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Math416Complex Variables (Lecture 22)

Math416 Lecture 22

Chapter 9: Generalized Cauchy Theorem

Winding numbers

Definition:

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

12πiΔ(arg(zz0),γ)\frac{1}{2\pi i}\Delta(arg(z-z_0),\gamma)

where Δ(arg(zz0),γ)\Delta(arg(z-z_0),\gamma) is the change in the argument of zz0z-z_0 along γ\gamma.

Interior of curve

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

intγ(z)={zC12πiΔ(arg(zz0),γ)0}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 Γ\Gamma around zz is the sum of the winding numbers of the contours γj\gamma_j around zz.

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

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

Separation lemma

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

KintΓ(Γ)ΩK\subset int_\Gamma(\Gamma)\subset \Omega

Proof:

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

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

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

Let S1,S2,,SnS_1,S_2,\dots,S_n be the squares that intersect KK.

Let σj\sigma_j be the boundary of SjS_j traversed in counterclockwise direction.

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

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

We claim that ε\varepsilon forms a contour.

Proof of Claim:

Say a sequence of edges E1,E2,,EpE_1,E_2,\dots,E_p, EiεE_i\in \varepsilon. from a chain if terminal points of EkE_k is the initial point of Ek+1E_{k+1} for 1kp11\leq k\leq p-1.

Say it forms a cycle if inaddition the terminal points of EpE_p is the initial point of E1E_1.

Any cycle is a piecewise continuous closed curve.

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

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

Let γ=(E1,E2,,Ep)\gamma=(E_1,E_2,\dots,E_p) be a maximal cycle in ε\varepsilon. (Maximal means that we cannot add another edge to it while still having a cycle.)

Then γ\gamma is a cycle.

Look at the terminal point of EpE_p, This is initial point for some edge EE', where EE' is one of the edges of γ\gamma.

If EE' is not E1E_1, then we can add EE' to γ\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=E1E'=E_1.

Thus γ\gamma is a cycle.

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

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

Second, we show that intΓ(Γ)Ωint_\Gamma(\Gamma)\subset \Omega.

Let z0int(Sj)z_0\in int(S_j) for some 1jq1\leq j\leq q.

indSk(z0)={1k=j0kjind_{S_k}(z_0)=\begin{cases} 1 & k=j\\ 0 & k\neq j \end{cases}

Thus

k=1qindSk(z0)=k=1q12πiSk1zz0dz=1=indΓ(z0)\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 z0int(j=1qSj)z_0\in int(\bigcup_{j=1}^q S_j), then indΓ(z0)=1ind_\Gamma(z_0)=1.

And j=1qSjK\bigcup_{j=1}^q S_j\supset K, so z0intΓ(K)z_0\in int_\Gamma(K).

Let z1C(j=1qSjΓ)z_1\in\mathbb{C}\setminus\left(\bigcup_{j=1}^q S_j\cup\Gamma\right).

Then indSk(z1)=0ind_{S_k}(z_1)=0 for all 1kq1\leq k\leq q.

Thus k=1qindSk(z1)=0=indΓ(z1)\sum_{k=1}^q ind_{S_k}(z_1)=0=ind_\Gamma(z_1).

QED

Continue on Generalized Cauchy Theorem next time!!

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