The modulus of the complex number frac{(1+i)^ | vedclass

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(C) Let $z = \frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$.Using the property of modulus $|\frac{z_1 z_2}{z_3 z_4}| = \frac{|z_1| |z_2|}{|z_3| |z_4|}$,we hav

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$\frac{\sqrt{2}}{4}$

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(C) Let $z = \frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$.Using the property of modulus $|\frac{z_1 z_2}{z_3 z_4}| = \frac{|z_1| |z_2|}{|z_3| |z_4|}$,we have:$|z| = \frac{|1+i|^2 |1+3i|}{|2-6i| |2-2i|}$$|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,so $|1+i|^2 = 2$.$|1+3i| = \sqrt{1^2+3^2} = \sqrt{10}$.$|2-6i| = \sqrt{2^2+(-6)^2} = \sqrt{4+36} = \sqrt{40} = 2\sqrt{10}$.$|2-2i| = \sqrt{2^2+(-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.Substituting these values:$|z| = \frac{2 	imes \sqrt{10}}{2\sqrt{10} 	imes 2\sqrt{2}} = \frac{2}{4\sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$.

The modulus of the complex number $\frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$ is

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