Question

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

Answer 1

$x-iy=\sqrt{\frac{a-ib}{c-id}}$

$=\sqrt{\frac{a-ib}{c-id}\times \frac{c+id}{c+id}}$

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

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

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

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)

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

$={\left(\frac{ac+bd}{c^2+d^2}\right)}^2+{\left(\frac{ad-bc}{c^2+d^2}\right)}^2\left[using1\right]$

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

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

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

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

Hence, proved