If $ x+iy = \sqrt{\frac{a+ib}{c+id} } ,$Show that$ (x^2+y^2)^2 = \frac {a^2+b^2}{c^2+d^2} $ $ x+iy = \sqrt{\frac{a+ib}{c+id} } , $Show that $ ({x^2+y^2})^2 = \frac {a^2+b^2}{c^2+d^2} $
How do i do this ?I tried squaring both sides but x+iy expansion becomes difficult when squaring the next time .I also tried conjugating the R.H.S
 A: $$x^2+y^2=|x+iy|^2=\left|\sqrt{\frac{a+ib}{c+id}}\right|^2=\left|\frac{a+ib}{c+id}\right|$$
Therefore,
$$(x^2+y^2)^2=\frac{|a+ib|^2}{|c+id|^2}=\frac{a^2+b^2}{c^2+d^2}$$
A: 
Notice, when $q\space\wedge\space s\space\wedge\space z\in\mathbb{C}$, so we set $q=x+yi,s=a+bi,z=c+di$:

*

*$$\left(x^2+y^2\right)^2=\left(\Re^2[q]+\Im^2[q]\right)^2=|q|^4$$

*$$a^2+b^2=\Re^2[s]+\Im^2[s]=|s|^2$$

*$$c^2+d^2=\Re^2[z]+\Im^2[z]=|z|^2$$

So:
$$x+yi=\sqrt{\frac{a+bi}{c+di}}\Longleftrightarrow q=\sqrt{\frac{s}{z}}$$
Now, we can find the absolute value:
$$\left|q\right|=\left|\sqrt{\frac{s}{z}}\right|\Longleftrightarrow$$
$$\left|q\right|=\left|\left(\frac{s}{z}\right)^{\frac{1}{2}}\right|\Longleftrightarrow$$
$$\left|q\right|=\left|\frac{s}{z}\right|^{\frac{1}{2}}\Longleftrightarrow$$
$$\left|q\right|=\left(\frac{\left|s\right|}{\left|z\right|}\right)^{\frac{1}{2}}\Longleftrightarrow$$
$$\left|q\right|^4=\left(\frac{\left|s\right|}{\left|z\right|}\right)^2\Longleftrightarrow$$
$$\left|q\right|^4=\frac{\left|s\right|^2}{\left|z\right|^2}$$
