We have, $\quad x-i y=\sqrt{\frac{a-i b}{c-i d}} \ldots$ (i)
Replacing $i$ by $-i$ both side, we have
$x+i y=\sqrt{\frac{a+i b}{c+i d}} \quad \ldots \text { (ii) }$
Multiply eqns. (i) and (ii)
$\begin{aligned}
&(x-i y)(x+i y)=\sqrt{\frac{a-i b}{c-i d}} \times \sqrt{\frac{a+i b}{c+i d}} \\
&x^2-i^2 y^2=\sqrt{\frac{a^2-i^2 b^2}{c^2-i^2 d^2}} \\
&\Rightarrow x^2+y^2=\sqrt{\frac{a^2+b^2}{c^2+d^2}}
\end{aligned}$
$\Rightarrow\left(x^2+y^2\right)^2=\left(\sqrt{\frac{a^2+b^2}{c^2+d^2}}\right)^2$
(squaring on both side)
$\therefore\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}, \quad$ Hence proved