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I have been studying the 'general' version of the residue theorem involving winding numbers and I am having a silly trouble with one step of the proof presented in Conway's Functions of One Complex Variable.

For the sake of completeness, I enunciate the theorem below.

Residue Theorem: Let $f:U\rightarrow \mathbb{C}$ be an analytic function in an open set $U\subset \mathbb{C}$ with the exception of finite isolated singularities $a_1,\dots,a_n$. If $\gamma$ is a retifiable closed curve in $U$ that does not pass through the points $a_1,\dots,a_n$ and if $n(\gamma,a)=0$ for all $a\notin U$, then $$\oint_{\gamma} f(z)dz = 2\pi i \sum\limits_{k=1}^n Res(f;a_k)n(\gamma;a_k).$$

Here, $n(\gamma,a)$ denotes the winding number.

The proof starts with the construction of open balls in $U$ around the singularities that do not intersect themselves, nor $\gamma$. Then it defines the boundary of the balls as the curves

$$\gamma_k(t) = a_k +r_k \exp(-2\pi i n(\gamma;a_k) t),$$ where $r_k$ denotes the radius of the ball around $a_k$.

Claim 1: for $1\leq j \leq m$, $$n(\gamma;a_j) + \sum\limits_{k=1}^n n(\gamma_k;a_j) = 0.$$ I assume this is obtained by calculating the winding numbers.

Then it goes on claiming that $n(\gamma;a) + \sum\limits_{k=1}^n n(\gamma_k;a) = 0$ for $a\notin U\setminus \{a_1,\dots,a_n\}$, uses Cauchy Theorem and Laurent expansion to conclude that

$$\oint_{\gamma} f(z)dz + 2\pi i \sum\limits_{k=1}^n Res(f;a_k) n(\gamma_k,a_k) = 0,$$ all very straightforward and easy to follow. But then it states that the proof is complete.

I do not understand how from the last expression we obtain the result. Why can we 'simply' replace $n(\gamma_k,a_k)$ by $-n(\gamma,a_k)$?

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    $\begingroup$ By construction, $\gamma_k$ goes around $a_k$ "clockwise" $n(\gamma,a_k)$ times, so $n(\gamma_k,a_k)=-n(\gamma,a_k)$. $\endgroup$
    – user1266745
    Jan 9 at 21:16
  • $\begingroup$ Is it simple as that? I mean, what is Claim 1 for in the proof then? $\endgroup$
    – MsWynfled
    Jan 9 at 21:19
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    $\begingroup$ Claim $1$ together with the fact that $n(\gamma_k,a_j)=0$ for $k\neq j$ (since $\gamma_k$ doesn't go around $a_nj$) is equivalent to $n(\gamma,a_k)=-n(\gamma_k,a_k)$ for all $k$. $\endgroup$
    – user1266745
    Jan 9 at 21:23

1 Answer 1

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Your main question has already been answered in the comments: $n(\gamma_k,a_k)=-n(\gamma,a_k)$ by construction.

To answer your question in the comments what Claim $1$ is for in the proof then (which isn’t satisfactorily answered by the response in the comments since Claim $1$ would be an unnecessary complication if all it was used for was to deduce $n(\gamma_k,a_k)=-n(\gamma,a_k)$): Claim $1$ is used as a premise in applying Cauchy’s theorem (referred to as Theorem IV.$5$.$7$ in the book).

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