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

Sum

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

#### 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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