Problem #80
The path of integration is any contour from z = -1 to z = 1 that lies above the real axis.
\[ \displaylines{\int_{-1}^1z^{i}dz=\frac{1+e^{-\pi}}{2}\left(1-i\right)\\ \text{where }z^{i}\text{ denotes the principal branch}\\ z^{i}=\exp\left(i\log z\right)\text{ }\left(\left|z\right|\gt0,-\pi\lt Arg\left(z\right)\lt\pi\right)} \]