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  • If a+ib= (x+i)^2/2x^2+1, prove that a^2+b^2=(x^2+1)^2/(2x^2+1)^2

Q : 11      If   \small a+ib=\frac{(x+i)^2}{2x^2+1} , prove that  \small a^2+b^2=\frac{(x^2+1)^2}{(2x^2+1)^2}.

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best_answer

It is given that
\small a+ib=\frac{(x+i)^2}{2x^2+1}
Now, we will reduce it into

\small a+ib=\frac{(x+i)^2}{2x^2+1} = \frac{x^2+i^2+2xi}{2x^2+1}=\frac{x^2-1+2xi}{2x^2+1}=\frac{x^2-1}{2x^2+1}+i\frac{2x}{2x^2+1}
On comparing real and imaginary part. we will get
a=\frac{x^2-1}{2x^2+1}\ and \ b=\frac{2x}{2x^2+1}
Now,
a^2+b^2=\left ( \frac{x^2-1}{2x^2+1} \right )^2+\left ( \frac{2x}{2x^2+1} \right )^2
                = \frac{x^4+1-2x^2+4x^2}{(2x^2+1)^2}
                = \frac{x^4+1+2x^2}{(2x^2+1)^2}
                = \frac{(x^2+1)^2}{(2x^2+1)^2}
Hence proved 

Posted by

Gautam harsolia

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