Show that `e^(2m itheta)((icottheta+1)/(i cottheta-1))^m=1.`

Explanation

Let `cot^(-1) p = theta`. Then `cot theta = p` Now,
L.H.S `= e^(2mi theta)((icot theta + 1)/(i cot theta -1))^(m)`
`=e^(2mi theta)[(i(cot theta -i))/(i(cot theta +i))]^(m)`
`= e^(2m i theta)((cot theta -i)/(cot theta + i))^(m)`
`= e^(2 mitheta)((cos theta - i sintheta)/(cos theta + i sin theta))^(m)`
`= e^(2 mitheta)((e^(-itheta))/(e^(itheta)))^(m)`
`= e^(2mi theta)(e^(-2itheta))^(m)`
`=e^(2mitheta)(e^(-2itheta))^(m) = e^(0) = 1= R.H.S`