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Question: If \(m\)and \(x\) are two real numbers, then \(e^{2mi\cot^{- 1}x}\left( \frac{xi + 1}{xi - 1} \right...

If mmand xx are two real numbers, then e2micot⁡−1x(xi+1xi−1)me^{2mi\cot^{- 1}x}\left( \frac{xi + 1}{xi - 1} \right)^{m}is equal to

A

cos⁡x+isin⁡x\cos x + i\sin x

B

m2\frac{m}{2}

C

1

D

(m+1)2\frac{(m + 1)}{2}

Answer

1

Explanation

Solution

Sol. Let cot⁡−1x=θ,\cot^{- 1}x = \theta, then cot⁡θ=x\cot\theta = x or tan⁡θ=1x\tan\theta = \frac{1}{x}.

We have e2icot⁡−1x=e2iθ=cos⁡(2θ)+isin⁡(2θ)e^{2i\cot^{- 1}x} = e^{2i\theta} = \cos(2\theta) + i\sin(2\theta)

= 1−tan⁡2θ1+tan⁡2θ+i(2tan⁡θ1+tan⁡2θ)=1−1/.x21+1/x2+i(2/x1+1/x2)\frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} + i\left( \frac{2\tan\theta}{1 + \tan^{2}\theta} \right) = \frac{1 - 1/.x^{2}}{1 + 1/x^{2}} + i\left( \frac{2/x}{1 + 1/x^{2}} \right)

x2−1x2+1+2ixx2+1=(x+i)2(x−i)(x+i)=x+ix−i=ix+i2ix−i=ix−1ix+1\frac{x^{2} - 1}{x^{2} + 1} + \frac{2ix}{x^{2} + 1} = \frac{(x + i)^{2}}{(x - i)(x + i)} = \frac{x + i}{x - i} = \frac{ix + i^{2}}{ix - i} = \frac{ix - 1}{ix + 1}

⇒ (e2icot⁡−1x)m=(ix−1ix+1)m\left( e^{2i\cot^{- 1}x} \right)^{m} = \left( \frac{ix - 1}{ix + 1} \right)^{m}

⇒ e2micot⁡−1x(ix+1ix−1)m=1e^{2mi\cot^{- 1}x}\left( \frac{ix + 1}{ix - 1} \right)^{m} = 1