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The modulus of $\frac{(1+i \sqrt{3})(2+2 i)}{(\sqrt{3}-i)}$ is
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$2 \sqrt{2}$
$\begin{aligned} & \frac{(1+i \sqrt{3})(2+2 i)}{\sqrt{3}-i}=\frac{2+2 \sqrt{3} i+2 i-2 \sqrt{3}}{\sqrt{3}-i} \\ & =\frac{(2-2 \sqrt{3})+(2 \sqrt{3}+2) i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} \\ & =\frac{2 \sqrt{3}-6+2 i-2 \sqrt{3} i+6 i+2 \sqrt{3} i-2 \sqrt{3}-2}{3+1} \\ & =\frac{8 i-8}{4}=-2+2 i \\ & \therefore \text { Modulus }=\sqrt{(-2)^2+(2)^2}=2 \sqrt{2}\end{aligned}$
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