# Complex Analysis/Integration over chains

Definition (continuously differentiable 1-chain):
A continuously differentiable 1-chain is an element of the free $\mathbb {Z}$-module over the set of all continuously differentiable curves $\gamma :[0,1]\to \mathbb {C}$.
Definition (image):
Let $\Gamma$ be a continuously differentiable 1-chain. Then the image of $\Gamma$ is defined to be
    $\bigcup _{k=1}^{n}\gamma _{k}([0,1])$,
where $\gamma _{1},\ldots ,\gamma _{n}$ are precisely the continuous differentiable curves $\gamma :[0,1]\to \mathbb {C}$ such that $\Gamma (\gamma )\neq 0$.

## Argument and winding numbers

Assume we are given a closed contour supported in $\mathbb {C}$, and suppose that we are an observer located at the origin. Suppose we want to measure how often a moving object rotates about us (ie. passes through a point which is chosen fixed and has a fixed angle with respect to us). The resulting number is called the winding number of the given closed contour. Note though that it is signed; that is, if we contour were to travel (regarding angular distance) first round the circle, and then again but in reverse direction, the winding number is supposed to be zero.
To make this precise,
argument definition to circle and lift to standard covering
homotopy invariance of the latter
Theorem (Cauchy's formula):
Let $U\subseteq \mathbb {C}$ be open, and let $\Gamma$ be a cycle which is contained within $U$, and which is nullhomologous in $U$. Let also $f:U\to \mathbb {C}$ be a holomorphic function. Then we have
    $\int _{\Gamma }{\frac {f(z)}{z-z_{0}}}dz=\operatorname {ind} _{\Gamma }(z_{0})f(z_{0})$
for all $z_{0}$ that are not in the image of $\Gamma$
Proof: Define a function on $U\times U$ by
    $G(z,w):={\begin{cases}{\frac {f(z)-f(w)}{z-w}}&z\neq w\\f'(z)&z=w.\end{cases}}$
When one variable is kept fixed, this function is holomorphic in the other. Hence, upon considering the function
    $g(w):={\begin{cases}\int _{\Gamma }G(z,w)dz&w\in U\\0&w\notin U\end{cases}}$,
we find that $g$ is holomorphic in $U$ by exchanging integration and differentiation. But it is in fact holomorphic on $\mathbb {C}$, because we assumed the cycle $\Gamma$ to be nullhomologous in $U$. By shrinking $U$ if necessary, we may assume that $U$ is bounded, since the image of a curve is compact and finite unions of compact sets are compact. Then $g$ becomes a bounded function by a Weierstraß-type theorem and by Liouville's theorem it is then constant, and hence equal to zero. In particular, inserting $w=z_{0}$, we get
    $\int _{\Gamma }{\frac {f(z)-f(z_{0})}{z-z_{0}}}dz=0$,
that is,
    $\int _{\Gamma }{\frac {f(z)}{z-z_{0}}}dz=\int _{\Gamma }{\frac {f(z_{0})}{z-z_{0}}}dz=f(z_{0})\operatorname {ind} _{\Gamma }(z_{0})$. $\Box$
Definition (chain integral):
Let $U\subseteq \mathbb {C}$ be open and let $f:U\to \mathbb {C}$ be holomorphic. Let $\Gamma$ be a continuously differentiable 1-chain whose image is contained within $U$. Then the integral over $\Gamma$ is defined to be
    $\int _{\Gamma }f(z)dz:=\sum _{k=1}^{n}a_{k}\int _{\gamma _{k}}f(z)dz$,
where
    $\Gamma =\sum _{k=1}^{n}a_{k}\gamma _{k}$.
Proposition (homologous chains induce equal integrals):
Let $U\subseteq \mathbb {C}$ be open, and let $f:U\to \mathbb {C}$ be a holomorphic function. Suppose that $\Gamma _{1},\Gamma _{2}$ are continuously differentiable 1-chains, whose image is contained within $U$, such that $\Gamma _{1}-\Gamma _{2}=\delta \Sigma$ for some 2-chain $\Sigma$ in the [[singular chain complex over $\mathbb {Z}$]]. Then
    $\int _{\Gamma _{1}}f(z)dz=\int _{\Gamma _{2}}f(z)dz$.
Proof: By definition of integration over a chain, it suffices to prove that whenever $\Sigma$ is a 2-chain, then
    $\int _{\delta \Sigma }f(z)dz=0$.
Moreover, by linearity we may restrict to the case where $\Sigma :\Delta ^{2}\to U$ is a simplex. But $\delta \Sigma$ is nullhomologous, so that
    $\int _{\delta \Sigma }f(z)dz=0$
by Cauchy's theorem. $\Box$
Theorem (Residue theorem):
Let $U\subseteq \mathbb {C}$ be an open, bounded subset. Let $\Gamma$ be a cycle whose image is contained within $U$, and let $f:U\to \mathbb {C}$ be meromorphic, so that no singularity of $f$ is contained within the image of $\Gamma$. Then
    $\int _{\Gamma }f(z)dz=2\pi i\sum _{k=1}^{n}\operatorname {ind} _{\Gamma }(s_{k})\operatorname {res} _{s_{k}}(f)$,
where $s_{1},\ldots ,s_{n}\in U$ are the singularities of $f$.
Proof: Note that the image of any continuous 1-chain of $\mathbb {C}$ is compact, hence closed since $\mathbb {C}$ is Hausdorff. Hence, for each singularity $s_{j}$ of $f$, choose a radius $r_{j}>0$ such that the image of $\Gamma$ does not intersect ${\overline {B_{r_{j}}(s_{j})}}$, and the latter set shall also be contained in $U$ (which is open, after all). Moreover, set $\gamma _{j}:=\operatorname {ind} _{\Gamma }(s_{k})\partial B_{r_{j}}(s_{j})$, where the latter boundary path is traversed once and counterclockwise (so that its winding number is one). Then define a new continuously differentiable 1-chain by
    $\Delta :=\Gamma -\sum _{j=1}^{n}\gamma _{j}$.
Then $\Delta$ will be nullhomologous, so by Cauchy's theorem and Cauchy's formula
    $0=\int _{\Delta }f(z)dz=\int _{\Gamma }f(z)dz-2\pi i\sum _{k=1}^{n}\operatorname {ind} _{\Gamma }(s_{k})\operatorname {res} _{s_{k}}(f)$. $\Box$