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Show that I Log ( X − I X + I ) = π − 2 Tan 6 − 1 X - Applied Mathematics 1

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Sum

Show that `ilog((x-i)/(x+i))=pi-2tan6-1x`

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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)`

Concept: Logarithmic Functions
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