### 数学代写|复分析作业代写Complex function代考|KMA152

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|复分析作业代写Complex function代考|Counting Zeros and Poles

Suppose that $f: U \rightarrow \mathbb{C}$ is a holomorphic function on a connected, open set $U \subseteq \mathbb{C}$ and that $\bar{D}(P, r) \subseteq U$. We know from the Cauchy integral formula that the values of $f$ on $D(P, r)$ are completely determined by the values of $f$ on $\partial D(P, r)$. In particular, the number and even the location of the zeros of $f$ in $D(P, r)$ are determined in principle by $f$ on $\partial D(P, r)$. But it is nonetheless a pleasant surprise that there is a simple formula for the number of zeros of $f$ in $D(P, r)$ in terms of $f$ (and $f^{\prime}$ ) on $\partial D(P, r)$. In order to construct this formula, we shall have to agree to count zeros in a particular fashion. This method of counting will in fact be a generalization of the notion of counting the zeros of a polynomial according to multiplicity. We now explain the precise idea.

Let $f: U \rightarrow \mathbb{C}$ be holomorphic as before, and assume that $f$ has zeros but that $f$ is not identically zero. Fix $z_{0} \in U$ such that $f\left(z_{0}\right)=0$. Since the zeros of $f$ are isolated, there is an $r>0$ such that $\bar{D}\left(z_{0}, r\right) \subseteq U$ and such that $f$ does not vanish on $\bar{D}\left(z_{0}, r\right) \backslash\left{z_{0}\right}$.

Now the power series expansion of $f$ about $z_{0}$ has a first nonzero term determined by the least positive integer $n$ such that $f^{(n)}\left(z_{0}\right) \neq 0$. [Note that $n \geq 1$ since $f\left(z_{0}\right)=0$ by hypothesis.] Thus the power series expansion of $f$ about $z_{0}$ begins with the $n^{\text {th }}$ term:
$$f(z)=\sum_{j=n}^{\infty} \frac{1}{j !} \frac{\partial^{j} f}{\partial z^{j}}\left(z_{0}\right)\left(z-z_{0}\right)^{j} .$$
Under these circumstances we say that $f$ has a zero of order $n$ (or multiplicity $n$ ) at $z_{0}$. When $n=1$, then we say that $z_{0}$ is a simple zero of $f$.

## 数学代写|复分析作业代写Complex function代考|The Local Geometry of Holomorphic Functions

The argument principle for holomorphic functions (the formula of Proposition 5.1.2) has a consequence which is one of the most important facts about holomorphic functions considered as geometric mappings:

Theorem 5.2.1 (The open mapping theorem). If $f: U \rightarrow \mathbb{C}$ is a nonconstant holomorphic function on a connected open set $U$, then $f(U)$ is an open set in $\mathbb{C}$.

Before beginning the proof of the theorem, we discuss its significance. The theorem says, in particular, that if $U \subseteq \mathbb{C}$ is connected and open and if $f: U \rightarrow \mathbb{C}$ is holomorphic, then either $f(U)$ is a connected open set (the nonconstant case) or $f(U)$ is a single point. There is no analogous result for $C^{\infty}$, or even real analytic functions from $\mathbb{C}$ to $\mathbb{C}$ (or from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$ ). As an example, consider the function
\begin{aligned} g: \mathbb{C} & \rightarrow \mathbb{C} \ z & \mapsto|z|^{2} . \end{aligned}
The domain of $g$ is the entire plane $\mathbb{C}$, which is certainly open and connected. The set $g(\mathbb{C})$, however, is ${x+i 0: \mathbb{R} \ni x \geq 0}$ which is not open as a subset of $\mathbb{C}$. The function $g$ is in fact real analytic, but of course not holomorphic.
Note, by contrast, that the holomorphic function
\begin{aligned} g: \mathbb{C} & \rightarrow \mathbb{C} \ z & \mapsto z^{2} \end{aligned}
has image the entire complex plane (which is, of course, an open set). More significantly, every open subset of $\mathbb{C}$ has image under $g$ which is open.
In the subject of topology, a function $f$ is defined to be continuous if the inverse image of any open set under $f$ is also open. In contexts where the $\epsilon-\delta$ definition makes sense, the $\epsilon-\delta$ definition is equivalent to the inverse-image-of-open-sets definition.

## 数学代写|复分析作业代写Complex function代考|Counting Zeros and Poles

$$f(z)=\sum_{j=n}^{\infty} \frac{1}{j !} \frac{\partial^{j} f}{\partial z^{j}}\left(z_{0}\right)\left(z-z_{0}\right)^{j} .$$

## 数学代写|复分析作业代写Complex function代考|The Local Geometry of Holomorphic Functions

$$g: \mathbb{C} \rightarrow \mathbb{C} z \quad \mapsto|z|^{2} .$$

$$g: \mathbb{C} \rightarrow \mathbb{C} z \quad \mapsto z^{2}$$

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## MATLAB代写

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