If a + ib \( \dfrac{(x+i)^2}{2x^2+1}\) Prove that \( a^2+b^2= \dfrac{(x^2+1)^2}{(2x^2+1)^2}\)

Asked by Pragya Singh | 2 years ago |  85

1 Answer

Solution :-

Expression

\(a + ib = \dfrac{(x+i)^2}{2x^2+1}\)

\( \dfrac{x^2+i^2+2xi}{2x^2+1}\)

\( \dfrac{x^2-1+i2x}{2x^2+1}\)

\( \dfrac{x^2-1}{2x^2+1}+i (\dfrac{2x}{2x^2+1})\)

On comparing

\( a^2+b^2= \dfrac{(x^2+1)^2}{(2x^2+1)^2}+(\dfrac{2x}{2x^2+1})^2\)

\( \dfrac{x^4+1-2x^2+4x^2}{(2x+1)^2}\)

\( \dfrac{x^4+1+2x^2}{(2x^2+1)^2}\)

\( \dfrac{(x^2+1)^2}{(2x^2+1)^2}\)

\( a + ib = \dfrac{(x^2+1)^2}{(2x^2+1)^2}\)

Hence, proved

Answered by Abhisek | 2 years ago

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