Calculating Residium of Complex Integral |z|=1

nhrock3
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\int_{|z|=1}^{nothing } \frac{1}{z}e^{\frac{1}{z}}
in this integral there is no upper bound
its around |z|=1

there are no poles here
only singular significant
what to do here
when calclating the residium
??
 
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I believe, and don't trust me on this, that it's asking you to calculate a path integral around the unit circle on the complex plane. There's no "upper bound" because the integral is describing a path, not just a starting point and an ending point. Incidentally, I believe that starting and ending points are the same.
 
Start by expanding \exp (1/z) as a power series, multiply by 1/z and look for the z^{-1} term. That will be your residue.

Mat
 
A change of variables w = 1/z will also do the trick.
 
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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