show that the integral of z^i dz from -1 to 1 is equal to (1+e^(-pi))/x * (1-i)
where the path of integration is any contour from -1 to 1 above the x axis
The Attempt at a Solution
I wrote z^i as e^iLogz and I think all I really need to do is find the antiderivitive of e^i*logz where -pi/2<argz<3pi/2
this antiderivative is giving me trouble I tried to break it up into e^i*lnr-theta
and then get -e^(i*lnr)*[e^(3pi/2)-e^(-pi/2)] but I can't get it to work
can anyone help?