Cauchy Integral Theorem and the complex logarithm function I am given the following integral:
$\int_C {e^{(-1+i)\log(z)}}$
with 
$C:|z|=1$
$\operatorname{Log} (z): 0\le \operatorname{arg} (z)\le 2\pi$
Is it possible to resolve this integral using Cauchy theorem? The function is not analytic in a line inside C so my guess would be that it is not possible.
Thanks for the help    

Here is a little trick using the contour integral formula
$$\int_{|z|=1}e^{(-1+i)\log{z}}dz=\lim_{\epsilon\rightarrow 0}\int_{\epsilon}^{2\pi-\epsilon}e^{(-1+i)\log{e^{i\phi}}}\cdot ie^{i\phi}d\phi$$
The contour isolates the point where the function is not analytic

