I : The modulus of $\frac{\sqrt{3} +\mathrm{i}}{\left(1+\mathrm{i}\right)\left(1+\sqrt{3}\mathrm{i}\right)}$is $\frac{1}{\sqrt{2}}$

II : The least positive value of n for which ${\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)}^{2\mathrm{n}}=1 \mathrm{is} 2$

Which of the statements are true

$$\begin{gathered} \mathrm{I} : \left|\frac{\sqrt{3}+\mathrm{i}}{(1+\mathrm{i})(1+\mathrm{i}\sqrt{3})}\right| =\frac{\sqrt{3+1}}{\sqrt{1+1}\sqrt{1+3}}=\frac{1}{\sqrt{2}} \\ \mathrm{II} : {\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)}^{2\mathrm{n}}={\left(\frac{(1-\mathrm{i}) (1-\mathrm{i})}{(1+\mathrm{i}) (1-\mathrm{i})}\right)}^{2\mathrm{n}}={\left(\frac{{(1-\mathrm{i})}^2}{1+1}\right)}^{2\mathrm{n}} \\ ={\left(\frac{1+{\mathrm{i}}^2-2\mathrm{i}}{2}\right)}^{2\mathrm{n}} \\ {\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)}^{2\mathrm{n}}={(-\mathrm{i})}^{2\mathrm{n}} = {(-1)}^{\mathrm{n}}=1 \mathrm{if} \mathrm{n}=2, 4, 6, . . . \\ \mathrm{Least} \mathrm{value} \mathrm{of} \mathrm{n}=2. \\ \mathrm{Both} \mathrm{I} \& \mathrm{II} \mathrm{are} \mathrm{true}. \end{gathered}$$