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

\(x-iy=√\frac{a-ib}{c-id}\) 

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

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

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

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

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.