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# 数学代写|复分析代写Complex analysis代考|Math4100

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## 数学代写|复分析代写Complex analysis代考|The Cauchy Theorem for a Triangle

At the end of the nineteenth century, amongst many different versions of Cauchy’s Theorem, a most ingenious proof for a triangular contour was conceived by Eliakim Hastings Moore. Earlier proofs usually insisted that the function $f$ should have a continuous derivative $f^{\prime}$. By restricting the contour to a triangle, Moore’s proof requires only that $f^{\prime}$ exists throughout $D$. It therefore provides a suitable basis for the development of the theory for all differentiable functions.

For $z_1, z_2, z_3 \in \mathbb{C}$, let $T\left(z_1, z_2, z_3\right)$ be the set of points inside and on the triangle with vertices $z_1, z_2, z_3$. Formally,
$$T\left(z_1, z_2, z_3\right)=\left{z \in \mathbb{C}: z=\lambda_1 z_1+\lambda_2 z_2+\lambda_3 z_3, \lambda_j \in \mathbb{R}, \lambda_j \geq 0, \lambda_1+\lambda_2+\lambda_3=1\right}$$
where $j=1,2,3$.
The boundary contour of the triangle, composed of the three line segments that form its sides, is
$$\partial T\left(z_1, z_2, z_3\right)=\left[z_1, z_2\right]+\left[z_2, z_3\right]+\left[z_3, z_1\right]$$
Whenever there is no confusion we denote the triangle by $T$ and its boundary by $\partial T$.
THEOREM 8.1 (Cauchy’s Theorem for a Triangle). Letf be a differentiable function in a domain D. If the triangle T lies in D, as in Figure 8.4, then $\int_{\partial T} f=0$.
Proof. Let $\left|\int_{\partial T} f\right|=c \geq 0$.
We prove that $c=0$ by an indirect argument. First we subdivide $T$ into four smaller triangles $T^{(1)}, T^{(2)}, T^{(3)}, T^{(4)}$ by joining the midpoints of the sides as in Figure 8.5.
We know that
$$\int_{\partial T} f=\sum_{r=1}^4 \int_{\partial T^{(r)}} f$$
Therefore
$$c=\left|\int_{\partial T} f\right| \leq \sum_{r=1}^4\left|\int_{\partial T^{(r)}} f\right|$$
so we must be able to choose $r$ such that
$$\left|\int_{\partial T^{(r)}} f\right| \geq \frac{c}{4}$$
(If more than one $r$ satisfies this inequality, choose any of those – say the one with smallest $r$.) Define $T_1=T^{(r)}$. Then
$$\left|\int_{\partial T_1} f\right| \geq \frac{c}{4} \quad \text { and } \quad L\left(\partial T_1\right)=\frac{1}{2} L(\partial T)$$
Repeat this process of subdivision to get a sequence of triangles
$$T \supseteq T_1 \supseteq T_2 \supseteq \cdots \supseteq T_n \cdots$$
satisfying
$$\left|\int_{\partial T_n} f\right| \geq\left(\frac{1}{4}\right)^n c \quad \text { and } \quad L\left(\partial T_1\right)=\left(\frac{1}{2}\right)^n L(\partial T)$$

## 数学代写|复分析代写Complex analysis代考|Existence of an Antiderivative in a Star Domain

We begin with a formal definition, previewed in the introduction to this chapter:
DEFINITION 8.2. A domain $D$ is a star domain if there exists $z_* \in D$, called a star centre, such that for all $z \in D$ the straight line segment $\left[z_*, z\right]$ lies in $D$.
(A star centre need not be unique. For example, a disc is a star domain and every point in the disc is a star centre.)

In a star domain there is an obvious candidate for an antiderivative of a function $f$, namely the integral $F(z)=\int_{\left[z_*, z\right]} f$. We now show that this is indeed an antiderivative, by applying Theorem 8.1 .

THEOREM 8.3. Iff is differentiable in a star domain $D$ with star centre $z_$, then $F(z)=$ $\int_{\left[z_, z\right]} f$ is an antiderivative off in $D$.
Proof. The domain $D$ is open, so for any $z_1 \in D$ there exists $\varepsilon_1>0$ such that $N_{\varepsilon_1}\left(z_1\right) \subseteq D$. If $|h|<\varepsilon_1$, the triangle $T\left(z_, z_1, z_1+h\right)$ lies entirely in $D$, Figure 8.6. Now Theorem 8.1 gives $$\int_{\left[z_, z_1\right]} f+\int_{\left[z_1, z_1+h\right]} f+\int_{\left[z_1+h, z_*\right]} f=0$$
This can be written as
$$F\left(z_1\right)+\int_{\left[z_1, z_1+h\right]} f-F\left(z_1+h\right)=0$$
or
$$\frac{F\left(z_1\right)-F\left(z_1+h\right)}{h}=\frac{1}{h} \int_{\left[z_1, z_1+h\right]} f$$
The proof now proceeds in the same manner as Theorem 6.44. For a constant $c \in \mathbb{C}$,
$$\int_{\left[z_1, z_1+h\right]} c \mathrm{~d} z=c h$$
hence
$$\frac{F\left(z_1\right)-F\left(z_1+h\right)}{h}-f\left(z_1\right)=\int_{\left[z_1, z_1+h\right]} \frac{f(z)-f\left(z_1\right)}{h} \mathrm{~d} z$$

## 数学代写|复分析代写Complex analysis代考|The Cauchy Theorem for a Triangle

19世纪末，在柯西定理的许多不同版本中，埃利亚基姆·黑斯廷斯·摩尔(Eliakim Hastings Moore)提出了一个最巧妙的证明三角形轮廓的方法。早期的证明通常坚持认为函数$f$应该有一个连续的导数$f^{\prime}$。通过将等高线限制为三角形，摩尔的证明只需要$f^{\prime}$存在于$D$。因此，它为所有可微函数的理论发展提供了一个合适的基础。

$$T\left(z_1, z_2, z_3\right)=\left{z \in \mathbb{C}: z=\lambda_1 z_1+\lambda_2 z_2+\lambda_3 z_3, \lambda_j \in \mathbb{R}, \lambda_j \geq 0, \lambda_1+\lambda_2+\lambda_3=1\right}$$

$$\partial T\left(z_1, z_2, z_3\right)=\left[z_1, z_2\right]+\left[z_2, z_3\right]+\left[z_3, z_1\right]$$

$$\int_{\partial T} f=\sum_{r=1}^4 \int_{\partial T^{(r)}} f$$

$$c=\left|\int_{\partial T} f\right| \leq \sum_{r=1}^4\left|\int_{\partial T^{(r)}} f\right|$$

$$\left|\int_{\partial T^{(r)}} f\right| \geq \frac{c}{4}$$
(如果不止一个$r$满足这个不等式，选择其中任何一个——比如最小的$r$。)定义$T_1=T^{(r)}$。然后
$$\left|\int_{\partial T_1} f\right| \geq \frac{c}{4} \quad \text { and } \quad L\left(\partial T_1\right)=\frac{1}{2} L(\partial T)$$

$$T \supseteq T_1 \supseteq T_2 \supseteq \cdots \supseteq T_n \cdots$$

$$\left|\int_{\partial T_n} f\right| \geq\left(\frac{1}{4}\right)^n c \quad \text { and } \quad L\left(\partial T_1\right)=\left(\frac{1}{2}\right)^n L(\partial T)$$

## 数学代写|复分析代写Complex analysis代考|Existence of an Antiderivative in a Star Domain

8.2.定义如果存在$z_* \in D$，则域$D$是星型域，称为星型中心，因此对于所有$z \in D$，直线段$\left[z_*, z\right]$位于$D$。
(一个明星中心不一定是唯一的。例如，圆盘是一个星域，圆盘上的每个点都是一个星中心。)

$$F\left(z_1\right)+\int_{\left[z_1, z_1+h\right]} f-F\left(z_1+h\right)=0$$

$$\frac{F\left(z_1\right)-F\left(z_1+h\right)}{h}=\frac{1}{h} \int_{\left[z_1, z_1+h\right]} f$$

$$\int_{\left[z_1, z_1+h\right]} c \mathrm{~d} z=c h$$

$$\frac{F\left(z_1\right)-F\left(z_1+h\right)}{h}-f\left(z_1\right)=\int_{\left[z_1, z_1+h\right]} \frac{f(z)-f\left(z_1\right)}{h} \mathrm{~d} z$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。