If $(x + iy) = \sqrt{\frac{1 + 2i}{3 + 4i}}$, then $\left( x^{2} + y^{2} \right)^{2}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $x + iy = \left( \frac{1 + 2i}{3 + 4i} \right)^{\frac{1}{2}}$
$\Rightarrow \ \ \ \ \ (x + iy)^{2} = \frac{1 + 2i}{3 + 4i}$
Taking modulus from both sides we get
$|x + iy|^{2} = \left| \frac{1 + 2i}{3 + 4i} \right|$
$\Rightarrow \ \ \ \ \ \ x^{2} + y^{2} = \sqrt{\frac{1 + 4}{9 + 16}}$
$\Rightarrow \ \ \ \ \ \ \left( x^{2} + y^{2} \right)^{2} = \frac{5}{25} = \frac{1}{5}$

Q. If $(x + iy) = \sqrt{\frac{1 + 2i}{3 + 4i}}$, then $\left( x^{2} + y^{2} \right)^{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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