If $x+i y=\frac{(3+2 i)(4-7 i)(12+13 i)}{(13-12 i)(2-3 i)(11+3 i)}$, then $x^2+y^2=$

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Aicharya · Accepted Answer

Given, $\begin{aligned} z & =x+i y=\frac{(3+2 i)(4-7 i)(12+13 i)}{(13-12 i)(2-3 i)(11+3 i)} \ |z| & =\frac{|3+2 i| \cdot|4-7 i| \cdot|12+13 i|}{|13-12 i| \cdot|2-3 i| \cdot|11+3 i|} \ & =\frac{\left(\sqrt{3^2+2^2}\right)\left(\sqrt{4^2+7^2}\right)\left(\sqrt{\left(12^2+13^2\right.}\right)}{\left(\sqrt{13^2+12^2}\right) \cdot\left(\sqrt{2^2+3^2}\right)\left(\sqrt{11^2+3^2}\right)} \end{aligned}$ $\begin{aligned} = & \frac{\sqrt{4^2+7^2}}{\sqrt{11^2+3^2}}=\frac{\sqrt{65}}{\sqrt{130}}=\frac{1}{\sqrt{2}} \ \Rightarrow \quad \sqrt{x^2+y^2} & =\frac{1}{\sqrt{2}} \text { or } x^2+y^2=\frac{1}{2} \end{aligned}$

If \(x+i y=\frac{(3+2 i)(4-7 i)(12+13 i)}{(13-12 i)(2-3 i)(11+3 i)}\), then \(x^2+y^2=\)

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