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## 复分析代考_Complex analysis代考_MTH7054 Notations and Definitions

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## 复分析代考_Complex analysis代考_Notations and Definitions

The letters $a, b, c$, and $d$ are fixed complex numbers, and the letters $Z$ and $W$ represent complex variables. The letter $\mathbb{C}$ stands for the complex plane and the notation $\mathbb{S}=\mathbb{C} \cup(\infty)$ represents the complex plane with the “point” at infinity added. This can be realized as the closed unit sphere, but we shall not need that information in our work.

The words “fractional linear maps” stand for the family of mappings $f(Z)=\frac{a Z+b}{c Z+d}=W$. These are also known as Mobius mappings. There are certain special cases (choices of $a, b, c$, or $d$ ) that we wish to avoid and let us say a word about this matter. First, if $c$ is zero and $d$ is zero we can see there will be serious questions about the meaning of such an expression. To avoid this we assume (the determining number) $a d-b c \neq 0$.
For if this were not the case we observe that for different values $Z_1$ and $Z_2$ the expression
$$f\left(Z_1\right)-f\left(Z_2\right)=\frac{\left(Z_2-Z_1\right)(b c-a d)}{\left(c Z_1+d\right)\left(c Z_2+d\right)}=0$$
implying that $\mathrm{f}$ is a constant mapping. So for example the choice of $a=d=1$ and $b=\frac{\imath}{2}$ and $c=2 \imath$ would yield
$$f(Z)=\frac{Z+\frac{\imath}{2}}{2 \imath Z+1}=\frac{\imath}{2}$$

## 复分析代考_Complex analysis代考_Some Geometric Considerations

Assume $Z_1, Z_2$, and $Z_3$ are three distinct complex numbers. They determine a circle in our sense. Then the element $T \in \mathbb{L}$ of the form
$$f(Z)=\frac{\left(Z-Z_1\right)\left(Z_2-Z_3\right)}{\left(Z-Z_3\right)\left(Z_2-Z_1\right)}=W$$
has some interesting properties. Note that it maps $Z_1$ to zero, and $Z_2$ to the number 1 , and $Z_3$ to infinity. Thus the circle containing $Z_1, Z_2$, and $Z_3$ is mapped to the circle containing 0,1 , and the point at infinity. This $f(Z)=\frac{a Z+b}{c Z+d}=W$ is known as the “cross ratio” and is denoted by the form $\left(Z, Z_1, Z_2, Z_3\right)$. Now it follows from these properties if $W_1, W_2$, and $W_3$ are three distinct complex numbers the equation
$$f(Z)=\left(Z, Z_1, Z_2, Z_3\right)=\left(W, W_1, W_2, W_3\right)=g(W)$$
maps the circle determined by $Z_1, Z_2, Z_3$, one to one and onto the circle determined by the points $W_1, W_2, W_3$. This follows easily since setting $Z=Z_1$ gives the value number zero in the left terms and since $W_1$ is the unique point in the righthand term giving the value zero it must corresponding to $Z_1$, etc., for the two remaining terms.

## 复分析代考_Complex analysis代考_Notations and Definitions

“分数线性映射”一词代表映射族 $f(Z)=\frac{a Z+b}{c Z+d}=W$. 这些也称为莫比乌斯映射。有一些特殊情况（选择 $a, b, c$ ，或者 $d$ ) 我们希望避免，让我们炏谈这件事。首先，如果 $c$ 为零且 $d$ 为零，我们可以看到对这种表达式的 含义会有严重的疑问。为了避免这种情况，我们假设 (确定数) $a d-b c \neq 0$.

$$f\left(Z_1\right)-f\left(Z_2\right)=\frac{\left(Z_2-Z_1\right)(b c-a d)}{\left(c Z_1+d\right)\left(c Z_2+d\right)}=0$$

$$f(Z)=\frac{Z+\frac{\imath}{2}}{2 \imath Z+1}=\frac{\imath}{2}$$

## 复分析代考_Complex analysis代考_Some Geometric Considerations

$$f(Z)=\frac{\left(Z-Z_1\right)\left(Z_2-Z_3\right)}{\left(Z-Z_3\right)\left(Z_2-Z_1\right)}=W$$

$$f(Z)=\left(Z, Z_1, Z_2, Z_3\right)=\left(W, W_1, W_2, W_3\right)=g(W)$$

## 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。

Posted on Categories:Complex analys, 复分析, 数学代写

## 复分析代考_Complex analysis代考_MTH7054 Szegő kernel and Cauchy Kernel Under Transformation

avatest复分析Complex analysis代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest™， 最高质量的复分析Complex analysis作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此复分析Complex analysis作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

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## 复分析代考_Complex analysis代考_Szegő kernel and Cauchy Kernel Under Transformation

In general, we assume $\Omega \subset \mathbb{C}$ is a bounded region with $C^{\infty}$ smooth boundary. Most of the discussion applies to more general regions, too, as will be apparent in examples.
2.2.1 Szegő kernel
To begin, the Lebesgue space $L^2(\partial \Omega)$ is defined with respect to arc length measure using the Hermitian inner product

$$(g, h){\partial \Omega}=\int{\partial \Omega} g \bar{h} d s .$$
The Hardy space $H^2(\partial \Omega)$ is loosely the subspace of functions that extend holomorphically to the region $\Omega$. Such extensions are unique by the Cauchy integral formula. To be precise, $H^2(\partial \Omega)$ is the closure in $L^2(\partial \Omega)$ of the space of functions $A^{\infty}(\Omega)$ that arise as boundary values of functions holomorphic in $\Omega$ and smooth in $\bar{\Omega}$. Since $H^2(\partial \Omega)$ is closed, one has an orthogonal projection $\mathscr{S}: L^2(\partial \Omega) \rightarrow H^2(\partial \Omega)$ called the Szegő projector.

## 复分析代考_Complex analysis代考_Cauchy kernel

In contrast to the Szegő projector, there is the explicit Cauchy operator defined initially for $g \in L^1(\partial \Omega)$ according to
$$\mathscr{C} g(z)=\frac{1}{2 \pi \mathrm{i}} \int_{\partial \Omega} \frac{g(w) d w}{w-z}$$
for $z \in \Omega$. From this definition, one sees that $\mathscr{C} g$ is holomorphic in $\Omega$. By restricting to functions $g \in C^{\infty}(\partial \Omega)$, one can show that the holomorphic function $\mathscr{C} g$ extends smoothly to boundary. That is, $\mathscr{C}$ restricts to an operator $C^{\infty}(\partial \Omega) \rightarrow C^{\infty}(\partial \Omega)$. In fact, $\mathscr{C}$ is bounded on $L^2(\partial \Omega)$ and therefore extends continuously to a bounded projector $\mathscr{C}: L^2(\partial \Omega) \rightarrow H^2(\partial \Omega)$. (The reason that $\mathscr{C}$ reproduces functions in the Hardy space is the Cauchy integral formula.) In this way, the Cauchy projector arises as an explicitly defined alternative to the Szegő projector.

For further comparison with the Szegö projector, observe that the Cauchy projector can be expressed via $\mathscr{C} g(z)=\left(g, C_z\right)_{\partial \Omega}$ where
$$C_z(w)=C(w, z)=\overline{\frac{1}{2 \pi \mathrm{i}} \frac{T(w)}{w-z}}$$
and $T(w)$ is the positively oriented unit tangent vector at $w \in \partial \Omega$.

## 复分析代考_Complex analysis代考_Poles, Residues, and All That

10.1. 残留物。一个点 $z_0$ 是函数的奇异点 $f$ 如果 $f$ 不分析 $z_0$ ，但是在的每个邻域的某个点上是解析的 $z_0$. 奇异点 $z_0$ 的 $f$ 如果有一 个邻域，则据兑是孤立的 $z_0$ 其中不包含奇异点 $f$ 节省 $z_0$.换句话说， $f$ 是对某个区域的解析 $0<\left|z-z_0\right|<\varepsilon$.

$$f(z)=\frac{1}{z\left(z^2+4\right)}$$

## 复分析代考_Complex analysis代考_Poles and other singularities

10.2. 极点和其他奇点。为了让留数定理对计算积分有很大邦助，需要有一些更好的方法来计算留数一找到关于每个孤立奇异点 的 Laurent 展开是一件苦差事。我们现在将看到，在一种恃殊但常见的奇点类型的情况下，留数很容易找到。认为 $z_0$ 是一个孤立 的奇点 $f$ 并假设 Laurent 系列 $f$ 在 $z_0$ 仅包含有限数量的涉及负菂的项 $z-z_0$. 因此，
$$f(z)=\frac{c_{-n}}{\left(z-z_0\right)^n}+\frac{c_{-n+1}}{\left(z-z_0\right)^{n-1}}+\ldots+\frac{c_{-1}}{\left(z-z_0\right)}+c_0+c_1\left(z-z_0\right)+\ldots$$

$$\phi(z)=\left(z-z_0\right)^n f(z)=c_{-n}+c_{-n+1}\left(z-z_0\right)+\ldots+c_{-1}\left(z-z_0\right)^{n-1}+\ldots$$

## 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。

Posted on Categories:Complex analys, 复分析, 数学代写

## 复分析代考_Complex analysis代考_MTH7054 (0, 1)-forms with holomorphic coefficients

avatest复分析Complex analysis代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest™， 最高质量的复分析Complex analysis作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此复分析Complex analysis作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 复分析代考_Complex analysis代考_(0, 1)-forms with holomorphic coefficients

Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ and let $A_{(0,1)}^2(\Omega)$ denote the space of all $(0,1)$-forms with holomorphic coefficients belonging to $L^2(\Omega)$. With the same proof as in Section 8.2, one shows that the canonical solution operator
$$S: A_{(0,1)}^2(\Omega) \longrightarrow L^2(\Omega)$$
has the form
$$S(g)(z)=\int_{\Omega} K(z, w)\langle g(w), z-w\rangle d \lambda(w),$$
where $K$ denotes the Bergman kernel of $\Omega$,
$$g(z)=\sum_{j=1}^n g_j(z) d \bar{z}j \in A{(0,1)}^2(\Omega)$$
and
$$\langle g(w), z-w\rangle=\sum_{j=1}^n g_j(w)\left(\bar{z}j-\bar{w}_j\right),$$ for $z=\left(z_1, \ldots, z_n\right)$ and $w=\left(w_1, \ldots, w_n\right)$. Let $v(z)=\sum{j=1}^n \bar{z}j g_j(z)$. Then it follows that $$\bar{\partial} v=\sum{j=1}^n \frac{\partial v}{\partial \bar{z}j} d \bar{z}_j=\sum{j=1}^n g_j d \bar{z}_j=g .$$

## 复分析代考_Complex analysis代考_General properties of Fréchet spaces

Assuming basic knowledge of general topology, we collect important facts about topological vector spaces.

A topological vector space $X$ is a vector space endowed with a topology such that addition $+: X \times X \longrightarrow X$ and scalar multiplication $\cdot: \mathbb{C} \times X \longrightarrow X$ are continuous. $X$ is a normed vector space if there is a norm $|\cdot|$ on $X$; each open set of $X$ can be written as a union of open balls $\left{\chi \in X:\left|x-x_0\right|<r\right}$.
$X$ is a metric topological vector space if there is a metric $d: X \times X: \longrightarrow \mathbb{R}_{+}$on $X$, each open set of $X$ can be written as a union of open balls $\left{x \in X: d\left(x, x_0\right)<r\right}$; we will also suppose that the metric is translation invariant, i.e. $d(x+u, y+u)=d(x, y)$, for all $x, y, z \in X$.

A subset $M$ of a vector space $X$ is called absolutely convex, if $\lambda x+\mu y \in M$ for each $x, y \in M$ and $\lambda, \mu \in \mathbb{C}$ with $|\lambda|+|\mu| \leq 1$.

A locally convex vector space $X$ is a topological vector space for which each point has neighborhood basis consisting of absolutely convex sets.

Let $X$ be a locally convex vector space and let $U$ be an absolutely convex 0-neighborhood in $X$. Then $|\cdot|_U: x \mapsto \inf {t>0: x \in t U}$ is a continuous seminorm on $X$; we call $|\cdot|_U$ the Minkowski functional of $U$.

One can explain the topology of a locally convex vector space $X$ in a different way: a family $\mathcal{U}$ of 0 -neighborhoods is a fundamental system of 0-neighborhoods, if for each 0-neighborhood $U$ there exist $V \in \mathcal{U}$ and $\epsilon>0$ such that $\epsilon V \subset U$. A family $\left(p_\alpha\right){\alpha \in A}$ of seminorms is called a fundamental system of seminorms, if the sets $U\alpha={\chi \in X$ : $\left.p_\alpha(x)<1\right}$ constitute a fundamental system of 0 -neighborhoods of $X$. We will write $\left(X,\left(p_\alpha\right)_{\alpha \in A}\right)$ to refer to that.

Let $X$ and $Y$ be locally convex vector spaces with fundamental systems $\left(p_\alpha\right){\alpha \in A}$ and $\left(q\beta\right){\beta \in B}$ of seminorms. A linear mapping $T: X \longrightarrow Y$ is continuous, if and only if for each $\beta \in B$ there exist $\alpha \in A$ and a constant $C>0$ such that $q\beta(T x) \leq C p_\alpha(x)$, for all $x \in X$.

A linear functional $x^{\prime}$ on $X$ is continuous, if and only if there exist $\alpha \in A$ and a constant $C>0$ such that $\left|x^{\prime}(x)\right| \leq C p_\alpha(x)$ for all $x \in X$.

## 复分析代考Complex analysis代考(0,1)-forms with holomorphic coefficients

$$S: A_{(0,1)}^2(\Omega) \longrightarrow L^2(\Omega)$$

$$S(g)(z)=\int_{\Omega} K(z, w)\langle g(w), z-w\rangle d \lambda(w)$$

$$g(z)=\sum_{j=1}^n g_j(z) d \bar{z} j \in A(0,1)^2(\Omega)$$

$$\langle g(w), z-w\rangle=\sum_{j=1}^n g_j(w)\left(\bar{z} j-\bar{w}j\right),$$ 为了 $z=\left(z_1, \ldots, z_n\right)$ 和 $w=\left(w_1, \ldots, w_n\right)$. 让 $v(z)=\sum j=1^n \bar{z} j g_j(z)$. 然后就是 $$\bar{\partial} v=\sum j=1^n \frac{\partial v}{\partial \bar{z} j} d \bar{z}_j=\sum j=1^n g_j d \bar{z}_j=g .$$

## 复分析代考_Complex analysis代考_General properties of Fréchet spaces

\left 缺少或无法识别的分隔符 $\quad$; 我们还将假设度量是平移不变的，即 $d(x+u, y+u)=d(x, y)$ ，对所有人 $x, y, z \in X$.

## 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。

Posted on Categories:Complex analys, 复分析, 数学代写

## 数学代写|复分析代写Complex analysis代考|MTH7054 Natural logarithm and the exponential functions

avatest复分析Complex analysis代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest™， 最高质量的复分析Complex analysis作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此复分析Complex analysis作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

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## 数学代写|复分析代写Complex analysis代考|Natural logarithm and the exponential functions

The function that takes a non-zero $x$ to $1 / x$ is continuous everywhere on its domain since it is a rational function. Thus by Theorem 7.3.1 and Notation 7.3.3, for all $x>0$, $\int_1^x \frac{1}{x} d x$ is well-defined. This function has a familiar name:
Definition 7.6.1. The natural logarithm is the function
$$\ln x=\int_1^x \frac{1}{t} d t$$
for all $x>0$.
We prove below all the familiar properties of this familiar function.
Remark 7.6.2.
(1) $\ln 1=\int_1^1 \frac{1}{t} d t=0$.
(2) By geometry, for $x>1, \ln x=\int_1^x \frac{1}{t} d t>0$, and for $x \in(0,1), \ln x=\int_1^x \frac{1}{t} d t=$ $-\int_x^1 \frac{1}{t} d t<0$.
(3) By the Fundamental theorem of calculus (Theorem 7.4.3), for all $b \in \mathbb{R}^{+}$, $\ln$ is differentiable on $(0, b)$, so that $\ln$ is differentiable on $\mathbb{R}^{+}$. Furthermore, $\ln ^{\prime}(x)=\frac{1}{x}$.
(4) $\ln$ is continuous (since it is differentiable) on $\mathbb{R}^{+}$.
(5) The derivative of $\ln$ is always positive. Thus by Theorem $6.3 .5$, $\ln$ is everywhere increasing.
(6) Let $c \in \mathbb{R}^{+}$, and set $g(x)=\ln (c x)$. By the chain rule, $g$ is differentiable, and $g^{\prime}(x)=\frac{1}{c x} c=\frac{1}{x}=\ln ^{\prime}(x)$. Thus the function $g-\ln$ has constant derivative 0 . It follows by Theorem $6.3 .5$ that $g-\ln$ is a constant function. Hence for all $x \in \mathbb{R}^{+}$,
$$\ln (c x)-\ln (x)=g(x)-\ln (x)=g(1)-\ln (1)=\ln (c)-0=\ln (c) .$$
This proves that for all $c, x \in \mathbb{R}^{+}$,
$$\ln (c x)=\ln (c)+\ln (x) .$$
(7) By the previous part, for all $c, x \in \mathbb{R}^{+}$,
$$\ln \left(\frac{c}{x}\right)=\ln (c)-\ln (x) .$$
(8) For all non-negative integers $n$ and all $c \in \mathbb{R}^{+}, \ln \left(c^n\right)=n \ln (c)$. We prove this by mathematical induction. If $n=0$, then $\ln \left(c^n\right)=\ln (1)=0=0 \ln c=n \ln c$. Now suppose that equality holds for some $n-1$. Then $\ln \left(c^n\right)=\ln \left(c^{n-1} c\right)=$ $\ln \left(c^{n-1}\right)+\ln (c)$ by what we have already established, so that by the induction assumption $\ln \left(c^n\right)=(n-1) \ln (c)+\ln (c)=n \ln (c)$.
(9) For all rational numbers $r$ and all $c \in \mathbb{R}^{+}, \ln \left(c^r\right)=r \ln (c)$. Here is a proof. We have proved this result if $r$ is a non-negative integer. If $r$ is a negative integer, then $-r$ is a positive integer, so that by the previous case, $\ln \left(c^r\right)=\ln \left(1 / c^{-r}\right)=$ $\ln (1)-\ln \left(c^{-r}\right)=0-(-r) \ln (c)=r \ln (c)$, which proves the claim for all integers.
Now write $r=\frac{m}{n}$ for some integers $m, n$ with $n \neq 0$. Then $n \ln \left(c^r\right)=n \ln \left(c^{m / n}\right)=$ $\ln \left(c^m\right)=m \ln (c)$, so that $\ln \left(c^r\right)=\frac{m}{n} \ln (c)=r \ln (c)$.
(10) The range of $\ln$ is $\mathbb{R}=(-\infty, \infty)$. Here is a proof. By geometry, $\ln (0.5)<0<\ln (2)$. Let $y \in \mathbb{R}^{+}$. By Theorem 2.10.3, there exists $n \in \mathbb{N}^{+}$such that $y<n \ln (2)$. Hence $\ln 1=0<y<n \ln (2)=\ln \left(2^n\right)$, so that since $\ln$ is continuous, by the Intermediate value theorem (Theorem 5.3.1), there exists $x \in\left(1,2^n\right)$ such that $\ln (x)=y$. If $y \in \mathbb{R}^{-}$, then by the just proved we have that $-y=\ln (x)$ for some $x \in \mathbb{R}^{+}$, so that $y=-\ln (x)=\ln \left(x^{-1}\right)$. Finally, $0=\ln (1)$. Thus every real number is in the range of $\ln$.
(11) Thus $\ln : \mathbb{R}^{+} \rightarrow \mathbb{R}$ is a strictly increasing continuous and surjective function. Thus by Theorem 2.9.4, $\ln$ has an inverse $\ln ^{-1}: \mathbb{R} \rightarrow \mathbb{R}^{+}$. By Theorem 5.3.4, $\ln ^{-1}$ is increasing and continuous.
(12) By Theorem 6.2.7, the derivative of $\ln ^{-1}$ is
$$\left(\ln ^{-1}\right)^{\prime}(x)=\frac{1}{\ln ^{\prime}\left(\ln ^{-1}(x)\right)}=\ln ^{-1}(x) .$$

## 数学代写|复分析代写Complex analysis代考|Applications of integration

The Fundamental theorems of calculus relate integration with differentiation. In particular, to compute $\int_a^b f$, if we know an antiderivative $g$ of $f$, then the integral is easy. However, as already mentioned after the Fundamental theorem of calculus I, many functions do not have a “closed-form” antiderivative. One can still compute definite integrals up to a desired precision, however: we take finer and finer partitions of $[a, b]$, and when $U(f, P)$ and $L(f, P)$ are within a specified distance from each other, we know that the true integral is somewhere in between, and hence up to the specified precision either $L(f, P)$ or $U(f, P)$ stands for $\int_a^b f$. In applications, such as in science and engineering, many integrals have to be and are computed in this way because of the lack of closed-form antiderivatives.

In this section we look at many applications that exploit the original definition of integrals via sums over finer and finer partitions. For many concrete examples we can then solve the integral via antiderivatives, but for many we have to make do with numerical approximation.
7.7.1 Length of a curve
Let $f:[a, b] \rightarrow \mathbb{R}$ be a continuous function. If the graph of $f$ is a line, then by the Pythagorean theorem the length of the curve is $\sqrt{(b-a)^2+(f(b)-f(a))^2}$. For a general curve it is harder to determine its length from $(a, f(a))$ to $(b, f(b))$. But we can do the standard calculus trick: let $P=\left{x_0, x_1, \ldots, x_n\right}$ be a partition of $[a, b]$; on each subinterval $\left[x_{k-1}, x_k\right]$ we “approximate” the curve with the line $\left(x_{k-1}, f\left(x_{k-1}\right)\right)$ to $\left(x_k, f\left(x_k\right)\right)$, compute the length of that line as $\sqrt{\left(x_k-x_{k-1}\right)^2+\left(f\left(x_k\right)-f\left(x_{k-1}\right)\right)^2}$, and sum up all the lengths:
$$\sum_{k=1}^n \sqrt{\left(x_k-x_{k-1}\right)^2+\left(f\left(x_k\right)-f\left(x_{k-1}\right)\right)^2} .$$

## 数学代写|复分析代写Complex analysis代考|Natural logarithm and the exponential functions

$$\ln x=\int_1^x \frac{1}{t} d t$$

(1) $\ln 1=\int_1^1 \frac{1}{t} d t=0$
(2) 相搱几何, 对于 $x>1, \ln x=\int_1^x \frac{1}{t} d t>0$ ，并恥于 $x \in(0,1), \ln x=\int_1^x \frac{1}{t} d t=-\int_x^1 \frac{1}{t} d t<0$.
(4) $\ln$ 是连封的 (因为它是可微. $) \mathbb{R}^{+}$.
(6) 让 $c \in \mathbb{R}^{+}$，并设置 $g(x)=\ln (c x)$. 根倨軒法则, $g$ 是可铂, 并且 $g^{\prime}(x)=\frac{1}{c x} c=\frac{1}{x}=\ln ^{\prime}(x)$. 因此函数 $g-\ln$ 有㸓数
$$\ln (c x)-\ln (x)=g(x)-\ln (x)=g(1)-\ln (1)=\ln (c)-0=\ln (c) .$$

$$\ln (c x)=\ln (c)+\ln (x) .$$
(7) 由上部分，对于所有 $c, x \in \mathbb{R}^{+}$，
$$\ln \left(\frac{c}{x}\right)=\ln (c)-\ln (x) .$$
$\ln \left(c^n\right)=\ln (1)=0=0 \ln c=n \ln c$. 现在徦纯等恜对桌些人成立 $n-1$. 然咞 $\ln \left(c^n\right)=\ln \left(c^{n-1} c\right)=$
$\ln (1)-\ln \left(c^{-r}\right)=0-(-r) \ln (c)=r \ln (c)$ ，这证明了所有整数的声明。

$\ln \left(c^r\right)=\frac{m}{n} \ln (c)=r \ln (c)$.
(10) 范围珀是 $\mathbb{R}=(-\infty, \infty)$. 这是一个证明。 通过几何, $\ln (0.5)<0<\ln (2)$. 让 $y \in \mathbb{R}^{+}$. 由定理 2.10.3，存在 $n \in \mathbb{N}^{+}$这
(12) 由定理 6.2.7，导数 $\ln ^{-1}$ 是
$$\left(\ln ^{-1}\right)^{\prime}(x)=\frac{1}{\ln ^{\prime}\left(\ln ^{-1}(x)\right)}=\ln ^{-1}(x) .$$

## 数学代写复分析代写Complex analysis代考|Applications of integration

7.7.1曲线的长度
Let $f:[a, b] \rightarrow \mathbb{R}$ 是 个连续函数. 如果图 $f$ 是一条线, 那 根据股股定，曲线的苌是 $\sqrt{(b-a)^2+(f(b)-f(a))^2}$. 对
〈left 的分隔符缽失或无法识别 $\quad$ 成为一个分区 $[a, b]$; 在每个子区间 $\left[x_{k-1}, x_k\right]$ ]戈们用线”近似”曲线
$\left(x_{k-1}, f\left(x_{k-1}\right)\right)$ 至 $\left(x_k, f\left(x_k\right)\right)$ ，计算那各线得度为 $\sqrt{\left(x_k-x_{k-1}\right)^2+\left(f\left(x_k\right)-f\left(x_{k-1}\right)\right)^2}$ ，并总狜所有长度:
$$\sum_{k=1}^n \sqrt{\left(x_k-x_{k-1}\right)^2+\left(f\left(x_k\right)-f\left(x_{k-1}\right)\right)^2} .$$

## 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。

Posted on Categories:Complex analys, 复分析, 数学代写

## 数学代写|复分析代写Complex analysis代考|MTH7054 The inhomogeneous Cauchy formula

avatest复分析Complex analysis代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest™， 最高质量的复分析Complex analysis作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此复分析Complex analysis作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 数学代写|复分析代写Complex analysis代考|The inhomogeneous Cauchy formula

Now we apply Stokes’ Theorem to prove a more general version of Cauchy’s formula, which will be a useful tool for the study of the inhomogeneous Cauchy-Riemann equations.

Theorem 2.47. Let $G$ be a bounded domain in $\mathbb{C}$ with piecewise smooth positively oriented boundary $\partial G$. Let $f \in \mathcal{C}^1(\bar{G})$. Then for $z \in G$ we have
$$f(z)=\frac{1}{2 \pi i} \int_{\partial G} \frac{f(\zeta)}{\zeta-z} d \zeta+\frac{1}{2 \pi i} \int_G \frac{(\partial f / \partial \bar{\zeta})(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta} .$$
Remark. If $f \in \mathcal{H}(G)$, we have $(\partial f / \partial \bar{\zeta})(\zeta)=0 \forall \zeta \in G$ and hence
$$f(z)=\frac{1}{2 \pi i} \int_{\partial G} \frac{f(\zeta)}{\zeta-z} d \zeta .$$
In this sense, Theorem $2.47$ yields a generalization of Cauchy’s formula.
Proof. Fix $z \in G$ and choose $r>0$ such that $D_r(z) \subseteq G$. We remove the disc $D_r(z)$ from $G$ and define $G_r=G \backslash \overline{D_r(z)}$, the boundary of $G_r$ consists of the positively oriented boundary of $G$ and of the negatively oriented circle $\kappa_r=\partial D_r(z)$. Walking on $\partial G_r$, the domain $G_r$ lies always on the left-hand side.

For $\zeta \in G_r$ we define the 1-form
$$\omega(\zeta)=\frac{1}{2 \pi i} \frac{f(\zeta)}{\zeta-z} d \zeta,$$
we can apply Stokes’ Theorem (see Theorem 2.46) and obtain from Theorem $2.45$
$$\frac{1}{2 \pi i} \int_{\partial G_r} \frac{f(\zeta)}{\zeta-z} d \zeta=-\frac{1}{2 \pi i} \int_{G_r} \frac{(\partial f / \partial \bar{\zeta})(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta}$$

## 数学代写|复分析代写Complex analysis代考|General versions of Cauchy’s Theorem and Cauchy’s formula

It will be convenient to consider integrals over sums of paths. This leads to the concepts of chains and cycles.

Definition 2.48. Let $\Omega \subseteq \mathbb{C}$ be open and let $y_1, \ldots, y_n$ be paths in $\Omega$. Let $\Gamma^=\bigcup_{j=1}^n y_j^$ and define
$$\tilde{\gamma}j(f):=\int{y_j} f(z) d z,$$
for $f \in \mathcal{C}\left(\Gamma^\right) . \tilde{\gamma}j: \mathcal{C}\left(\Gamma^\right) \longrightarrow \mathbb{C}$ can be seen as a linear functional on $\mathcal{C}\left(\Gamma^\right)$. We set $\tilde{\Gamma}=\tilde{y}_1+\cdots+\tilde{y}_n$, and denote by $\Gamma=y_1+\cdots+y_n$ the formal sum of the paths $y_1, \ldots, \gamma_n$. We define $$\int{\Gamma} f(z) d z=\tilde{\Gamma}(f)=\sum_{j=1}^n \int_{y_j} f(z) d z,$$
for $f \in \mathcal{C}\left(\Gamma^\right) . \Gamma$ is called a chain in $\Omega$. If all paths $\gamma_1, \ldots, \gamma_n$ are closed, we call $\Gamma$ a cycle in $\Omega$.

Remark. (a) Chains and cycles can be represented as sums of paths in many ways.
(b) By $-\Gamma$ we denote the cycle, where each path $\gamma_j, j=1, \ldots, n$ is replaced by its opposite path, for $f \in \mathcal{C}\left(\Gamma^\right)$ we have $$\int_{-\Gamma} f(z) d z=-\int_{\Gamma} f(z) d z$$ (c) If $\Gamma_1$ and $\Gamma_2$ are chains or cycles, we can form the sum $\Gamma=\Gamma_1+\Gamma_2$ and have $$\int_{\Gamma} f(z) d z=\int_{\Gamma_1} f(z) d z+\int_{\Gamma_2} f(z) d z, \quad f \in \mathcal{C}\left(\Gamma_1^ \cup \Gamma_2^*\right)$$

## 数学代写|复分析代写Complex analysis代考|The inhomogeneous Cauchy formula

$$f(z)=\frac{1}{2 \pi i} \int_{\partial G} \frac{f(\zeta)}{\zeta-z} d \zeta+\frac{1}{2 \pi i} \int_G \frac{(\partial f / \partial \bar{\zeta})(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta} .$$

$$f(z)=\frac{1}{2 \pi i} \int_{\partial G} \frac{f(\zeta)}{\zeta-z} d \zeta .$$

$$\omega(\zeta)=\frac{1}{2 \pi i} \frac{f(\zeta)}{\zeta-z} d \zeta,$$

$$\frac{1}{2 \pi i} \int_{\partial G_r} \frac{f(\zeta)}{\zeta-z} d \zeta=-\frac{1}{2 \pi i} \int_{G_r} \frac{(\partial f / \partial \bar{\zeta})(\zeta)}{\zeta-z} d \zeta \wedge d \bar{\zeta}$$

## 数学代写|复分析代写Complex analysis代考|General versions of Cauchy’s Theorem and Cauchy’s formula

$$\tilde{\gamma} j(f):=\int y_j f(z) d z,$$

$$\int \Gamma f(z) d z=\tilde{\Gamma}(f)=\sum_{j=1}^n \int_{y_j} f(z) d z,$$

$$\int_{\Gamma} f(z) d z=-\int_{\Gamma} f(z) d z$$
(c) 如果 $\Gamma_1$ 和 $\Gamma_2$ 是链或循环，我们可以形成总和 $\Gamma=\Gamma_1+\Gamma_2$ 并且有
$$\int_{\Gamma} f(z) d z=\int_{\Gamma_1} f(z) d z+\int_{\Gamma_2} f(z) d z, \quad f \in \mathcal{C}\left(\Gamma_1^{\cup} \Gamma_2^*\right)$$

## 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。