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If $\;a+ib =\large\frac{(x+i)^{2}}{2x^{2}+1}\;$ , find $\;a^{2} +b^{2}$

$(a)\;\large\frac{(x^{2}+1)^{2}}{(2x+1)^{2}}\qquad(b)\;\large\frac{(x+1)^{2}}{(2x+1)^{2}}\qquad(c)\;\large\frac{(x^{2}+1)^{2}}{(2x+1)}\qquad(d)\;\large\frac{(x^{2}+1)}{(2x+1)}$

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Answer : $\;\large\frac{(x^{2}+1)^{2}}{(2x+1)^{2}}$
Explanation :
$a+ib =\large\frac{(x+i)^{2}}{2x^{2}+1}$
$=\large\frac{x^2+i^{2}+2 xi}{2x^{2} +1}$
$=\large\frac{x^2-1+2 xi}{2x^{2} +1}$
$= \large\frac{x^{2}-1}{2x^{2}+1} + i \large\frac{2x}{2x^{2}+1}$
On comparing real and imaginary parts , we obtain
$a= \large\frac{x^{2}-1}{2x^{2}+1} \;and \; b= \large\frac{2x}{2x^{2}+1} $
$a^{2}+b^{2} = (\large\frac{x^{2}-1}{2x^{2}+1})^{2} + (\large\frac{2x}{2x^{2}+1})^{2}$
$ = \large\frac{x^{4}+1-2\;.x^{2}\;.1+4x^{2}}{(2x^{2}+1)^{2}}$
$ = \large\frac{x^{4}+1+2x^{2}}{(2x^{2}+1)^{2}}$
$=\;\large\frac{(x^{2}+1)^{2}}{(2x+1)^{2}}$
Therefore , $\;a^{2} +b^{2} = \large\frac{(x^{2}+1)^{2}}{(2x+1)^{2}}$
Hence proved .
answered Apr 11, 2014 by yamini.v
 

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