Question

$if\, x-iy=√\frac{a-ib}{c-id}$ prove that $(x^2y^2)^2=\frac{a^2+b^2}{c^2+d^2}.$

Answer 1

$x-iy=√\frac{a-ib}{c-id}$

$=\sqrt\frac{a-ib}{c-id}×\frac{c-ib}{c-id}$ $[ On\, multiplaying numerator\,and\,denominator \,by(c+id)]$

$=\sqrt\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$

∴$(x-iy)^2=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$

$⇒x^2-y^2-2ixy=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$

on comparing real and imaginary parts, we obtain

$x^2=y^2=\frac{ac+bd}{c^2+d^2},-2xy=\frac{ad-bc}{c^2+d^2}$ $(1)$

$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2$

$=(\frac{ac+bd}{c^2+d^2})+(\frac{ad-bc}{c^2+d^2})$ $[Using\,(1)]$

$=\frac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)}$

$=\frac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)}$

$=\frac{a^2(c^2+d^2+b)+b^2(c^2+d^2)}{(c^2+d^2)^2}$

$\frac{(c^2+d^2+b)(c^2+d^2)}{(c^2+d^2)^2}$

$=\frac{a^2+b^2}{c^2+b^2}$

Hence, proved.