# 数学代写|复分析作业代写Complex function代考|MATH3979

#### Doug I. Jones

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## 数学代写|复分析作业代写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
$$\sum_{j=-\infty}^{+\infty} a_{j}(z-P)^{j} \text {. }$$
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
$$\sum_{-\infty}^{+\infty} \alpha_{j}=\left(\sum_{j=0}^{+\infty} \alpha_{j}\right)+\left(\sum_{j=1}^{+\infty} \alpha_{-j}\right) .$$

## 数学代写|复分析作业代写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
$$0=\oint_{|\zeta-P|=s_{2}} g_{z}(\zeta) d \zeta-\oint_{|\zeta-P|=s_{1}} g_{z}(\zeta) d \zeta$$

# 复分析代写

## 数学代写|复分析作业代写Complex function代考|Expansion around Singular Points

$$\sum_{j=-\infty}^{+\infty} a_{j}(z-P)^{j}$$

$$\sum_{-\infty}^{+\infty} \alpha_{j}=\left(\sum_{j=0}^{+\infty} \alpha_{j}\right)+\left(\sum_{j=1}^{+\infty} \alpha_{-j}\right) .$$

## 数学代写|复分析作业代写Complex function代考|Existence of Laurent Expansions

$$f(z)=\frac{1}{2 \pi i} \oint_{|\zeta-P|=s_{2}} \frac{f(\zeta)}{\zeta-z} d \zeta-\frac{1}{2 \pi i} \oint_{|\zeta-P|=s_{1}} \frac{f(\zeta)}{\zeta-z} d \zeta$$

$$g_{z}(\zeta)=\left{\frac{f(\zeta)-f(z)}{\zeta-z} \quad \zeta \neq z f^{\prime}(z) \quad \zeta=z\right.$$

$$\oint_{|\zeta-P|=s_{1}} g_{z}(\zeta) d \zeta$$

$$\oint_{|\zeta-P|=s_{2}} g_{z}(\zeta) d \zeta$$

$$0=\oint_{|\zeta-P|=s_{2}} g_{z}(\zeta) d \zeta-\oint_{|\zeta-P|=s_{1}} g_{z}(\zeta) d \zeta$$

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