Conway's proof of Residue Theorem

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 $$-n(\gamma,a_k)$$?

• By construction, $\gamma_k$ goes around $a_k$ "clockwise" $n(\gamma,a_k)$ times, so $n(\gamma_k,a_k)=-n(\gamma,a_k)$.
– user1266745
Jan 9 at 21:16
• Is it simple as that? I mean, what is Claim 1 for in the proof then? Jan 9 at 21:19
• 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 $k$.
– user1266745
Jan 9 at 21:23

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).