### 数学代写|复变函数作业代写Complex function代考|Holomorphic Functions, the Cauchy-Riemann

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## 数学代写|复变函数作业代写Complex function代考|Equations, and Harmonic Functions

Functions $f$ which satisfy $(\partial / \partial \bar{z}) f \equiv 0$ are the main concern of complex analysis. We make a precise definition:

Definition 1.4.1. A continuously differentiable $\left(C^{1}\right)$ function $f: U \rightarrow \mathbb{C}$ defined on an open subset $U$ of $\mathbb{C}$ is said to be holomorphic if
$$\frac{\partial f}{\partial \bar{z}}=0$$
at every point of $U$.
Remark: Some books use the word “analytic” instead of “holomorphic.” Still others say “differentiable” or “complex differentiable” instead of “holomorphic.” The use of “analytic” derives from the fact that a holomorphic function has a local power series expansion about each point of its domain. The use of “differentiable” derives from properties related to the CauchyRiemann equations and conformality. These pieces of terminology, and their significance, will all be sorted out as the book develops.

If $f$ is any complex-valued function, then we may write $f=u+i v$, where $u$ and $v$ are real-valued functions. For example,
$$z^{2}=\left(x^{2}-y^{2}\right)+i(2 x y)$$
in this example $u=x^{2}-y^{2}$ and $v=2 x y$. The following lemma reformulates Definition $1.4 .1$ in terms of the real and imaginary parts of $f$ :

Lemma 1.4.2. A continuously differentiable function $f: U \rightarrow \mathbb{C}$ defined on an open subset $U$ of $\mathbb{C}$ is holomorphic if, writing $f(z)=u(x, y)+i v(x, y)$, with $z=x+i y$ and real-valued functions $u$ and $v$, we have that $u$ and $v$ satisfy the equations
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \quad \text { and } \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$
at every point of $U$.
Proof. The assertion follows immediately from the definition of holomorphic function and the formula
$$\frac{\partial f}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)$$
The equations
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \quad \text { and } \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$
are called the Cauchy-Riemann equations. The proof of the following easy result is left as an exercise for you:

Proposition 1.4.3. If $f: U \rightarrow \mathbb{C}$ is $C^{1}$ and if $f$ satisfies the CauchyRiemann equations, then
$$\frac{\partial f}{\partial z} \equiv \frac{\partial f}{\partial x} \equiv-i \frac{\partial f}{\partial y}$$
on $U$.
The Cauchy-Riemann equations suggest a further line of investigation which is of considerable importance. Namely, suppose that $u$ and $v$ are $C^{2}$ functions which satisfy the Cauchy-Riemann equations. Then
$$\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\partial v}{\partial y}\right)$$
and
$$\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial y}\right)=-\frac{\partial}{\partial y}\left(\frac{\partial v}{\partial x}\right)$$
Exploiting the standard theorem on the equality of mixed partial derivatives $\left(\right.$ that $\left.\partial^{2} v / \partial x \partial y=\partial^{2} v / \partial y \partial x\right)$, we obtain
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0$$
A similar calculation shows that
$$\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}=0$$
You should check this last equation as an exercise.

## 数学代写|复变函数作业代写Complex function代考|Real and Holomorphic Antiderivatives

In this section we want to treat in greater generality the question of whether a real-valued harmonic function $u$ is the real part of a holomorphic function $F$. Notice that if we write $F=u+i v$, then the Cauchy-Riemann equations say that
\begin{aligned} &\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} \ &\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x} \end{aligned}
In short, once $u$ is given, then $\partial v / \partial x$ and $\partial v / \partial y$ are completely determined. These in turn determine $v$ up to an additive constant. Thus determining the existence of $v$ (and hence of $F$ ) amounts to solving a familiar problem of multivariable calculus: Given two functions $f$ and $g$ (in this case $-\partial u / \partial y$ and $\partial u / \partial x$, respectively), can we find a function $v$ such that $\partial v / \partial x=f$ and $\partial v / \partial y=g$ ?

A partial solution to this problem is given by the following theorem. We shall see later that the practice, begun in this theorem, of restricting consideration to functions defined on rectangles is not simply a convenience. In fact, the next theorem would actually be false if we considered functions defined on arbitrary open sets in $\mathbb{C}$ (see Exercise 52 ).
Theorem 1.5.1. If $f, g$ are $C^{1}$ functions on the rectangle
$$\mathcal{R}=\left{(x, y) \in \mathbb{R}^{2}:|x-a|<\delta,|y-b|<\epsilon\right}$$
and if
$$\frac{\partial f}{\partial y} \equiv \frac{\partial g}{\partial x} \quad \text { on } \mathcal{R}$$
then there is a function $h \in C^{2}(\mathcal{R})$ such that
$$\frac{\partial h}{\partial x} \equiv f \quad \text { and } \quad \frac{\partial h}{\partial y} \equiv g$$
on $\mathcal{R}$. If $f$ and $g$ are real-valued, then we may take $h$ to be real-valued also.
Proof. For $(x, y) \in \mathcal{R}$, set
$$h(x, y)=\int_{a}^{x} f(t, b) d t+\int_{b}^{y} g(x, s) d s .$$
By the fundamental theorem of calculus,
$$\frac{\partial h}{\partial y}(x, y)=g(x, y)$$

## 数学代写|复变函数作业代写Complex function代考|This is half of our resul

This is half of our result. To calculate $\partial h / \partial x$, notice that, by the fundamental theorem of calculus,
$$\frac{\partial}{\partial x} \int_{a}^{x} f(t, b) d t=f(x, b) .$$
Moreover, since $g$ is $C^{1}$, the theorem on differentiation under the integral sign (see Appendix A) guarantees that
$$\frac{\partial}{\partial x} \int_{b}^{y} g(x, s) d s=\int_{b}^{y} \frac{\partial}{\partial x} g(x, s) d s,$$
which by (1.5.1.1)
\begin{aligned} &=\int_{b}^{y} \frac{\partial}{\partial y} f(x, s) d s \ &=f(x, y)-f(x, b) \end{aligned}
(by the fundamental theorem of calculus). Now (1.5.1.2)-(1.5.1.4) give that $\partial h / \partial x=f$. Since
\begin{aligned} &\frac{\partial h}{\partial x}=f \in C^{1}(\mathcal{R}) \ &\frac{\partial h}{\partial y}=g \in C^{1}(\mathcal{R}) \end{aligned}
we see that $h \in C^{2}(\mathcal{R})$. It is clear from (1.5.1.2) that $h$ is real-valued if $f$ and $g$ are.

It is worth noting that, while we constructed $h$ using integrals beginning at $(a, b)$ (the coordinates of the center of the square), we could have used any $\left(a_{0}, b_{0}\right) \in \mathcal{R}$ as our base point. This changes $h$ only by an additive constant. Note also that Theorem $1.5 .1$ holds for $\mathcal{R}$ an open disc: The only special property needed for the proof is that for some fixed point $P_{0} \in \mathcal{R}$ and for any point $Q \in \mathcal{R}$ the horizontal-vertical path from $P_{0}$ to $Q$ lies in $\mathcal{R}$. This property holds for the disc if we choose $P_{0}$ to be the center.

Corollary 1.5.2. If $\mathcal{R}$ is an open rectangle (or open disc) and if $u$ is a real-valued harmonic function on $\mathcal{R}$, then there is a holomorphic function $F$ on $\mathcal{R}$ such that $\operatorname{Re} F=u$.
Proof. Notice that
$$f=-\frac{\partial u}{\partial y}, \quad g=\frac{\partial u}{\partial x}$$
satisfy
$$\frac{\partial f}{\partial y}=\frac{\partial g}{\partial x} \quad \text { on } \mathcal{R}$$

## 数学代写|复变函数作业代写Complex function代考|Equations, and Harmonic Functions

∂F∂和¯=0

∂在∂X=∂在∂是 和 ∂在∂是=−∂在∂X

∂F∂和¯=12(∂在∂X−∂在∂是)+一世2(∂在∂X+∂在∂是)

∂在∂X=∂在∂是 和 ∂在∂是=−∂在∂X

∂F∂和≡∂F∂X≡−一世∂F∂是

Cauchy-Riemann 方程提出了一个相当重要的进一步研究方向。即，假设在和在是C2满足 Cauchy-Riemann 方程的函数。然后
∂∂X(∂在∂X)=∂∂X(∂在∂是)

∂∂是(∂在∂是)=−∂∂是(∂在∂X)

∂2在∂X2+∂2在∂是2=0

∂2在∂X2+∂2在∂是2=0

## 数学代写|复变函数作业代写Complex function代考|Real and Holomorphic Antiderivatives

∂在∂X=−∂在∂是 ∂在∂是=∂在∂X

\mathcal{R}=\left{(x, y) \in \mathbb{R}^{2}:|xa|<\delta,|yb|<\epsilon\right}\mathcal{R}=\left{(x, y) \in \mathbb{R}^{2}:|xa|<\delta,|yb|<\epsilon\right}

∂F∂是≡∂G∂X 在 R

∂H∂X≡F 和 ∂H∂是≡G

H(X,是)=∫一种XF(吨,b)d吨+∫b是G(X,s)ds.

∂H∂是(X,是)=G(X,是)

## 数学代写|复变函数作业代写Complex function代考|This is half of our resul

∂∂X∫一种XF(吨,b)d吨=F(X,b).

∂∂X∫b是G(X,s)ds=∫b是∂∂XG(X,s)ds,

=∫b是∂∂是F(X,s)ds =F(X,是)−F(X,b)
（由微积分基本定理）。现在 (1.5.1.2)-(1.5.1.4) 给出∂H/∂X=F. 自从
∂H∂X=F∈C1(R) ∂H∂是=G∈C1(R)

F=−∂在∂是,G=∂在∂X

∂F∂是=∂G∂X 在 R

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