$\text{ If }\mathrm{m}\text{ and }\mathrm{x}\text{ are two real numbers and }i=\sqrt{-1}\text{ , then }{e}^{2mi{\cot }^{-1}x}{\left(\frac{xi+1}{xi-1}\right)}^m=$

$\text{ Let }{Cot}^{-1}\mathrm{x}=\theta ,\text{ then }\mathrm{x}=\cot \theta$

$\therefore e^{2mi{\cot }^{-1}x}{\left(\frac{xi+1}{xi-1}\right)}^m=e^{2mi\theta }{\left(\frac{i\cot \theta +1}{i\cot \theta -1}\right)}^m$

                                      $$\begin{gathered} =e^{2mi\theta }{\left(\frac{i\left(\cot \theta -i\right)}{i\left(\cot \theta +i\right)}\right)}^m \\ =e^{2mi\theta }{\left(\frac{\cos \theta -i\sin \theta }{\cos \theta +i\sin \theta }\right)}^m \end{gathered}$$

                                      $$\begin{gathered} =e^{2mi\theta }{\left(\frac{e^{-i\theta }}{e^{i\theta }}\right)}^m \\ =e^{2mi\theta }{e}^{-2i\theta m} \\ =1 \end{gathered}$$