Question

If $x-i y=\sqrt{\frac{a-i b}{c-i d}}$, then $\left(x^2+y^2\right)^2$ is equal to

• A. $\frac{a^2+b^2}{c^2+d^2}$
• B. $\frac{c^2+d^2}{a^2+b^2}$
• C. $\frac{a^2-b^2}{c^2-d^2}$
• D. $\frac{c^2-d^2}{a^2-b^2}$

Answer 1

Answer: (A)

Given, $x-i y=\sqrt{\frac{a-i b}{c-i d}}$
$\Rightarrow x+i(-y)=\left(\frac{a-i b}{c-i d}\right)^{1 / 2}$
Taking modulus both sides, we get
$|x+i(-y)| =\left|\left(\frac{a-i b}{c-i d}\right)^{\frac{1}{2}}\right| $
$\Rightarrow \sqrt{x^2+(-y)^2} =\left|\frac{a-i b}{c-i d}\right|^{\frac{1}{2}} $ $\left(\because|x+i y|=\sqrt{x^2+y^2} and \left|z^n\right|=|z|^n\right)$
$\Rightarrow \sqrt{x^2+y^2} =\left|\frac{a-i b}{c-i d}\right|^{\frac{1}{2}}$
Squaring both sides, we get
$x^2+y^2 =\left|\frac{a-i b}{c-i d}\right| $
$\Rightarrow x^2+y^2 =\frac{|a-i b|}{|c-i d|} \left(\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right) $
$\Rightarrow \left(x^2+y^2\right) =\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\left(\because|x-i y|=\sqrt{x^2+y^2}\right)$
Squaring both sides, we get
$\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}$