If x+iy=sqrt{frac{3+i}{1+3i}},then left(x^2+y | vedclass

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(B) Given $x+iy=\sqrt{\frac{3+i}{1+3i}}$.Taking the modulus on both sides,we have $|x+iy| = \left|\sqrt{\frac{3+i}{1+3i}}\right|$.$|x+iy| = \sqrt{\lef

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(B) Given $x+iy=\sqrt{\frac{3+i}{1+3i}}$.Taking the modulus on both sides,we have $|x+iy| = \left|\sqrt{\frac{3+i}{1+3i}}\right|$.$|x+iy| = \sqrt{\left|\frac{3+i}{1+3i}\right|}$.Since $|z_1/z_2| = |z_1|/|z_2|$,we get $|x+iy| = \sqrt{\frac{|3+i|}{|1+3i|}} = \sqrt{\frac{\sqrt{3^2+1^2}}{\sqrt{1^2+3^2}}} = \sqrt{\frac{\sqrt{10}}{\sqrt{10}}} = \sqrt{1} = 1$.We know that $|x+iy| = \sqrt{x^2+y^2}$,so $\sqrt{x^2+y^2} = 1$.Squaring both sides,$x^2+y^2 = 1$.Therefore,$\left(x^2+y^2\right)^2 = 1^2 = 1$.

If $x+iy=\sqrt{\frac{3+i}{1+3i}}$,then $\left(x^2+y^2\right)^2=$

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