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Question: \(i\log\left( \frac{x - i}{x + i} \right)\) is equal to...

ilog⁡(x−ix+i)i\log\left( \frac{x - i}{x + i} \right) is equal to

A

π+2tan⁡−1x\pi + 2\tan^{- 1}x

B

π−2tan⁡−1x\pi - 2\tan^{- 1}x

C

−π+2tan⁡−1x- \pi + 2\tan^{- 1}x

D

−π−2tan⁡−1x- \pi - 2\tan^{- 1}x

Answer

π−2tan⁡−1x\pi - 2\tan^{- 1}x

Explanation

Solution

Sol. Let z=ilog⁡(x−ix+i)z = i\log\left( \frac{x - i}{x + i} \right)⇒ zi=log⁡(x−ix+i)\frac{z}{i} = \log\left( \frac{x - i}{x + i} \right)

⇒ zi=log⁡[x−ix+i×x−ix−i]=log⁡[x2−1−2ixx2+1]\frac{z}{i} = \log\left\lbrack \frac{x - i}{x + i} \times \frac{x - i}{x - i} \right\rbrack = \log\left\lbrack \frac{x^{2} - 1 - 2ix}{x^{2} + 1} \right\rbrack

⇒ zi=log⁡[x2−1x2+1−i2xx2+1]\frac{z}{i} = \log\left\lbrack \frac{x^{2} - 1}{x^{2} + 1} - i\frac{2x}{x^{2} + 1} \right\rbrack .....(i)

∵ log⁡(a+ib)=log⁡(reiθ)=log⁡r+iθ\log(a + ib) = \log(re^{i\theta}) = \log r + i\theta =log⁡a2+b2+itan⁡−1(ba)= \log\sqrt{a^{2} + b^{2}} + i\tan^{- 1}\left( \frac{b}{a} \right)

Hence zi=log⁡(x2−1x2+1)2+(−2xx2+1)2+itan⁡−1(−2xx2−1)\frac{z}{i} = \log\sqrt{\left( \frac{x^{2} - 1}{x^{2} + 1} \right)^{2} + \left( \frac{- 2x}{x^{2} + 1} \right)^{2}} + i\tan^{- 1}\left( \frac{- 2x}{x^{2} - 1} \right)

(By equation (i))

zi=log⁡x4+1−2x2+4x2(x2+1)2+itan⁡−1(2x1−x2)=log⁡1+i(2tan⁡−1x)=0+i(tan⁡−1x)\frac{z}{i} = \log\frac{\sqrt{x^{4} + 1 - 2x^{2} + 4x^{2}}}{(x^{2} + 1)^{2}} + i\tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right) = \log 1 + i(2\tan^{- 1}x) = 0 + i(\tan^{- 1}x) ∴ z=i22tan⁡−1x=−2tan⁡−1x=π−2tan⁡−1x.z = i^{2}2\tan^{- 1}x = - 2\tan^{- 1}x = \pi - 2\tan^{- 1}x.