# Line integrals: 3. Applications to complex analysis


## Preliminaries

Let $$F$$ be a complex Banach space, let $$U\subseteq\bbc$$ be an open set, and let $$f:U\to F$$. Recall that the complex Fréchet derivative of $$f$$ at $$z\in U$$, if it exists, is a $$\bbc$$-linear map $$Df(z):\bbc\to F$$. Not all real differentiable functions on $$\bbc$$ are complex differentiable: for example, the (real) derivative of $$f(z)=\overline{z}$$ (i.e. $$f(x+yi)=x-yi$$) at any point is represented by the matrix $$\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix},$$ but $$Df(z)$$ is clearly not a $$\bbc$$-linear map.

Since $$\bbc$$ is a one-dimensional vector space over itself, $$Df(z)$$ is completely determined by the value $$Df(z)(1)\in F$$: $$Df(z)w=wDf(z)(1).$$ We can therefore identify $$Df(z)$$ with $$Df(z)(1)$$, and from now on we will use the notation $$f'(z)=Df(z)(1)\in F$$. For example, if $$f(z)=z^2$$ then the Fréchet derivative is $$Df(z)w=2zw$$, and $$f'(z)=Df(z)(1)=2z$$ as usual.

Theorem 18. Let $$U\subseteq\bbc$$ be an open set. A function $$f:U\to F$$ is complex differentiable at $$z\in U$$ if and only if the limit $$c=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$$ exists. In that case, $$c=f'(z)$$.

Proof. We have $$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h} = c \Leftrightarrow \lim_{h\to 0}\frac{f(z+h)-f(z)-ch}{|h|} = 0.$$ $$\square$$

If $$f:U\to F$$ is complex differentiable at all $$z\in U$$, then we say that $$f$$ is holomorphic on $$U$$ or simply holomorphic. If $$A\subseteq\bbc$$ is any set, we say that $$f$$ is holomorphic on $$A$$ if it is holomorphic on an open set containing $$A$$. If $$z\in\bbc$$ and $$f$$ is holomorphic on a neighborhood of $$z$$, we say that $$f$$ is holomorphic at $$z$$.

Using the basic properties of the derivative, we have:

Theorem 19 (Chain rule). Let $$U,V\subseteq\bbc$$ be open sets. Let $$f:U\to\bbc$$ and $$g:V\to F$$ with $$f(U)\subseteq V$$. If $$f$$ is complex differentiable at $$z$$ and $$g$$ is complex differentiable at $$f(z)$$, then $$g\circ f$$ is complex differentiable at $$z$$ and $$(g\circ f)'(z)=g'(f(z))f'(z).$$

Theorem 20. Let $$U\subseteq\bbc$$ be an open set, let $$F_1,F_2$$ be complex Banach spaces, and let $$f:U\to F_1$$ and $$f:U\to F_2$$ be complex differentiable at $$z\in U$$.

1. If $$f$$ is constant then $$f'(z)=0$$.
2. If $$F_1=F_2$$ then $$(f+g)'(z)=f'(z)+g'(z)$$.
3. $$(cf)'(z)=cf'(z)$$ for all $$c\in\bbc$$.
4. If $$F_1=\bbc$$ or $$F_2=\bbc$$, then $$(fg)'(z)=f'(z)g(z)+f(z)g'(z)$$.
5. If $$F_2=\bbc$$ and $$g(z)\ne 0$$ then $$(f/g)'(z)=[f'(z)g(z)-f(z)g'(z)]/g(z)^2$$.

Theorem 21. Let $$U\subseteq\bbc$$ be an open set. A function $$f:U\to\bbc$$ is complex differentiable at $$z\in U$$ if and only if it is real differentiable at $$z$$ and the real derivative $$Df(z)$$ is represented by a matrix of the form $$\begin{bmatrix}a & -b \\ b & a\end{bmatrix}.$$ In that case, $$Df(z)$$ is the matrix representation of the complex number $$f'(z)$$. Furthermore, $$\det Df(z)=|f'(z)|^2$$.

The preceding theorem shows that a holomorphic function $$f:U\to\bbc$$ is simply a real differentiable function with the property that its derivative is a scalar times a rotation matrix at every point of $$U$$. (That is, the derivative is $$\bbc$$-linear.) Suppose that $$f(x+yi)=u(x,y)+v(x,y)i$$ where $$u,v$$ are real valued functions. If $$f$$ is holomorphic then the theorem implies that $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad\mathrm{and}\quad\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$ These are known as the Cauchy-Riemann equations.

## Integration

We now define complex line integrals as in part 1, taking $$E=\bbc$$. If $$U\subseteq\bbc$$ is an open set and $$f:U\to F$$ is continuous, then we define its associated form $$\omega_f:U\to L(\bbc,F)$$ by $$\omega_f(z)w=wf(z).$$ If $$\gamma$$ is a curve in $$U$$ then the integral of $$f$$ along $$\gamma$$ is defined by $$\int_\gamma f = \int_\gamma f(z)\,dz = \int_\gamma \omega_f.$$ Note that if $$\gamma:[a,b]\to U$$ is a curve with partition $$\{a_0,\dots,a_k\}$$, then $$\int_\gamma f = \sum_{i=1}^k \int_{a_{i-1}}^{a_i} f(\gamma(t))\gamma'(t)\,dt.$$ The usual properties in Theorem 1 apply. A holomorphic function $$g:U\to F$$ satisfying $$f=g’$$ is called a primitive of $$f$$. It is easy to check that any potential function for $$\omega_f$$ (a function $$g$$ such that $$\omega_f=Dg$$) is a primitive for $$f$$.

Example 22. Let $$n$$ be an integer and define a curve $$\gamma:[0,2\pi]\to\bbc$$ by $$\gamma(t)=e^{it}$$. Then \begin{align}
\int_\gamma z^n\,dz &= i\int_0^{2\pi} e^{(n+1)it}\,dt \\
&= \begin{cases}2\pi i, & n=-1, \\ 0, & n\ne -1.\end{cases}
\end{align}

Suppose $$f:U\to F$$ is holomorphic. Since $$\omega_f=\ell\circ f$$ where $$\ell:F\to L(\bbc,F)$$ is the linear map given by $$\ell(x)h=hx$$, we have \begin{align}
D\omega_f(z)(u,v) &= (D\ell(f(z))Df(z)u)(v) \\
&= \ell(Df(z)u)(v) \\
&= uvf'(z).
\end{align} Clearly, $$D\omega_f(z)$$ is symmetric for all $$z\in U$$. Thus $$\omega_f$$ is closed, and Goursat’s theorem (Corollary 8) shows that $$\omega_f$$ is locally exact. As a consequence, we can define the integral of a holomorphic function $$f$$ along any path (see Lemma 9).

Theorem 23 (Cauchy’s theorem, local version). Let $$U\subseteq\bbc$$ be an open set, let $$\gamma_1,\gamma_2$$ be paths in $$U$$ that are homotopic, and let $$f$$ be holomorphic on $$U$$. Then $$\int_{\gamma_1} f = \int_{\gamma_2} f.$$ In particular, if $$U$$ is simply connected then $$\int_\gamma f=0$$ for any closed path $$\gamma$$ in $$U$$.

Proof. Apply Theorem 10. $$\square$$

If $$C$$ is a circle, we write $$\int_C f$$ for the integral of $$f$$ along $$C$$, taken counterclockwise.

Theorem 24 (Cauchy’s integral formula, local version). Let $$D$$ be a closed disc and let $$f$$ be holomorphic on $$D$$. Then $$f(z)=\frac{1}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{\zeta-z}\,d\zeta$$ for every $$z\in\Int D$$.

Proof. Let $$U$$ be an open set containing $$D$$ on which $$f$$ is holomorphic. For small $$r > 0$$, the circle $$C_r$$ of radius $$r$$ around $$z$$ is contained in $$D$$. Note that $$C_r$$ is homotopic to $$\partial D$$ in $$U\setminus\{z\}$$, so using Example 22 and Theorem 23 we have \begin{align}
\left\vert \frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-z}\,d\zeta-f(z)\right\vert &= \left\vert \frac{1}{2\pi i}\int_{C_r}\frac{f(\zeta)}{\zeta-z}\,d\zeta – \frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{\zeta-z}\,d\zeta \right\vert \\
&= \left\vert \frac{1}{2\pi i}\int_{C_r}\frac{f(\zeta)-f(z)}{\zeta-z}\,d\zeta\right\vert \\
&\le \frac{1}{2\pi} 2\pi r \sup_{\zeta\in C_r} \left\vert \frac{f(\zeta)-f(z)}{\zeta-z} \right\vert \\
&\to 0
\end{align} as $$r\to 0$$ since $$f$$ is complex differentiable at $$z$$. $$\square$$

The open disc $$D_r(z_0)$$ is the set $$\{z\in\bbc:|z-z_0| < r\}$$, and the closed disc $$\overline{D}_r(z_0)$$ is the set $$\{z\in\bbc:|z-z_0| \le r\}$$.

Theorem 25. Let $$D=\overline{D}_r(z_0)$$ be a closed disc and let $$f$$ be holomorphic on $$D$$. Then $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n \tag{*}$$ for every $$z\in\Int D$$, where $$a_n = \frac{1}{2\pi i} \int_{\partial D} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta = \frac{1}{n!} f^{(n)}(z_0).$$ We have $$|a_n| \le \frac{1}{r^n} \sup_{\zeta\in\partial D} |f(\zeta)|,$$ so the power series in (*) has a radius of convergence of at least $$r$$.

Proof. By Theorem 24, we have $$f(z)=\frac{1}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{\zeta-z}\,d\zeta.$$ Let $$0 < s < r$$ and let $$D'=\overline{D}_s(z_0)$$. For all $$z\in D'$$ and $$\zeta\in\partial D$$ we have \begin{align} \frac{1}{\zeta-z} &= \frac{1}{\zeta-z_0}\left(\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}\right) \\ &= \frac{1}{\zeta-z_0}\sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n, \end{align} where the geometric series converges absolutely and uniformly for $$\zeta\in\partial D$$ since $$\left\vert\frac{z-z_0}{\zeta-z_0}\right\vert \le \frac{s}{r} < 1.$$ Therefore \begin{align} f(z) &= \frac{1}{2\pi i}\int_{\partial D} \frac{f(\zeta)}{\zeta-z_0} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n\,d\zeta \\ &= \sum_{n=0}^\infty \left[ \frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta \right] (z-z_0)^n \end{align} for all $$z\in D'$$. $$\square$$ A function $$f:\bbc\to F$$ is called entire if it is holomorphic on $$\bbc$$. From Theorem 25 we can see that a function is entire if and only if it is represented by a power series with infinite radius of convergence.

Theorem 26. Let $$f$$ be an entire function. If there is a constant $$c$$ and a positive integer $$k$$ such that $$\sup_{|z|=r} |f(z)| \le cr^k$$ for all $$r > 0$$, then $$f$$ is a polynomial of degree $$k$$ or less (with coefficients in $$F$$).

Proof. Write $$f(z)=\sum_{n=0}^\infty a_n z^n$$ where $$a_n\in F$$. By Theorem 25, we have $$|a_n| \le \frac{1}{r^n} \sup_{|z|=r} |f(z)| \le cr^{k-n}.$$ If $$n > k$$, we can take $$r\to\infty$$ to deduce that $$a_n=0$$. $$\square$$

Corollary 27 (Liouville’s theorem). Any bounded entire function is constant.

As a simple application, we show that the spectrum of any element of a complex Banach algebra is nonempty. Let $$A$$ be a complex unital Banach algebra and let $$x\in A$$. We define the spectrum of $$x$$, denoted by $$\sigma(x)$$, to be the set of numbers $$\lambda\in\bbc$$ such that $$x-\lambda 1$$ is not invertible (where $$1$$ is the unit in $$A$$).

Theorem 28. The spectrum of any $$x\in A$$ is nonempty.

Proof. Suppose that $$\sigma(x)$$ is empty. The map $$f:\bbc\to A$$ given by $$z\mapsto (x-z1)^{-1}$$ is entire, since \begin{align}
Df(z)w &= -(x-z1)^{-1}(-w1)(x-z1)^{-1} \\
&= w(x-z1)^{-2}.
\end{align} (See this result.) If $$|z| > |x|$$ then $$x-z1=-z(1-x/z)$$ is invertible and $$|(x-z1)^{-1}|=|z|^{-1}|(1-x/z)^{-1}|\le\frac{|z|^{-1}}{1-|x/z|} \to 0$$ as $$|z|\to\infty$$. By Liouville’s theorem, $$f=0$$. But $$f(z)=(x-z1)^{-1}\ne 0$$ for any $$z\in\bbc$$, which is a contradiction. $$\square$$

## Cauchy’s theorem and winding numbers

An important consequence of Theorem 17 is the following result, which is the global version of Cauchy’s theorem (Theorem 23):

Theorem 29 (Cauchy’s theorem). Let $$U\subseteq\bbc$$ be an open set, let $$\gamma_1,\gamma_2$$ be 1-cycles in $$U$$ that are homologous, and let $$f$$ be holomorphic on $$U$$. Then $$\int_{\gamma_1} f = \int_{\gamma_2} f.$$

Usually, Cauchy’s theorem is stated in terms of winding numbers (which will be defined shortly). Our goal is to prove the following:

Theorem 30 (Cauchy’s theorem with winding numbers). Let $$U\subseteq\bbc$$ be an open set, let $$\gamma_1,\gamma_2$$ be 1-cycles in $$U$$ such that $$W(\gamma_1,z)=W(\gamma_2,z)$$ for all $$z\in\bbc\setminus U$$, and let $$f$$ be holomorphic on $$U$$. Then $$\int_{\gamma_1} f = \int_{\gamma_2} f.$$

We recall some concepts and theorems from algebraic topology. For any topological space $$X$$, we define a loop in $$X$$ to be a continuous map $$\gamma:[0,1]\to X$$ with $$\gamma(0)=\gamma(1)$$, and we say that $$\gamma$$ is based at $$\gamma(0)$$. Let $$\mathbb{S}^1=\{z\in\bbc:|z|=1\}$$ be the circle. The map $$q:\mathbb{R}\to\mathbb{S}^1$$ given by $$s\mapsto e^{2\pi is}$$ is a universal covering of $$\mathbb{S}^1$$. Let $$f:[0,1]\to\mathbb{S}^1$$ be a loop based at a point $$z_0\in\mathbb{S}^1$$. We define the winding number of $$f$$ by $$\widetilde{f}(1)-\widetilde{f}(0)$$, where $$\widetilde{f}:[0,1]\to\mathbb{R}$$ is any lift of $$f$$. Since any two lifts of $$f$$ differ by a constant, the winding number is well-defined. Since $$\widetilde{f}(1)$$ and $$\widetilde{f}(0)$$ are both elements of the fiber $$q^{-1}(\{z_0\})$$, they differ by an integer; thus the winding number of a loop is always an integer.

Theorem 31. Let $$f,g$$ be loops in $$\mathbb{S}^1$$ based at the same point. Then $$f$$ and $$g$$ are (path) homotopic if and only if they have the same winding number.

Now let $$z_0\in\bbc$$ and let $$\gamma:[0,1]\to\bbc\setminus\{z_0\}$$ be a closed path. Define a retraction $$r:\bbc\setminus\{z_0\}\to\mathbb{S}^1$$ by $$r(z)=\frac{z-z_0}{|z-z_0|}.$$ Then $$r\circ\gamma$$ is a loop in $$\mathbb{S}^1$$, and we can define the winding number of $$\gamma$$ with respect to $$z_0$$ to be the winding number of $$r\circ\gamma$$. We denote this integer by $$W(\gamma,z_0)$$. If we consider $$\gamma$$ as a singular 1-cycle in the homology group $$H_1(\bbc\setminus\{z_0\})\cong\mathbb{Z}$$, then $$[\gamma]=W(\gamma,z_0)[\alpha]$$ where $$\alpha$$ is the generator of $$H_1(\bbc\setminus\{z_0\})$$ defined by $$\alpha(s)=z_0+e^{2\pi is}$$. Therefore, for any 1-cycle $$\gamma$$ in $$\bbc\setminus\{z_0\}$$ we define the winding number of $$\gamma$$ with respect to $$z_0$$ to be the unique integer $$W(\gamma,z_0)$$ such that $$[\gamma]=W(\gamma,z_0)[\alpha]$$.

Theorem 32.

1. If $$\gamma$$ is homologous to $$\eta$$ in $$\bbc\setminus\{z_0\}$$, then $$W(\gamma,z_0)=W(\eta,z_0)$$.
2. If $$\gamma_1,\dots,\gamma_k$$ are closed paths and $$n_1,\dots,n_k$$ are integers, then $$W(n_1\gamma_1+\cdots+n_k\gamma_k,z_0)=n_1 W(\gamma_1,z_0)+\cdots+n_k W(\gamma_k,z_0).$$

Note that we have a convenient expression for the winding number of a 1-cycle as an integral:

Theorem 33. For every 1-cycle $$\gamma$$ in $$\bbc\setminus\{z_0\}$$, we have $$W(\gamma,z_0)=\frac{1}{2\pi i}\int_\gamma \frac{1}{z-z_0}\,dz.$$

Proof. We first prove the result for closed paths in $$\bbc\setminus\{z_0\}$$. By Theorem 12, we may assume that $$\gamma$$ is a closed curve. By linearity, we may also assume that $$\gamma$$ is a $$C^1$$ path. Let $$\widetilde{\gamma}:[0,1]\to\mathbb{R}$$ be a lift of $$r\circ\gamma$$; then $$\widetilde{\gamma}$$ is $$C^1$$ and $$e^{2\pi i\widetilde{\gamma}(s)} = \frac{\gamma(s)-z_0}{f(s)},$$ where $$f(s)=|\gamma(s)-z_0|$$. We compute \begin{align}
\frac{1}{2\pi i}\int_\gamma \frac{1}{z-z_0}\,dz &= \frac{1}{2\pi i}\int_0^1 \frac{\gamma'(s)}{\gamma(s)-z_0}\,ds \\
&= \frac{1}{2\pi i}\int_0^1 \frac{2\pi if(s)\widetilde{\gamma}'(s)e^{2\pi i\widetilde{\gamma}(s)}+f'(s)e^{2\pi i\widetilde{\gamma}(s)}}{f(s)e^{2\pi i\widetilde{\gamma}(s)}}\,ds \\
&= \frac{1}{2\pi i}\int_0^1 \left(2\pi i\widetilde{\gamma}'(s)+\frac{f'(s)}{f(s)}\right)\,ds \\
&= \frac{1}{2\pi i}[2\pi i\widetilde{\gamma}(s)+\log f(s)]_0^1 \\
&= \widetilde{\gamma}(1)-\widetilde{\gamma}(0) \\
&= W(\gamma,z_0).
\end{align} Now let $$\gamma$$ be a 1-cycle in $$\bbc\setminus\{z_0\}$$. By Theorem 15, $$\gamma$$ is homologous to a sum $$\sum_{j=1}^k c_j \gamma_j$$ where each $$\gamma_j$$ is a closed path. Then $$W(\gamma,z_0)=\sum_{j=1}^k c_j W(\gamma_j,z_0) = \sum_{j=1}^k c_j \frac{1}{2\pi i} \int_{\gamma_j} \frac{1}{z-z_0}\,dz = \frac{1}{2\pi i}\int_\gamma \frac{1}{z-z_0}\,dz.$$ $$\square$$

Lemma 34. Let $$\gamma:[a,b]\to\bbc$$ be a curve and let $$A=\gamma([a,b])$$. The function $$\alpha\mapsto\int_\gamma \frac{1}{z-\alpha}\,dz$$ is continuous on $$\bbc\setminus A$$.

Proof. Let $$\alpha_0\in\bbc\setminus A$$. The function $$t\mapsto|\alpha_0-\gamma(t)|$$ is positive and continuous on $$[a,b]$$, so it attains a minimum $$r > 0$$. For all $$|\alpha-\alpha_0| < r/2$$ and $$t\in[a,b]$$ we have $$|\alpha-\gamma(t)| \ge |\alpha_0-\gamma(t)|-|\alpha-\alpha_0| \ge r/2,$$ so \begin{align} \left\vert \int_\gamma\left(\frac{1}{z-\alpha}-\frac{1}{z-\alpha_0}\right)\,dz \right\vert &\le L(\gamma) \sup_{t\in[a,b]} \left\vert\frac{\alpha-\alpha_0}{(\gamma(t)-\alpha)(\gamma(t)-\alpha_0)}\right\vert \\ &\le L(\gamma)\frac{4}{r^2}|\alpha-\alpha_0| \\ &\to 0 \end{align} as $$\alpha\to\alpha_0$$. $$\square$$ Corollary 35. Let $$\gamma:[a,b]\to\bbc$$ be a closed curve and let $$A=\gamma([a,b])$$. If $$E$$ is a connected subset of $$\bbc\setminus A$$, then $$z\mapsto W(\gamma,z)$$ is constant on $$E$$. If $$E$$ is unbounded, then $$W(\gamma,z)=0$$ for all $$z\in E$$.

Proof. The first claim is clear. Let $$n$$ be the winding number of $$\gamma$$ with respect to any point of $$E$$. We have $$n=\frac{1}{2\pi i}\int_\gamma\frac{1}{\zeta-z}\,d\zeta$$ for all $$z\in E$$, so $$n=0$$ since $$\left\vert \int_\gamma\frac{1}{\zeta-z}\,d\zeta \right\vert \to 0$$ as $$|z|\to\infty$$. $$\square$$

We now come to the fundamental theorem that links singular homology and winding numbers. We will provide a proof later.

Theorem 36. Let $$U\subseteq\bbc$$ be an open set and let $$\gamma$$ be a 1-cycle in $$U$$. If $$W(\gamma,z)=0$$ for all $$z\in\bbc\setminus U$$, then $$\gamma$$ is a boundary, i.e. $$\gamma=\partial b$$ for some 2-chain $$b$$.

Corollary 37. Let $$\gamma,\eta$$ be 1-cycles in $$U$$. Then $$\gamma$$ and $$\eta$$ are homologous if and only if $$W(\gamma,z)=W(\eta,z)$$ for all $$z\in\bbc\setminus U$$.

Clearly, Theorem 30 (our goal) follows directly from Corollary 37.

If $$c=\sum_{i=1}^k c_i\sigma_i$$ is a $$p$$-chain where $$c_i\ne 0$$, we define the image of $$c$$ to be the set $$\bigcup_{i=1}^k \sigma_i(\triangle_p)$$.

Theorem 38 (Cauchy’s integral formula). Let $$U\subseteq\bbc$$ be an open set, let $$\gamma$$ be a 1-cycle in $$U$$ homologous to 0, and let $$f$$ be holomorphic on $$U$$. For all $$z\in U$$ not in the image of $$\gamma$$ we have $$W(\gamma,z)f(z)=\frac{1}{2\pi i}\int_\gamma\frac{f(\zeta)}{\zeta-z}\,d\zeta.$$

Proof. Write $$f(\zeta)=\sum_{n=0}^\infty a_n(\zeta-z)^n$$ in a neighborhood of $$z$$. Let $$C$$ be a small circle centered at $$z$$, contained in this neighborhood. By Theorem 30, \begin{align}
\frac{1}{2\pi i}\int_\gamma\frac{f(\zeta)}{\zeta-z}\,d\zeta &= \frac{1}{2\pi i}\int_{W(\gamma,z)C}\frac{f(\zeta)}{\zeta-z}\,d\zeta \\
&= \frac{1}{2\pi i}\sum_{n=0}^\infty \int_{W(\gamma,z)C} a_n(\zeta-z)^{n-1}\,d\zeta \\
&= a_0\frac{1}{2\pi i}\int_{W(\gamma,z)C} \frac{1}{\zeta-z}\,d\zeta \\
&= W(\gamma,z)f(z).
\end{align} $$\square$$

## Proof of the theorem

If $$\gamma:[a,b]\to\bbc$$ is a closed curve and there exists a partition $$\{a_0,\dots,a_k\}$$ of $$[0,1]$$ such that $$\gamma|_{[a_{j-1},a_j]}$$ is a horizontal or vertical line segment for each $$j$$, then we say that $$\gamma$$ is rectangular. A rectangular 1-cycle is a 1-cycle that can be written as a sum of rectangular closed curves. A grid is a union of finitely many vertical or horizontal lines in $$\bbc$$. Every grid partitions $$\bbc$$ into a finite number of rectangular regions, some bounded and some unbounded. Then it is clear that for any rectangular 1-cycle $$\gamma$$ there is a grid $$G$$ for which $$\gamma=\sum_{i=1}^k c_i\sigma_i$$, where each $$\sigma_i$$ is an edge of a bounded rectangle. We say that $$G$$ is a grid for $$\gamma$$.

Lemma 39. Let $$\gamma$$ be a rectangular 1-cycle in $$\bbc$$, let $$G$$ be a grid for $$\gamma$$, and let $$R_1,\dots,R_n$$ be the bounded rectangles. For each $$i$$, choose some $$p_i\in\Int R_i$$. Then $$\gamma=\sum_{i=1}^n W(\gamma,p_i)\partial R_i.$$ (Each $$\partial R_i$$ is oriented counterclockwise.)

Proof. Let $$\eta=\gamma-\sum_{i=1}^n W(\gamma,p_i)\partial R_i$$; it is clear that $$W(\eta,p)=0$$ for any $$p$$ not on the grid (i.e. not on the boundary of some bounded or unbounded rectangle). Suppose that $$\eta\ne 0$$ and write $$\eta=m\sigma+\eta’$$, where $$m\ne 0$$, $$\sigma$$ is an edge of a bounded rectangle $$R$$, and $$\eta’$$ is some 1-chain not containing $$\sigma$$. Then $$\sigma$$ is also an edge of exactly one other rectangle $$R’$$, which is either bounded or unbounded. Choose $$p\in\Int R$$ and $$p’\in\Int R’$$. Then $$W(\partial R,p)=1$$ and $$W(\partial R,p’)=0$$, so \begin{align}
W(\eta-m\partial R,p) &= W(\eta,p)-mW(\partial R,p)=-m, \\
W(\eta-m\partial R,p’) &= W(\eta,p’)-mW(\partial R,p’)=0.
\end{align} But the image $$E$$ of $$\eta-m\partial R$$ does not contain the edge $$\sigma$$, so $$p$$ and $$p’$$ are in the same connected component of $$\bbc\setminus E$$. Therefore $$W(\eta-m\partial R,p)=W(\eta-m\partial R,p’)$$ by Corollary 35, which is a contradiction. $$\square$$

Proof of Theorem 36. By using an argument similar to that of Theorem 12, we may assume that $$\gamma$$ is a rectangular 1-cycle in $$U$$. Let $$G$$ be a grid for $$\gamma$$, let $$R_1,\dots,R_n$$ be the bounded rectangles, and choose some $$p_i\in\Int R_i$$ for each $$i$$. By Lemma 39, we have $$\gamma=\sum_{i=1}^n W(\gamma,p_i)\partial R_i.$$ Suppose some $$R_i$$ contains a point $$p\in\bbc\setminus U$$; then $$W(\gamma,p)=0$$. If $$p\in\Int R_i$$, then $$W(\gamma,p_i)=W(\gamma,p)=0$$ since $$\Int R_i$$ is connected. If $$p\in\partial R_i$$ and $$p$$ is not in the image of $$\gamma$$, then again we have $$W(\gamma,p_i)=W(\gamma,p)=0$$. Note that $$p$$ cannot be in the image of $$\gamma$$. Therefore $$R_i\subseteq U$$ whenever $$W(\gamma,p_i)\ne 0$$, and $$\gamma$$ is the boundary of the 2-chain $$\sum_{i=1}^n W(\gamma,p_i) R_i.$$ $$\square$$

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