If x - iy = \sqrt{{a-ib}{c-id}} prove that ( x^2+y^2)^2={a^2+b^2}{c^2+d^2}

Expression

x - iy = \( \sqrt{\dfrac{a-ib}{c-id}}\)

\( \sqrt{\dfrac{a-ib}{c-id}}\times \dfrac{c+id}{c+id}\)

[On multiplying numerator and denominator by (c + id)]

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

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

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

On comparing

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

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

\(( \dfrac{ac+bd}{c^2+d^2})^2+\dfrac{ad-bc}{c^2+d^2}\)

\( \dfrac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}\)

\( \dfrac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}\)

\( \dfrac{a^2(c^2+d^2)b^2(c^2+d^2)}{(c^2+d^2)^2}\)

\( \dfrac{(c^2+d^2)(a^2+b^2)}{(c^2+d^2)^2}\)

\( \dfrac{a^2+b^2}{c^2+d^2}\)

Hence, proved