Quick complex analysis (integration) question

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SUMMARY

The integral from -1 to 1 of z^i can be evaluated as (1-i)(1+exp(-pi)/2, where the integration path is any contour above the real axis. The expression z^i can be rewritten using the exponential form, z^i = exp(i log(z)), with the condition that |z| > 0 and arg(z) is between -pi/2 and 3pi/2. To compute the integral, one effective method is to choose the contour z = exp(it) for t ranging from 0 to pi, allowing for a transformation into a t integration. Alternatively, the antiderivative of z^i can be utilized, specifically z^(i+1)/(i+1), since the integration does not cross any branch cuts.

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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.
 

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