>>7648367 look up the power series of the exponential function: exp(z). Replace z by 1/z and you've got the power series for the function we're discussing. According to Cauchy's residue theorem, every function that is differentiable in a closed domain C has a net integral value of 0. So every analytical function nets 0 when you integrate it over a closed region, EXCEPT for 1/(z-z0), where z0 is 0 for this case. If you look at the power series of this function, you can see that it contains exactly one "copy" of 1/z; the rest is 1/z^2, 1/z^3... : all differentiable! So if you were to integrate this function over a circle around 0, the answer would be 2*pi*i times the residu, which is the coefficient of the 1/z term: 1.
>>7648374 Thank you, complex analysis is really interesting. I love how you can use it to solve crazy rational function real integrals from -inf to inf by using jordans lemma or Ml inequality then using residue theorem or something else on the contour integral.
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