Question

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

• A. $\frac{{(p^2+1)}^2}{4p^2-1}$
• B. $\frac{{(p^2-1)}^2}{4p^2-1}$
• C. $\frac{{(p^2-1)}^2}{4p^2+1}$
• D. $\frac{{(p^2+1)}^2}{4p^2+1}$

Answer 1

The correct option is D $\frac{{(p^2+1)}^2}{4p^2+1}$

$\frac{(p+i{)}^2}{2p-i}=\mu +i\lambda$

$\frac{(|p+i|{)}^2}{|2p-i|}=|\mu +i\lambda |$

$\frac{(\sqrt{p^2+1}{)}^2}{\sqrt{4p^2+1}}=\sqrt{{\mu }^2+{\lambda }^2}$

Squaring on both sides

${\mu }^2+{\lambda }^2=\frac{(p^2+1{)}^2}{4p^2+1}$