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

(N/A) Given $x-iy = \sqrt{\frac{a-ib}{c-id}}$.
Taking the modulus on both sides,we have $|x-iy| = \left|\sqrt{\frac{a-ib}{c-id}}\right|$.
Since $|z_1/z_2| = |z_1|/|z_2|$ and $|\sqrt{z}| = \sqrt{|z|}$,we get $|x-iy| = \sqrt{\frac{|a-ib|}{|c-id|}}$.
We know that $|x-iy| = \sqrt{x^2+y^2}$,$|a-ib| = \sqrt{a^2+b^2}$,and $|c-id| = \sqrt{c^2+d^2}$.
Substituting these values,we get $\sqrt{x^2+y^2} = \sqrt{\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}}$.
Squaring both sides,we get $x^2+y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}}$.
Squaring again,we obtain $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$.
Hence,proved.