If (p + i) ^2 2p - i = + i ,then ^2 + ^2 is equal to - Vidyadip

MCQ

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
$\frac{{{{({p^2} - 1)}^2}}}{{4{p^2} + 1}}$
✓
$\frac{{{{({p^2} + 1)}^2}}}{{4{p^2} + 1}}$

✓

Answer

Correct option: D.

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

d
(d) $\mu + i\lambda = \frac{{{{(p + i)}^2}}}{{2p - i}} = \frac{{({p^2} - 1 + 2pi)(2p + i)}}{{(2p - i)(2p + i)}}$
$ = \frac{{2p({p^2} - 2) + i(5{p^2} - 1)}}{{4{p^2} + 1}}$
${\mu ^2} + {\lambda ^2} = \frac{{4{p^2}{{({p^2} - 2)}^2} + {{(5{p^2} - 1)}^2}}}{{{{(4{p^2} + 1)}^2}}}$
$ = \frac{{4{p^6} + 6{p^2} + 9{p^4} + 1}}{{{{(4{p^2} + 1)}^2}}}$
$ = \,\,\frac{{{p^4}(4{p^2} + 1) + 2{p^2}(4{p^2} + 1) + (4{p^2} + 1)}}{{{{(4{p^2} + 1)}^2}}}$
$ = \frac{{{p^4} + 2{p^2} + 1}}{{4{p^2} + 1}} = \frac{{{{({p^2} + 1)}^2}}}{{4{p^2} + 1}}$.

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