The value of left| frac{1+i sqrt{3}}{left(1+f | vedclass

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(D) Let $z = \frac{1+i \sqrt{3}}{\left(1+\frac{1}{i+1}\right)^{2}}$.First,simplify the denominator: $1+\frac{1}{i+1} = \frac{i+1+1}{i+1} = \frac{i+2}{

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$\frac{4}{5}$

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(D) Let $z = \frac{1+i \sqrt{3}}{\left(1+\frac{1}{i+1}\right)^{2}}$.First,simplify the denominator: $1+\frac{1}{i+1} = \frac{i+1+1}{i+1} = \frac{i+2}{i+1}$.Then,$z = \frac{1+i \sqrt{3}}{\left(\frac{i+2}{i+1}\right)^{2}} = \frac{(1+i \sqrt{3})(i+1)^{2}}{(i+2)^{2}}$.Since $(i+1)^{2} = i^{2}+1+2i = 2i$ and $(i+2)^{2} = i^{2}+4+4i = 3+4i$,we have $z = \frac{(1+i \sqrt{3})(2i)}{3+4i} = \frac{2i-2\sqrt{3}}{3+4i}$.Now,find the modulus: $|z| = \frac{|2i-2\sqrt{3}|}{|3+4i|} = \frac{\sqrt{(-2\sqrt{3})^{2} + 2^{2}}}{\sqrt{3^{2}+4^{2}}}$.$|z| = \frac{\sqrt{12+4}}{\sqrt{9+16}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}$.

The value of $\left| \frac{1+i \sqrt{3}}{\left(1+\frac{1}{i+1}\right)^{2}} \right|$ is

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