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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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 $\left(x^2+y^2\right)^2=$

Similar Questions

If $z = \frac{4}{1-i}$,then $\bar{z}$ is (where $\bar{z}$ is the complex conjugate of $z$).

Define $f: \mathbb{C} \rightarrow \mathbb{R}$ by $f(z) = |z|, \forall z \in \mathbb{C}$. Then,which of the following is false?

If $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$,then $a^2 + b^2 = \dots$

The modulus of the square root of the conjugate of $-7+24 \sqrt{-1}$ is ....

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module