Question

Let $(x+iy)=\sqrt{\frac{a+ib}{c+id}}$, then prove that $(x^2+y^2{)}^2=\frac{a^2+b^2}{c^2+d^2}$.

Answer 1

Given,

$(x+iy)=\sqrt{\frac{a+ib}{c+id}}$

Squaring both sides we get,

or, $(x+iy{)}^2=\frac{a+ib}{c+id}$

Now taking modulus both sides we get,

or, $|(x+iy{)}^2|=\left|\frac{a+ib}{c+id}\right|$

or, $|(x+iy){|}^2=\left|\frac{a+ib}{c+id}\right|$ [ Since $|z^2|=|z{|}^2$ for a complex number $z$].

or, $(x^2+y^2)=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}$

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