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Question: \(e^{2mi\cot^{- 1}p}\)·\(\left( \frac{pi + 1}{pi - 1} \right)^{m}\)=...

e2micot⁡−1pe^{2mi\cot^{- 1}p}·(pi+1pi−1)m\left( \frac{pi + 1}{pi - 1} \right)^{m}=

A

0

B

1

C

–1

D

None of these

Answer

1

Explanation

Solution

Sol. Let cot–1 p = q, then p = cot q

\ e2micot⁡−1pe^{2mi\cot^{- 1}p}· (pi+1pi−1)m\left( \frac{pi + 1}{pi - 1} \right)^{m}

= ·(icot⁡θ+1icot⁡θ−1)m\left( \frac{i\cot\theta + 1}{i\cot\theta - 1} \right)^{m}

=(i(cot⁡θ−i)i(cot⁡θ+i))m\left( \frac{i(\cot\theta - i)}{i(\cot\theta + i)} \right)^{m}

=· (cot⁡θ−icot⁡θ+i)m\left( \frac{\cot\theta - i}{\cot\theta + i} \right)^{m}

= · (cos⁡θ−isin⁡θcos⁡θ+isin⁡θ)m\left( \frac{\cos\theta - i\sin\theta}{\cos\theta + i\sin\theta} \right)^{m}

= · (e−iθeiθ)m\left( \frac{e^{- i\theta}}{e^{i\theta}} \right)^{m}

= e2miq (e–2iq)m = e0 = 1.