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

\(- \frac{3}{5} - \frac{1}{5}{\rm{i}}\)

\(- \frac{1}{5} - \frac{3}{5}{\rm{i}}\)

\(\frac{1}{5} - \frac{3}{5}{\rm{i}}\)

\(- \frac{1}{5} + \frac{3}{5}{\rm{i}}\)

If \({\rm{a}} > 0{\rm{\;and\;z}} = \frac{{{{(1 + {\rm{i}})}^2}}}{{{\rm

1.	If \({\rm{a}} > 0{\rm{\;and\;z}} = \frac{{{{(1 + {\rm{i}})}^2}}}{{{\rm{a}} - {\rm{i}}}}\), has magnitude \(\sqrt {\frac{2}{5}} ,{\rm{\;then\;\bar z}}\) is equal to:
A.	\(- \frac{1}{5} - \frac{3}{5}{\rm{i}}\)
B.	\(- \frac{3}{5} - \frac{1}{5}{\rm{i}}\)
C.	\(\frac{1}{5} - \frac{3}{5}{\rm{i}}\)
D.	\(- \frac{1}{5} + \frac{3}{5}{\rm{i}}\)
Answer» B. \(- \frac{3}{5} - \frac{1}{5}{\rm{i}}\)

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

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