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Question Number 19898 by Tinkutara last updated on 17/Aug/17

$$\mathrm{Simplify}:i\log \left(\frac{x-i}{x+i}\right).$$

Answered by ajfour last updated on 22/Aug/17

$$lett=i\ln \left(\frac{x-i}{x+i}\right)$$$$andletz=\frac{x-i}{x+i}$$$$\Rightarrow \mid z\mid =1$$$$soletz=e^{i\theta }here-\pi <\theta <0$$$$t=i\ln z=i\left[i\theta \right]=-\theta$$$$Nowlet\tan \mathrm{\varnothing }=\frac{1}{x}>0$$$$asx>0,0<\varphi <\frac{\pi }{2}$$$$\Rightarrow \theta =-2\varphi \left(seecomment\right)$$$$\Rightarrow t=i\ln \left(\frac{x-i}{x+i}\right)=-\theta =2\varphi$$$$={2\cot }^{-1}x.$$

Commented by Tinkutara last updated on 18/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by ajfour last updated on 18/Aug/17

Answered by Tinkutara last updated on 22/Aug/17

$$i[\log (x-i)-\log (x+i)]$$$$=i[\frac{1}{2}\log (x^2+1)+i{\tan }^{-1}\left(\frac{-1}{x}\right)-\frac{1}{2}\log (x^2+1)-i{\tan }^{-1}\left(\frac{1}{x}\right)]$$$$=i[-2i{\tan }^{-1}\left(\frac{1}{x}\right)]=2{\cot }^{-1}x$$

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