Question

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

Answer 1

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)`