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

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS4 Marks
Question
If $\text{x}-\text{iy}=\sqrt{\frac{\text{a}-\text{ib}}{\text{c}-\text{id}}}$ Prove that $(\text{x}^2+\text{y}^2)^2=\frac{\text{a}^2+\text{b}^2}{\text{c}^2+\text{d}^2}.$
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Answer

$\text{x}-\text{iy}=\sqrt{\frac{\text{a}-\text{ib}}{\text{c}-\text{id}}}$$=\sqrt{\frac{\text{a}-\text{ib}}{\text{c}-\text{id}}\times\frac{\text{c}+\text{id}}{\text{c}+\text{id}}}$ [On multiplying numerator and denominator by (c + id)]
$=\sqrt{\frac{(\text{ac}+\text{bd})+\text{i}(\text{ad}-\text{bc})}{\text{c}^2+\text{d}^2}}$
$\therefore\ (\text{x}-\text{iy})^2=\frac{(\text{ac}+\text{bd})+\text{i}(\text{ad}-\text{bc})}{\text{c}^2+\text{d}^2}$
$\Rightarrow\ \text{x}^2-\text{y}^2-2\text{ixy}=\frac{(\text{ac}+\text{bd})+\text{i}(\text{ad}-\text{bc})}{\text{c}^2+\text{d}^2}$
On comparing real and imaginary parts, we obtain
$\text{x}^2-\text{y}^2=\frac{\text{ac+bd}}{\text{c}^2+\text{d}^2},\ -2\text{xy}=\frac{\text{ad}-\text{bc}}{\text{c}^2+\text{d}^2}\ ....(1)$
$(\text{x}^2+\text{y}^2)^2=(\text{x}^2-\text{y}^2)^2+4\text{x}^2\text{y}^2$
$=\Big(\frac{\text{ac+bd}}{\text{c}^2+\text{d}^2}\Big)^2+\Big(\frac{\text{ad}-\text{bc}}{\text{c}^2+\text{d}^2}\Big)^2\ \ [\text{Using (1)}]$
$=\frac{\text{a}^2\text{c}^2+\text{b}^2\text{d}^2+2\text{acbd}+\text{a}^2\text{d}^2+\text{b}^2\text{c}^2-2\text{adbc}}{(\text{c}^2+\text{d}^2)^2}$
$=\frac{\text{a}^2\text{c}^2+\text{b}^2\text{d}^2+\text{a}^2\text{d}^2+\text{b}^2\text{c}^2}{(\text{c}^2+\text{d}^2)^2}$
$=\frac{\text{a}^2(\text{c}^2+\text{d}^2)+\text{b}^2(\text{c}^2+\text{d}^2)}{(\text{c}^2+\text{d}^2)^2}$
$=\frac{(\text{c}^2+\text{d}^2)(\text{a}^2+\text{b}^2)}{(\text{c}^2+\text{d}^2)^2}$
$=\frac{\text{a}^2+\text{b}^2}{\text{c}^2+\text{d}^2}$
Hence, proved.

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