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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!!