### 数学代写|复变函数作业代写Complex function代考|Meromorphic Functions and Residues

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## 数学代写|复变函数作业代写Complex function代考|The Behavior of a Holomorphic Function

In the proof of the Cauchy integral formula in Section 2.4, we saw that it is often important to consider a function that is holomorphic on a punctured open set $U \backslash{P} \subset \mathbb{C}$. The consideration of a holomorphic function with such an “isolated singularity” turns out to occupy a central position in much of the subject. These singularities can arise in various ways. Perhaps the most obvious way occurs as the reciprocal of a holomorphic function, for instance passing from $z^{j}$ to $1 / z^{j}, j$ a positive integer. More complicated examples can be generated, for instance, by exponentiating the reciprocals of holomorphic functions: for example, $e^{1 / z}, z \neq 0$.

In this chapter we shall study carefully the behavior of holomorphic functions near a singularity. In particular, we shall obtain a new kind of infinite series expansion which generalizes the idea of the power series expansion of a holomorphic function about a (nonsingular) point. We shall in the process completely classify the behavior of holomorphic functions near an isolated singular point.

Let $U \subseteq \mathbb{C}$ be an open set and $P \in U$. Suppose that $f: U \backslash{P} \rightarrow \mathbb{C}$ is holomorphic. In this situation we say that $f$ has an isolated singular point (or isolated singularity) at $P$. The implication of the phrase is usually just that $f$ is defined and holomorphic on some such “deleted neighborhood” of $P$. The specification of the set $U$ is of secondary interest; we wish to consider the behavior of $f$ “near $P^{\prime \prime}$.

There are three possibilities for the behavior of $f$ near $P$ that are worth distinguishing:
(i) $|f(z)|$ is bounded on $D(P, r) \backslash{P}$ for some $r>0$ with $D(P, r) \subseteq$ $U$; that is, there is some $r>0$ and some $M>0$ such that $|f(z)| \leq$ $M$ for all $z \in U \cap D(P, r) \backslash{P}$.
(ii) $\lim _{z \rightarrow P}|f(z)|=+\infty$.
(iii) Neither (i) nor (ii) applies.
Of course this classification does not say much unless we can find some other properties of $f$ related to (i), (ii), and (iii). We shall prove momentarily that if case (i) holds, then $f$ has a limit at $P$ which extends $f$ so that it is holomorphic on all of $U$. It is commonly said in this circumstance that $f$ has a removable singularity at $P$. In case (ii), we will say that $f$ has a pole at $P$. In case (iii), $f$ will be said to have an essential singularity at $P$. Our goal in this and the next section is to understand (i), (ii), and (iii) in some further detail.

Theorem 4.1.1 (The Riemann removable singularities theorem). Let $f$ : $D(P, r) \backslash{P} \rightarrow \mathbb{C}$ be holomorphic and bounded. Then
(1) $\lim {z \rightarrow P} f(z)$ exists; (2) the function $\widehat{f}: D(P, r) \rightarrow \mathbb{C}$ defined by $$\widehat{f}(z)=\left{\begin{array}{lll} f(z) & \text { if } & z \neq P \ \lim {\zeta \rightarrow P} f(\zeta) & \text { if } & z=P \end{array}\right.$$
is holomorphic.

## 数学代写|复变函数作业代写Complex function代考|Expansion around Singular Points

To aid in our further understanding of poles and essential singularities, we are going to develop a method of series expansion of holomorphic functions on $D(P, r) \backslash{P}$. Except for removable singularities, we cannot expect to expand such a function in a power series convergent in a neighborhood of $P$, since such a power series would define a holomorphic function on a whole neighborhood of $P$, including $P$ itself. A natural extension of the idea of power series is to allow negative as well as positive powers of $(z-P)$. This extension turns out to be enough to handle poles and essential singularities

both. That it works well for poles is easy to see; essential singularities take a bit more work. We turn now to the details.
A Laurent series on $D(P, r)$ is a (formal) expression of the form
$$\sum_{j=-\infty}^{+\infty} a_{j}(z-P)^{j}$$
Note that the individual terms are each defined for all $z \in D(P, r) \backslash{P}$.
To discuss Laurent series in terms of convergence, we must first make a general agreement as to the meaning of the convergence of a “doubly infinite” series $\sum_{j=-\infty}^{+\infty} \alpha_{j}$. We say that such a series converges if $\sum_{j=0}^{+\infty} \alpha_{j}$ and $\sum_{j=1}^{+\infty} \alpha_{-j}$ converge in the usual sense. In this case, we set
$$\sum_{-\infty}^{+\infty} \alpha_{j}=\left(\sum_{j=0}^{+\infty} \alpha_{j}\right)+\left(\sum_{j=1}^{+\infty} \alpha_{-j}\right)$$
You can check easily that $\sum_{-\infty}^{+\infty} \alpha_{j}$ converges to a complex number $\sigma$ if and only if for each $\epsilon>0$ there is an $N>0$ such that, if $\ell \geq N$ and $k \geq N$, then $\left|\left(\sum_{j=-k}^{\ell} \alpha_{j}\right)-\sigma\right|<\epsilon$. It is important to realize that $\ell$ and $k$ are independent here. [In particular, the existence of the limit $\lim {k \rightarrow+\infty} \sum{j=-k}^{+k} \alpha_{j}$ does not imply in general that $\sum_{-\infty}^{+\infty} \alpha_{j}$ converges. See Exercises 10 and 11.]

With these convergence ideas in mind, we can now present the analogue for Laurent series of Lemmas $3.2 .3$ and $3.2 .5$ for power series.

Lemma 4.2.1. If $\sum_{j=-\infty}^{+\infty} a_{j}(z-P)^{j}$ converges at $z_{1} \neq P$ and at $z_{2} \neq P$ and if $\left|z_{1}-P\right|<\left|z_{2}-P\right|$, then the series converges for all $z$ with $\left|z_{1}-P\right|<$ $|z-P|<\left|z_{2}-P\right|$.

Refer to Figure $4.1$ for an illustration of the situation described in the Lemma.

Proof of Lemma 4.2.1. If $\sum_{j=-\infty}^{+\infty} a_{j}\left(z_{2}-P\right)^{j}$ converges, then the definition of convergence of a doubly infinite sum implies that $\sum_{j=0}^{+\infty} a_{j}\left(z_{2}-P\right)^{j}$ converges. By Lemma $3.2 .3, \sum_{j=0}^{+\infty} a_{j}(z-P)^{j}$ then converges when $|z-P|<$ $\left|z_{2}-P\right|$. If $\sum_{j=-\infty}^{+\infty} a_{j}\left(z_{1}-P\right)^{j}$ converges, then so does $\sum_{j=1}^{+\infty} a_{-j}\left(z_{1}-P\right)^{-j}$. Since $0<\left|z_{1}-P\right|<|z-P|$, it follows that $|1 /(z-P)|<\left|1 /\left(z_{1}-P\right)\right|$. Hence Lemma 3.2.3 again applies to show that $\sum_{j=1}^{+\infty} a_{-j}(z-P)^{-j}$ converges. Thus $\sum_{-\infty}^{+\infty} a_{j}(z-P)^{j}$ converges when $\left|z_{1}-P\right|<|z-P|<\left|z_{2}-P\right|$.

## 数学代写|复变函数作业代写Complex function代考|Existence of Laurent Expansions

We turn now to establishing that convergent Laurent expansions of functions holomorphic on an annulus do in fact exist. We will require the following result.

Theorem 4.3.1 (The Cauchy integral formula for an annulus). Suppose that $0 \leq r_{1}<r_{2} \leq+\infty$ and that $f: D\left(P, r_{2}\right) \backslash \bar{D}\left(P, r_{1}\right) \rightarrow \mathbb{C}$ is holomorphic. Then, for each $s_{1}, s_{2}$ such that $r_{1}<s_{1}<s_{2}<r_{2}$ and each $z \in D\left(P, s_{2}\right) \backslash$ $\bar{D}\left(P, s_{1}\right)$, it holds that
$$f(z)=\frac{1}{2 \pi i} \oint_{|\zeta-P|=s_{2}} \frac{f(\zeta)}{\zeta-z} d \zeta-\frac{1}{2 \pi i} \oint_{|\zeta-P|=s_{1}} \frac{f(\zeta)}{\zeta-z} d \zeta .$$
Proof. Fix a point $z \in D\left(P, s_{2}\right) \backslash \bar{D}\left(P, s_{1}\right)$. Define, for $\zeta \in D\left(P, r_{2}\right) \backslash$ $\bar{D}\left(P, r_{1}\right)$,
$$g_{z}(\zeta)= \begin{cases}\frac{f(\zeta)-f(z)}{\zeta-z} & \zeta \neq z \ f^{\prime}(z) & \zeta=z\end{cases}$$
Then $g_{z}$ is a holomorphic function of $\zeta, \zeta \in D\left(P, r_{2}\right) \backslash \bar{D}\left(P, r_{1}\right)$ (by the Riemann removable singularities theorem).
Now we consider the integrals
$$\oint_{|\zeta-P|=s_{1}} g_{z}(\zeta) d \zeta$$
and
$$\oint_{|\zeta-P|=s_{2}} g_{z}(\zeta) d \zeta$$
By the considerations in Section 2.6, these two. integrals are equal. So
$$0=\oint_{|\zeta-P|=s_{2}} g_{z}(\zeta) d \zeta-\oint_{|\zeta-P|=s_{1}} g_{z}(\zeta) d \zeta$$

$$=\oint_{|\zeta-P|=s_{2}} \frac{f(\zeta)-f(z)}{\zeta-z} d \zeta-\oint_{|\zeta-P|=s_{1}} \frac{f(\zeta)-f(z)}{\zeta-z} d \zeta$$
Hence
\begin{aligned} &\oint_{|\zeta-P|=s_{2}} \frac{f(\zeta)}{\zeta-z} d \zeta-\oint_{|\zeta-P|=s_{2}} \frac{f(z)}{\zeta-z} d \zeta \ &=\oint_{|\zeta-P|=s_{1}} \frac{f(\zeta)}{\zeta-z} d \zeta-\oint_{|\zeta-P|=s_{1}} \frac{f(z)}{\zeta-z} d \zeta \end{aligned}
Now
$$\oint_{|\zeta-P|=s_{2}} \frac{f(z)}{\zeta-z} d \zeta=f(z) \oint_{|\zeta-P|=s_{2}} \frac{1}{\zeta-z} d \zeta=2 \pi i f(z)$$
by the Cauchy integral formula for the constant function 1 on $D\left(P, r_{2}\right)$ (or by direct calculation).
Also
$$\oint_{|\zeta-P|=s_{1}} \frac{f(z)}{\zeta-z} d \zeta=f(z) \oint_{|\zeta-P|=s_{1}} \frac{1}{\zeta-z} d \zeta=0 .$$
This can be seen from the Cauchy integral theorem (Theorem 2.4.3) since $1 /(\zeta-z)$ is holomorphic for $\zeta \in D(P,|z-P|)$ and $\left{\zeta:|\zeta-P| \leq s_{1}\right} \subseteq$ $D(P,|z-P|)$. See Figure $4.2$.
So
$$2 \pi i f(z)=\oint_{|\zeta-P|=s_{2}} \frac{f(\zeta)}{\zeta-z} d \zeta-\oint_{|\zeta-P|=s_{1}} \frac{f(\zeta)}{\zeta-z} d \zeta$$
as desired.

## 数学代写|复变函数作业代写Complex function代考|The Behavior of a Holomorphic Function

(i)|F(和)|有界D(磷,r)∖磷对于一些r>0和D(磷,r)⊆ 在; 也就是说，有一些r>0还有一些米>0这样|F(和)|≤ 米对全部和∈在∩D(磷,r)∖磷.
(二)林和→磷|F(和)|=+∞.
(iii) (i) 或 (ii) 均不适用。

（一）林和→磷F(和)存在；(2) 功能F^:D(磷,r)→C由 $$\widehat{f}(z)=\left{ 定义F(和) 如果 和≠磷 林G→磷F(G) 如果 和=磷\对。$$

## 数学代写|复变函数作业代写Complex function代考|Expansion around Singular Points

∑j=−∞+∞一种j(和−磷)j

∑−∞+∞一种j=(∑j=0+∞一种j)+(∑j=1+∞一种−j)

## 数学代写|复变函数作业代写Complex function代考|Existence of Laurent Expansions

F(和)=12圆周率一世∮|G−磷|=s2F(G)G−和dG−12圆周率一世∮|G−磷|=s1F(G)G−和dG.

G和(G)={F(G)−F(和)G−和G≠和 F′(和)G=和

∮|G−磷|=s1G和(G)dG

∮|G−磷|=s2G和(G)dG

0=∮|G−磷|=s2G和(G)dG−∮|G−磷|=s1G和(G)dG=∮|G−磷|=s2F(G)−F(和)G−和dG−∮|G−磷|=s1F(G)−F(和)G−和dG

∮|G−磷|=s2F(G)G−和dG−∮|G−磷|=s2F(和)G−和dG =∮|G−磷|=s1F(G)G−和dG−∮|G−磷|=s1F(和)G−和dG

∮|G−磷|=s2F(和)G−和dG=F(和)∮|G−磷|=s21G−和dG=2圆周率一世F(和)

∮|G−磷|=s1F(和)G−和dG=F(和)∮|G−磷|=s11G−和dG=0.

2圆周率一世F(和)=∮|G−磷|=s2F(G)G−和dG−∮|G−磷|=s1F(G)G−和dG

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