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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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The correct answer is:
$4$
Given, $(a+i b)^2=(c+i d)^2(x+i y)$
$\begin{array}{lc}\Rightarrow & |a+i b|^2=|c+i d|^2|x+i y| \\ \Rightarrow & a^2+b^2=\left(c^2+d^2\right)\left(\sqrt{x^2+y^2}\right) \\ \Rightarrow & 4=2\left(\sqrt{x^2+y^2}\right) \\ & \quad\left(\because a^2+b^2=4 \text { and } c^2+d^2=2 \text { given }\right) \\ \Rightarrow & \sqrt{x^2+y^2}=2 \\ \Rightarrow & x^2+y^2=4\end{array}$
$\begin{array}{lc}\Rightarrow & |a+i b|^2=|c+i d|^2|x+i y| \\ \Rightarrow & a^2+b^2=\left(c^2+d^2\right)\left(\sqrt{x^2+y^2}\right) \\ \Rightarrow & 4=2\left(\sqrt{x^2+y^2}\right) \\ & \quad\left(\because a^2+b^2=4 \text { and } c^2+d^2=2 \text { given }\right) \\ \Rightarrow & \sqrt{x^2+y^2}=2 \\ \Rightarrow & x^2+y^2=4\end{array}$
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