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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS3 Marks
Question
If a + ib $ = \frac{{{{(x + i)}^2}}}{{2{x^2} + 1}}$, prove that ${a^2} + {b^2} = \frac{{{{({x^2} + 1)}^2}}}{{{{(2{x^2} + 1)}^2}}}$.
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Answer

Here a + ib $ = \frac{{{{(x + i)}^2}}}{{2{x^2} + 1}} = \frac{{{x^2} + {i^2} + 2ix}}{{2{x^2} + 1}} = \frac{{{x^2} - 1}}{{2{x^2} + 1}} + i$$\frac{{2x}}{{2{x^2} + 1}}$
Comparing both sides, we have
$a = \frac{{{x^2} - 1}}{{2{x^2} + 1}}$ amd $b = \frac{{2x}}{{2{x^2} + 1}}$
$\therefore {a^2} + {b^2} = \left( {\frac{{{x^2} - 1}}{{2{x^2} + 1}}} \right)^2 + {\left( {\frac{{2x}}{{2{x^2} + 1}}} \right)^2}$
$ = \frac{{{{({x^2} - 1)}^2}}}{{{{(2{x^2} + 1)}^2}}} + \frac{{{{(2x)}^2}}}{{{{(2{x^2} + 1)}^2}}}$
$ = \frac{{{{({x^2} - 1)}^2} + {{(2x)}^2}}}{{{{(2{x^2} + 1)}^2}}}$$ = \frac{{{x^4} + 1 - 2{x^2} + 4{x^2}}}{{{{(2{x^2} + 1)}^2}}}$
$ = \frac{{{x^4} + 1 + 2{x^2}}}{{{{(2{x^2} + 1)}^2}}} = \frac{{{{({x^2} + 1)}^2}}}{{{{(2{x^2} + 1)}^2}}}$

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