### 数学代写|复分析作业代写Complex function代考|MATH307

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• (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|Definition and basic meanings of analyticity

Definition 1 (analyticity). A function $f(z)$ of a complex variable is holomorphic (a.k.a. complex-differentiable, analytic ${ }^1$ ) at $z$ if the limit
$$f^{\prime}(z):=\lim _{h \rightarrow 0} \frac{f(z+h)-f(z)}{h}$$
exists. In this case we call $f^{\prime}(z)$ the derivative of $f$ at $z$.
In the case when $f^{\prime}(z) \neq 0$, the existence of the derivative has a geometric meaning: if we write the polar decomposition $f^{\prime}(z)=r e^{i \theta}$ of the derivative, then for points $w$ that are close to $z$, we will have the approximate equality
$$\frac{f(w)-f(z)}{w-z} \approx f^{\prime}(z)=r e^{i \theta},$$
or equivalently
$$f(w) \approx f(z)+r e^{i \theta}(w-z)+\text { [lower order terms] },$$
where “lower order terms” refers to a quantity that is much smaller in magnitude that $|w-z|$. Geometrically, this means that to compute $f(w)$, we start from $f(z)$, and move by a vector that results by taking the displacement vector $w-z$, rotating it by an angle of $\theta$, and then scaling it by a factor of $r$ (which corresponds to a magnification if $r>1$, a shrinking if $0<r<1$, or doing nothing if $r=1$ ). This idea can be summarized by the slogan:
“Analytic functions behave locally as a rotation composed with a scaling.”

## 数学代写|复分析作业代写Complex function代考|The Cauchy-Riemann equations

In addition to the geometric picture associated with the definition of the complex derivative, there is yet another quite different but also extremely useful way to think about analyticity, that provides a bridge between complex analysis and ordinary multivariate calculus. Remembering that complex numbers are veclors that have real and imayinary components, we can denote $z=x+i y$, where $x$ and $y$ will denote the real and imaginary parts of the complex number $z$, and $f=u+i v$, where $u$ and $v$ are real-valued functions of $z$ (or equivalently of $x$ and $y$ ) that return the real and imaginary parts, respectively, of $f$. Now, if $f$ is analytic at $z$ then
\begin{aligned} f^{\prime}(z) & =\lim {h \rightarrow 0} \frac{f(z+h)-f(z)}{h} \ & =\lim {h \rightarrow 0, h \in \mathbb{R}} \frac{u(x+h+i y)-u(x+i y)}{h}+i \frac{v(x+h+i y)-v(x+i y)}{h} \ & =\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x} . \end{aligned}
On the other hand also
\begin{aligned} f^{\prime}(z) & =\lim {h \rightarrow 0} \frac{f(z+h)-f(z)}{h} \ & =\lim {h \rightarrow 0, h \in i \mathbb{R}} \frac{u(x+h+i y)-u(x+i y)}{h}+i \frac{v(x+h+i y)-v(x+i y)}{h} \ & =\lim _{h \rightarrow 0, h \in \mathbb{R}} \frac{u(x+i y+i h)-u(x+i y)}{i h}+i \frac{v(x+i y+i h)-v(x+i y)}{i h} \ & =-i \frac{\partial u}{\partial y}-i \cdot i \frac{\partial v}{\partial y}=\frac{\partial v}{\partial y}-i \frac{\partial u}{\partial y} . \end{aligned}
Since these limits are equal, by equating their real and imaginary parts we get a famous system of coupled partial differential equations, the CauchyRiemann equations:
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} .$$
We have proved that if $f$ is analytic at $z=x+i y$ then the components $u, v$ of $f$ satisfy the Cauchy-Riemann equations. Conversely, we now claim if $f=u+i v$ is continuously differentiable at $z=x+i y$ (in the sense that each of $u$ and $v$ is a continuously differentiable function of $x, y$ as defined in ordinary real analysis) and satisfies the Cauchy-Riemann equations there, $f$ is analytic at $z$.

# 复分析代写

## 数学代写|复分析作业代写Complex function代考|Definition and basic meanings of analyticity

$$f^{\prime}(z):=\lim _{h \rightarrow 0} \frac{f(z+h)-f(z)}{h}$$

$$\frac{f(w)-f(z)}{w-z} \approx f^{\prime}(z)=r e^{i \theta}$$

$$f(w) \approx f(z)+r e^{i \theta}(w-z)+[\text { lower order terms }]$$

“解析函数在局部表现为由缩放组成的旋转。”

## 数学代写|复分析作业代写Complex function代考|The Cauchy-Riemann equations

$$f^{\prime}(z)=\lim h \rightarrow 0 \frac{f(z+h)-f(z)}{h} \quad=\lim h \rightarrow 0, h \in \mathbb{R} \frac{u(x+h+i y)-u(x+i y)}{h}+i \frac{v}{h}$$

$$f^{\prime}(z)=\lim h \rightarrow 0 \frac{f(z+h)-f(z)}{h} \quad=\lim h \rightarrow 0, h \in i \mathbb{R} \frac{u(x+h+i y)-u(x+i y)}{h}+i$$

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} .$$

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