If \[\frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda ,\]then \[{{\mu }^{2}}+{{\lambda }^{2}}\] is equal to

\[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}-1}\]

\[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}-1}\]

\[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}+1}\]

\[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\]

If \[\frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda ,\]then \[{{\mu }^{2}}

1.	If \[\frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda ,\]then \[{{\mu }^{2}}+{{\lambda }^{2}}\] is equal to
A.	\[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}-1}\]
B.	\[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}-1}\]
C.	\[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}+1}\]
D.	\[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\]
Answer» E.

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If \[\frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda ,\]then \[{{\mu }^{2}}+{{\lambda }^{2}}\] is equal to

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