\(x-iy=√\frac{a-ib}{c-id}\)
\(=\sqrt\frac{a-ib}{c-id}×\frac{c-ib}{c-id}\) \(\text{[ On\, multiplaying numerator\,and\,denominator \,by\,(c+id)]}\)
\(=\sqrt\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}\)
\(∴(x-iy)^2=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}\)
\(⇒x^2-y^2-2ixy=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}\)
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.