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

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(B) Given $z = \frac{(1+i)^2}{a+i} = \frac{1+i^2+2i}{a+i} = \frac{2i}{a+i}$.Magnitude $|z| = \left| \frac{2i}{a+i} \right| = \frac{|2i|}{|a+i|} = \fra

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

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(B) Given $z = \frac{(1+i)^2}{a+i} = \frac{1+i^2+2i}{a+i} = \frac{2i}{a+i}$.Magnitude $|z| = \left| \frac{2i}{a+i} \right| = \frac{|2i|}{|a+i|} = \frac{2}{\sqrt{a^2+1}}$.Given $|z| = \frac{2}{\sqrt{5}}$,so $\frac{2}{\sqrt{a^2+1}} = \frac{2}{\sqrt{5}}$ $\Rightarrow a^2+1 = 5$ $\Rightarrow a^2 = 4$.Since $a > 0$,we have $a = 2$.Substituting $a = 2$ into $z$,we get $z = \frac{2i}{2+i} = \frac{2i(2-i)}{(2+i)(2-i)} = \frac{4i - 2i^2}{4+1} = \frac{2+4i}{5} = \frac{2}{5} + \frac{4}{5}i$.Therefore,$\bar{z} = \frac{2}{5} - \frac{4}{5}i$.

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

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