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My question is about the last equality in the following derivation:

f(z) is assumed to be a Laurent expansion about a singularity at $z_0$.

$\oint f(z)dz= \oint \sum_{n=-\infty}^{n=\infty} {a_n}(z-z_0)^n dz= \sum_{n=-\infty}^{n=\infty} {a_n} \oint(z-z_0)^n dz = a_{-1}2 \pi i$

and $a_{-1}$ is the residue.

-McMahon, David. Complex Variables Demystified. page 149.

This is my interpretation of the last step: It follows from Cauchy's Integral Formula: $\oint \frac{f(z)} {z-z_0} dz= f(z_0)2 \pi i$ for the particular case $\oint \frac{1} {z-z_0} dz= 2 \pi i$ But, for this to be the case, I must explain why the other $\oint(z-z_0)^n $ terms do not contribute. It seems that only $\oint(z-z_0)^{-1} $ does and this is where I'm stuck.

I've recently read a derivation of the residue theorem in an appendix of an excellent book titled Gamma that makes much more sense to me.The derivation in Gamma begins by deriving the explicit Laurent expansion with the terms (including the residue term) written out as complex contour integrals and then pointing out that the residue term in the expansion is $\oint f(w)dw$. Therefore, finding the residue term in the expansion of $f(z)$ gives you the integral $\oint f(w)dw$. That makes sense. This Complex Variables Demystified derivation seems to be missing a few steps (or possibly just wrong). I can't reconcile the two.

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    $\begingroup$ If $n\ne -1,$ then $(z-z_0)^n$ has an antiderivative, namely $(z-z_0)^{n+1}/(n+1).$ Any continuous function in a region that has a complex antiderivative integrates to $0$ on a closed contour; this is just the FTC. $\endgroup$ – zhw. Oct 13 '15 at 5:35
  • $\begingroup$ @zhw. could you tell me why the $2 \pi i$ isn't written as ln(z-z_0) if that's what the integral is? $\endgroup$ – MadScientist Oct 13 '15 at 6:17
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The fundamental theorem of line integrals states that if $g$ has an antiderivative, then $$ \oint_\gamma g(z) dz = 0$$ for all closed smooth paths $\gamma$. Really, this is the fundamental theorem of calculus applied to line integerals.

For every term except $(z - z_0)^{-1}$, we know that $(z-z_0)^n$ has an antiderivative. That's why those terms all vanish.

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  • $\begingroup$ Then my confusion seems to be that i'm assuming $ (z-z_0)^n $ isn't analytic for, negative n, because it blows up at $z_0$. Is that not the case? $\endgroup$ – MadScientist Oct 13 '15 at 5:43
  • $\begingroup$ @MadScientist Indeed, I spoke too soon. I was quoting one theorem (that functions with antiderivatives do not contribute) and calling it (and stating) another. $\endgroup$ – davidlowryduda Oct 13 '15 at 5:50
  • $\begingroup$ is it correct to say that it is because the antiderivative is analytic (or holomorphic)? (i want to connect this to analycity somehow) $\endgroup$ – MadScientist Oct 13 '15 at 5:53
  • $\begingroup$ No, because it is not analytic (nor holomorphic). That's where I realized I was conflating two theorems in my original answer. $\endgroup$ – davidlowryduda Oct 13 '15 at 5:54
  • $\begingroup$ could you tell me why the 2πi isn't written as ln(z-z_0) if that's what the integral is? $\endgroup$ – MadScientist Oct 13 '15 at 6:21

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