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Prove that for any $f:\mathbb{C} \rightarrow \mathbb{C}$ which is analytic everywhere except at a finite number of singularities $\{a_i \in \mathbb{C}|i = 1, \ldots N \}$, the sum of the residues at the singularities plus the residue at infinity is zero.

Is the proof below okay? This is one of the last questions in the complex analysis section of a book I'm reading and I'm not sure if I've understood correctly that the residue at infinity is the same as the clockwise integral of $f$ around a contour which encloses all singularities of $f$.

Proof:

Consider a circle $C$ centred at the origin which has a radius $r$ large enough to enclose all singularities within $C$. From the residue theorem

$$ \frac{1}{2 \pi i}\oint_C f(z) \, dz = \sum_{i} \underset{z=a_i}{\rm Res}(f(z)),$$

where the contour integral is computed in the anticlockwise direction. Moving integral to the right hand side and changing the direction of the integration to be clockwise around the circle C we obtain

\begin{align} \sum_{i} \underset{z=a_i}{\rm Res}(f(z)) - \frac{1}{2 \pi i}\oint_C f(z) \, dz &= 0, \\ \sum_{i} \underset{z=a_i}{\rm Res}(f(z)) + \frac{1}{2 \pi i}\oint_{C,\ \mathrm{clockwise}} f(z) \, dz &= 0. \end{align}

The integral on the right is the same as the residue at infinity and thus the sum of the residues at the singularities plus the residue at infinity is zero. $\square$

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    $\begingroup$ If you know that the last integral is the residue at $\infty$, it's good. $\endgroup$ – Daniel Fischer Nov 29 '16 at 23:04
  • $\begingroup$ I think by definition that the clockwise integral is the residue at infinity. I don't have to prove the last statement right? $\endgroup$ – Raj Nov 29 '16 at 23:08
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    $\begingroup$ Alternatively: Take a circle that contains none of the singularities of $f$, and integrate around that to get $0$. Redefine the inside to be the outside, and your circle now goes around each of the singularities of $f$, including the one at $\infty$ if that's a singularity, exactly once, without changing the value of the integral. $\endgroup$ – Arthur Nov 29 '16 at 23:11
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    $\begingroup$ There are several (equivalent) definitions. If yours is the integral, you're home and dry. If it's "the residue of $-z^{-2}f(z^{-1})$ at $0$", one needs to show the equivalence (which is easy, but needs to be done). $\endgroup$ – Daniel Fischer Nov 29 '16 at 23:12
  • $\begingroup$ @DanielFischer Thanks! I saw in several places that people were using the "the residue of $-z^{-2}f(z^{-1})$" definition and I wasn't sure whether I needed to prove the clockwise integral definition. I didn't realize that there were several equivalent definitions and that was what was confusing me. The book uses the clockwise integral definition as the starting point so I don't need to prove it. $\endgroup$ – Raj Nov 30 '16 at 3:34

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