Quick complex analysis (integration) question

indigogirl
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I want to show that the integral from -1 to 1 of z^i = (1-i)(1+exp(-pi)/2

where the path of integration is any contour from z=-1 to z=1 that lies above the real axis (except for its endpoints).

So, I know that z^i=exp(i log(z)) and the problem states that |z|>0, and arg(z) is between -pi/2 and 3pi/2. But we didn't study how to integrate z^(complex number) in class, and I"m really confused on how to do this.

So, how do I integrate this?
 
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You could pick a sample contour like z=exp(it) for t=0 to pi. Then log(z)=it, dz=i*exp(it)*dt. Change it into a t integration. You could also just use the antiderivative of z^i=z^(i+1)/(i+1) since you aren't passing over any branch cuts.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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