MHB How can the complex exponential product be proven for all real p and m?

[Greg]

Show that, for all real $$p$$ and $$m$$,

$$e^{2mi\cot^{-1}(p)}\left(\dfrac{pi+1}{pi-1}\right)^m=1$$

 

[Nono713]

Spoiler

Notice that, for all real $u$, we have:
$$\frac{i \cot{u} + 1}{i \cot{u} - 1} = \frac{i \frac{\cos{u}}{\sin{u}} + 1}{i \frac{\cos{u}}{\sin{u}} - 1} = \frac{\sin{u} + i \cos{u}}{i \cos{u} - \sin{u}} = \cos{2u} - i \sin{2u} = e^{-2iu}$$
Therefore it follows that:
$$e^{2iu} \left ( \frac{i \cot{u} + 1}{i \cot{u} - 1} \right ) = e^{2iu} e^{-2iu} = 1$$
And so for all real $m$ we have:
$$e^{2miu} \left ( \frac{i \cot{u} + 1}{i \cot{u} - 1} \right )^m = 1^m = 1$$
Setting $p = \cot{u}$ so that $u = \cot^{-1}{p}$ completes the proof.