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