I am reading James Ward Brown's Complex Variables and Applications and I am stuck on this problem.
Problem:
Show that $$\int_{-1}^{1} z^i dz=\frac{1+e^{-\pi}}{2}(1-i),$$ where the integrand denotes he principal branch $$z^i=exp(i\space Log\space z),\space\space (|z|>0,\space -\pi<Arg\space z<\pi)$$ of $z^i$ and where the path of integration is any contour from $z=-1$ to $z=1$ that, except for its end points, lies above the real axis.
The problem gives a hint that I should evaluate the antiderivative over the new branch where $arg\space z$ ranges from $-\dfrac{\pi}{2}$ to $\dfrac{3\pi}{2}$.
My questions are:
Why is it necessary to change a branch, since the contour never crosses the branch cut of the principal branch of $z^i$?
I'd also like to know how the antiderivative is found, does it follow directly from real calculus to be $F(x)=\dfrac{z^{i+1}}{i+1}$, which does yield the correct result?