|z| = \frac{|2i|}{|a - i|} = \frac{2}{\sqrt{a^2 + 1}}$.Given $|z| = \sqrt{\frac{2}{5}}$,so $\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}} \implies \frac{4}{a^2 + 1} = \frac{2}{5} \implies a^2 + 1 = 10 \implies a^2 = 9$.Since $a > 0$,we have $a = 3$.Substituting $a = 3$ into $z$,we get $z = \frac{2i(3 + i)}{3^2 + 1} = \frac{6i - 2}{10} = -\frac{1}{5} + \frac{3}{5}i$.The conjugate $\bar{z}$ is $-\frac{1}{5} - \frac{3}{5}i$.

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

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