# 复分析代考_Complex function代考_MATH3401

## 复分析代考_Complex function代考_Simple Connectivity

If $\gamma$ is a closed curve in $\mathbb{C}$, then the winding number, or index, $\operatorname{Ind}\gamma(a)$ is a continuous function of the point $a \in \mathbb{C} \backslash \gamma$. In particular, if $\gamma:[0,1] \rightarrow U$ is a closed curve in an open and connected set $U$, then $\operatorname{Ind}\gamma(a)$ is continuous on $\mathbb{C} \backslash U$. It follows that the winding number $\operatorname{Ind}\gamma(a)$ is constant on each connected component of $\mathbb{C} \backslash U$. In particular, if $C_1$ is an unbounded component of $\mathbb{C} \backslash U$, then $\operatorname{Ind}\gamma(a)=0$ for all $a \in C_1$ (cf. Exercise 9 ). If $U$ is bounded, then $\mathbb{C} \backslash U$ has exactly one unbounded component. Hence if $U$ is bounded and $\mathbb{C} \backslash U$ has only one component, then that component is unbounded and Ind $_\gamma(a) \equiv 0$ for all $a \in \mathbb{C} \backslash U$.

In Section 11.2, we saw that if $U$ is simply connected, then $\operatorname{Ind}\gamma(a)=0$ for all $a \in \mathbb{C} \backslash U$ (Lemma 11.2.3). It is natural to suspect that there is some relationship between the two possible reasons why $\operatorname{Ind}\gamma(a)=0$ for all $a \in$ $\mathbb{C} \backslash U$, one reason being that $U$ is simply connected, the other that $\mathbb{C} \backslash U$ has only one component. [This intuition is increased by noting that the annulus is not simply connected and its complement has two components. Moreover the curves which are not homotopic to constants are ones that, intuitively, go around the bounded component of the complement of the domain.] It is, in fact, true that these conditions are related and indeed equivalent for bounded domains. The following theorem makes the expectation precise:
Theorem 11.4.1. If $U$ is a bounded, open, connected subset of $\mathbb{C}$, then the following properties of $U$ are equivalent:
(a) $U$ is simply connected;
(b) $\mathbb{C} \backslash U$ is connected;
(c) for each closed curve $\gamma$ in $U$ and $a \in \mathbb{C} \backslash U, \operatorname{Ind}\gamma(a)=0$. We already know, from Section 11.2, that (a) implies (c). Also, (b) implies (c) by the reasoning already noted: If $\mathbb{C} \backslash U$ is connected, then Ind $\gamma(a)$, being continuous and integer-valued, is constant on $\mathbb{C} \backslash U$. But Ind $_\gamma(a)=0$ for all $a \in \mathbb{C} \backslash U$ with $|a|>\sup {|z|: z \in U}$ by Exercise 9 . Therefore (c) holds if (b) does.

## 复分析代考_Complex function代考_Multiply Connected Domains Revisited

We developed earlier the intuitive idea that a (topologically) simply connected domain was one with “no holes.” This chapter has given that notion a precise form: (i) the absence of a hole in the sense of there being no points in the complement for a closed curve in the domain to wrap around is exactly the same as (ii) the absence of a hole in the sense of there being no bounded component of the complement of the (bounded) domain. But what about domains that do have holes, that is, that do have nonempty bounded components of their complement?

It is natural, in case the complement $\mathbb{C} \backslash U$ of a bounded domain $U$ has a finite number $k$ of bounded components, to say that $U$ has $k$ holes. [As before, $\mathbb{C} \backslash U$ has exactly one unbounded component.] In more customary if less picturesque terminology, $U$ is said to have connectivity $k+1$ (i.e., the number of components of $\mathbb{C} \backslash U$ ). Then it is also tempting to ask whether the number of holes can be determined by looking at $\pi_1(U)$. For $k=0$, we know this to be true: $U$ has no holes, that is, $k=0$ if and only if $\pi_1(U)$ is trivial, that is, $\pi_1(U)$ equals the group with one element (the identity). But can we tell the difference between a domain with, say, two holes and a domain with three holes just by examining their fundamental groups? We know that the fundamental groups will then both be nontrivial, but do they somehow encode exactly how many holes there are?

The answer to this question is “yes.” But this assertion is not so easy to prove. The whole situation is really part of the subject of algebraic topology, not of complex analysis. Therefore we do not want to go into all the details. But, just so you will know the facts, we shall summarize the key points that topologists have established:

If $U$ is a bounded domain with $k$ holes (i.e., $\mathbb{C} \backslash U$ has $k$ bounded components), then $U$ is homeomorphic to the open disc with $k$ points removed. The disc with $k$ points removed is not homeomorphic to the disc with $\ell$ points removed if $k \neq \ell$. Indeed, if $k \neq \ell$, then the disc with $k$ points removed has a different fundamental group than the disc with $\ell$ points removed. [Here “different” means that the two groups are not isomorphic as groups.]

# 复分析代考

## 复分析代考_Complex function代考_Simple Connectivity

(a) $U$ 是简单连接的;
(乙) $\mathbb{C} \backslash U$ 已连接;
(c) 对于每条闭合曲线 $\gamma$ 在 $U$ 和 $a \in \mathbb{C} \backslash U$, Ind $\gamma(a)=0$. 从 $11.2$ 节我们已经知道，(a) 蕴含 (c)。此外，
(b) 通过已经提到的推理暗示 (c) : 如果 $\mathbb{C} \backslash U$ 连接，则 $\operatorname{lnd} \gamma(a)$ ，连续且整数值，在 $\mathbb{C} \backslash U$. 但印度 $\gamma(a)=0$ 对所有人 $a \in \mathbb{C} \backslash U$ 和 $|a|>\sup |z|: z \in U$ 通过练习 9。因此，如果 (b) 成立，则 (c) 成立。

## 复分析代考_Complex function代考_Multiply Connected Domains Revisited

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