$$ z = \frac{1}{1 + i \cot x} $$ — \begin{align} z & = \frac{1}{1 + i \cot x} \\[5pt] & = \frac{1}{1 + i \cdot \displaystyle \frac{\cos x}{\sin x}} \\[5pt] & = \frac{\sin x}{\sin x + i \cos x} \\[5pt] & = \frac{\sin x}{ \displaystyle \cos \left( \frac{\pi}{2} - x \right) + i \sin \left( \frac{\pi}{2} - x \right)} \\[5pt] & = \frac{\sin x}{ \displaystyle \mathrm{cis} \left( \frac{\pi}{2} - x \right) } \end{align} Assim, $$ \quad z = \frac{\sin x}{ \displaystyle \mathrm{cis} \left( \frac{\pi}{2} - x\right) } \ \text{.}$$ Extraindo o módulo de $ z $: $$ |z| = \left | \frac{\sin x}{\displaystyle \mathrm{cis} \left( \frac{\pi}{2} -x \right) } \right| = \frac{| \sin x |}{\underbrace{ \left| \displaystyle \mathrm{cis} \left( \frac{\pi}{2} - x\right) \right|}_{1} } = | \sin x | $$ Alternativa correta: $\boxed{\mathrm{E}} $ $$ \boxed{|z| = | \sin x |} $$