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

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(B) Given $z = \frac{(1 + i)^2}{a - i}$.Since $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$,we have $z = \frac{2i}{a - i}$.Multiply numerator and denom

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

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(B) Given $z = \frac{(1 + i)^2}{a - i}$.Since $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$,we have $z = \frac{2i}{a - i}$.Multiply numerator and denominator by the conjugate $(a + i)$:$z = \frac{2i(a + i)}{a^2 + 1} = \frac{2ai - 2}{a^2 + 1} = \frac{-2 + 2ai}{a^2 + 1}$.The magnitude is $|z| = \sqrt{\frac{(-2)^2 + (2a)^2}{(a^2 + 1)^2}} = \sqrt{\frac{4(1 + a^2)}{(a^2 + 1)^2}} = \sqrt{\frac{4}{a^2 + 1}} = \frac{2}{\sqrt{a^2 + 1}}$.Given $|z| = \sqrt{\frac{2}{5}}$,so $\frac{2}{\sqrt{a^2 + 1}} = \sqrt{\frac{2}{5}}$.Squaring both sides: $\frac{4}{a^2 + 1} = \frac{2}{5}$ $\Rightarrow 20 = 2(a^2 + 1)$ $\Rightarrow 10 = a^2 + 1$ $\Rightarrow a^2 = 9$.Since $a > 0$,$a = 3$.Substituting $a = 3$ into $z$: $z = \frac{-2 + 2(3)i}{3^2 + 1} = \frac{-2 + 6i}{10} = -\frac{1}{5} + \frac{3}{5}i$.Therefore,the conjugate $\overline{z} = -\frac{1}{5} - \frac{3}{5}i$.

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

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