If $a, b, c$ and $d \in R$ such that $a^2+b^2=4$ and $c^2+d^2=2$ and if $(a+i b)^2=(c+i d)^2(x+i y)$, then $x^2+y^2$ is equal to

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Aicharya · Accepted Answer

Given, $(a+i b)^2=(c+i d)^2(x+i y)$ $\begin{array}{lc}\Rightarrow & |a+i b|^2=|c+i d|^2|x+i y| \ \Rightarrow & a^2+b^2=\left(c^2+d^2ight)\left(\sqrt{x^2+y^2}ight) \ \Rightarrow & 4=2\left(\sqrt{x^2+y^2}ight) \ & \quad\left(\because a^2+b^2=4 	ext { and } c^2+d^2=2 	ext { given }ight) \ \Rightarrow & \sqrt{x^2+y^2}=2 \ \Rightarrow & x^2+y^2=4\end{array}$

If $a, b, c$ and $d \in R$ such that $a^2+b^2=4$ and $c^2+d^2=2$ and if $(a+i b)^2=(c+i d)^2(x+i y)$, then…

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