Question

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