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Question
Show that `ilog((x-i)/(x+i))=pi-2tan6-1x`
Sum
Solution
We have `log(x+i)=1/2log(x^2+1)+itan^(-1)(1/x)`
and `log(x-i)=1/2log(x^2+1)-itan^(-1)(1/x)`
`log((x-i)/(x+i))=log(x-i)-log(x+i)`
`=-2itan^(-1) 1/x=-2i(pi/2-tan^(-1)x)`
`therefore log((x-i)/(x+i))=-i(pi-2tan^(-1)x)`
`therefore ilog((x-i)/(x+i))=(pi-2tan^(-1)x)`
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Logarithmic Functions
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