$x-i y=\sqrt{\frac{a-i b}{c-i d}}$

$=\sqrt{\frac{a-i b}{c-i d} \times \frac{c+i d}{c+i d}}$   [On multiplying numerator and denominator by $(c+i d)]$

$=\sqrt{\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}}$

$\therefore(x-i y)^{2}=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}$

$\Rightarrow x^{2}-y^{2}-2 i x y=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}$

On comparing real and imaginary parts, we obtain

$x^{2}-y^{2}=\frac{a c+b d}{c^{2}+d^{2}},-2 x y=\frac{a d-b c}{c^{2}+d^{2}}$…..$(1)$

$\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$

$\left.=\left(\frac{a c+b d}{c^{2}+d^{2}}\right)^{2}+\left(\frac{a d-b c}{c^{2}+d^{2}}\right) \quad \text { [Using }(1)\right]$

$=\frac{a^{2} c^{2}+b^{2} d^{2}+2 a c b d+a^{2} d^{2}+b^{2} c^{2}-2 a d b c}{\left(c^{2}+d^{2}\right)^{2}}$

$=\frac{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}{\left(c^{2}+d^{2}\right)^{2}}$

$=\frac{a^{2}\left(c^{2}+d^{2}\right)+b^{2}\left(c^{2}+d^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}$

$=\frac{\left(c^{2}+d^{2}\right)\left(a^{2}+b^{2}\right)}{\left(c^{2}+d^{2}\right)^{2}}$

$=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$

Hence, proved.