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If $\;x-iy=\sqrt{\large\frac{a-ib}{c-id}}\;$ what is the value of $\;(x^{2}+y^{2})^{2}$?

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

1 Answer

Answer : $\;\large\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$
Explanation :
$\;x-iy=\sqrt{\large\frac{a-ib}{c-id}}\; = \sqrt{\large\frac{a-ib}{c-id} \times \large\frac{c+id}{c+id}} \qquad$ [on multiplying numerator and denominator by $\;c+id$ ]
$= \sqrt{\large\frac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}}$
$(x-iy)^{2} = \large\frac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}$
$x^{2}-y^{2} -2ixy = \large\frac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}$
On comparing real and imaginary parts , we obtain
$x^{2} -y^{2} = \large\frac{ac+bd}{c^{2} +d^{2}}\;\; ; -2xy = \large\frac{ad-bc}{c^{2}+d^{2}}$----(1)
$(x^{2} +y^{2})^{2} = (x^{2} -y^{2})^{2}+4x^{2}y^{2}$
$= (\large\frac{ac+bd}{c^{2} +d^{2}})^{2}+ ( \large\frac{ad-bc}{c^{2}+d^{2}})^{2} \qquad$ [using (1)]
$= \large\frac{a^2c^{2}+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^{2}+d^{2})^{2}}$
$= \large\frac{a^2c^{2}+b^2d^2+a^2d^2+b^2c^2}{(c^{2}+d^{2})^{2}}$
$= \large\frac{a^{2}(c^2+d^2)+b^{2}(c^2+d^2)}{(c^{2}+d^{2})^{2}}$
$=\large\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$
Hence proved .
answered Apr 11, 2014 by yamini.v
 
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