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

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detailed solution

Correct option is C

Let Cot-1x=θ, then x=cotθ∴e2m icot−1xxi+1xi−1m=e2m iθicotθ+1icotθ−1m =e2m iθicotθ−iicotθ+im =e2m iθcosθ−isinθcosθ+isinθm =e2m iθe−iθeiθm =e2mi θe−2iθm =1

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If m and x are two real numbers and i=-1 , then e2micot-1xxi+1xi-1m=

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