# Math416 Lecture 22

## Chapter 9: Generalized Cauchy Theorem

### Winding numbers

Definition:

Let $\gamma:[a,b]\to\mathbb{C}$ be a closed curve. The **winding number** of $\gamma$ around $z\in\mathbb{C}$ is defined as

$$
\frac{1}{2\pi i}\Delta(arg(z-z_0),\gamma)
$$

where $\Delta(arg(z-z_0),\gamma)$ is the change in the argument of $z-z_0$ along $\gamma$.

#### Interior of curve

The interior of $\gamma$ is the set of points $z\in\mathbb{C}$ such that the winding number of $\gamma$ around $z$ is non-zero.

$$
int_\gamma(z)=\{z\in\mathbb{C}|\frac{1}{2\pi i}\Delta(arg(z-z_0),\gamma)\neq 0\}
$$

#### Contour

The winding number of a contour $\Gamma$ around $z$ is the sum of the winding numbers of the contours $\gamma_j$ around $z$.

$$
ind_\Gamma(z)=\sum_{j=1}^nn_j ind_{\gamma_j}(z)
$$

A contour is simple if $ind_\gamma(z)=\{0,1\}$ for all $z\in\mathbb{C}\setminus\gamma([a,b])$.

#### Separation lemma

Let $\Omega\subseteq \mathbb{C}$ be open, let $K\subset \Omega$ be compact, then $\exists$ a simple contour $\Gamma\subset \Omega\setminus K$ such that

$$
K\subset int_\Gamma(\Gamma)\subset \Omega
$$

Proof:

First we show that $\exists$ a simple contour $\Gamma\subset \Omega\setminus K$

Let $0<\delta<dist(K,\partial\Omega)$.

We draw a grid fo horizontal ad vertical lines each separated from each other by $\delta$.

Let $S_1,S_2,\dots,S_n$ be the squares that intersect $K$.

Let $\sigma_j$ be the boundary of $S_j$ traversed in counterclockwise direction.

Let $\varepsilon$ be the set of edges with exactly one $s_j$ for $j=1,2,\dots,q$.

Note that $\varepsilon\subseteq \Omega\setminus K$.

We claim that $\varepsilon$ forms a contour.

Proof of Claim:

Say a sequence of edges $E_1,E_2,\dots,E_p$, $E_i\in \varepsilon$. from a chain if terminal points of $E_k$ is the initial point of $E_{k+1}$ for $1\leq k\leq p-1$.

Say it forms a cycle if inaddition the terminal points of $E_p$ is the initial point of $E_1$.

Any cycle is a piecewise continuous closed curve.

We want to show that $\varepsilon$ is a disjoint union of cycles.

We can prove that every terminal point of an edge in $\varepsilon$ is an initial point of an edge in $\varepsilon$. By case analysis for the state of the four square around the terminal point.

Let $\gamma=(E_1,E_2,\dots,E_p)$ be a maximal cycle in $\varepsilon$. (Maximal means that we cannot add another edge to it while still having a cycle.)

Then $\gamma$ is a cycle.

Look at the terminal point of $E_p$, This is initial point for some edge $E'$, where $E'$ is one of the edges of $\gamma$. 

If $E'$ is not $E_1$, then we can add $E'$ to $\gamma$ to form a larger cycle. Contradiction. (You can do this by case analysis. If there is three edges, then there must be four.)

Thus $E'=E_1$.

Thus $\gamma$ is a cycle.

We can now remove $\gamma$ from $\varepsilon$ to form a new set $\varepsilon'$.

We can repeat this process to form a disjoint union of cycles using induction.

Second, we show that $int_\Gamma(\Gamma)\subset \Omega$.

Let $z_0\in int(S_j)$ for some $1\leq j\leq q$.

$ind_{S_k}(z_0)=\begin{cases}
1 & k=j\\
0 & k\neq j
\end{cases}$

Thus 

$$
\begin{aligned}
\sum_{k=1}^q ind_{S_k}(z_0)&=\sum_{k=1}^q \frac{1}{2\pi i}\int_{\partial S_k}\frac{1}{z-z_0}dz\\
&=1\\
&=ind_\Gamma(z_0)
\end{aligned}
$$

So if $z_0\in int(\bigcup_{j=1}^q S_j)$, then $ind_\Gamma(z_0)=1$.

And $\bigcup_{j=1}^q S_j\supset K$, so $z_0\in int_\Gamma(K)$.

Let $z_1\in\mathbb{C}\setminus\left(\bigcup_{j=1}^q S_j\cup\Gamma\right)$.

Then $ind_{S_k}(z_1)=0$ for all $1\leq k\leq q$.

Thus $\sum_{k=1}^q ind_{S_k}(z_1)=0=ind_\Gamma(z_1)$.

QED

Continue on Generalized Cauchy Theorem next time!!