### 数学代写|复变函数作业代写Complex function代考|The Cauchy Integral Formula

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## 数学代写|复变函数作业代写Complex function代考|Cauchy Integral Theorem

In this section we shall study two results which are fundamental to the development of the rest of this subject: the Cauchy Integral Formula and the Cauchy integral theorem. The Cauchy integral formula gives the value of a holomorphic function at each point inside a circle in terms of the values of the function on the circle (later we shall learn that “circle” here may be replaced by a more general closed curve, subject to certain topological restrictions). We proceed now with our study of the Cauchy integral formula.
The Cauchy integral formula is so startling that one may at first be puzzled by it and find it unmotivated. However, as the theory develops, one can look at it in retrospect and see how natural it is. In any case it makes

possible the extremely rapid development of a very powerful theory, and a certain patience with the abruptness of its proof will be richly rewarded.
It will be useful to have at our disposal the following notation: If $P$ is a complex number, and $r$ a positive real number, then we let $D(P, r)={z \in$ $\mathbb{C}:|z-P|<r}$ and $\bar{D}(P, r)={z \in \mathbb{C}:|z-P| \leq r}$. These sets are called, respectively, open and closed discs in the complex plane. The boundary of the disc, $\partial D(P, r)$, is the set of $z$ such that $|z-P|=r$.

The boundary $\partial D(P, r)$ of the disc $D(P, r)$ can be parametrized as a simple closed curve $\gamma:[0,1] \rightarrow \mathbb{C}$ by setting
$$\gamma(t)=P+r e^{2 \pi i t} .$$
We say that this $\gamma$ is the boundary of a disc with counterclockwise orientation. The terminology is justified by the fact that, in the usual picture of $\mathbb{C}$, this curve $\gamma$ runs counterclockwise. Similarly, the curve $\sigma:[0,1] \rightarrow \mathbb{C}$ defined by
$$\sigma(t)=P+r e^{-2 \pi i t}$$
is called the boundary of the disc $D(P, r)$ with clockwise orientation. Recall that, as already explained, line integrals around $\partial D(P, r)$ do not depend on exactly which parametrization is chosen, but they do depend on the direction or orientation of the curve. Integration along $\sigma$ gives the negative of integration along $\gamma$. Thus we can think of integration counterclockwise around $\partial D(P, r)$ as a well-defined integration process – that is, independent of the choice of parametrization-provided that it goes around in the same direction as our specific parametrization $\gamma$.
We now turn to a fundamental lemma:
Lemma 2.4.1. Let $\gamma$ be the boundary of a disc $D\left(z_{0}, r\right)$ in the complex plane, equipped with counterclockwise orientation. Let $z$ be a point inside the circle $\partial D\left(z_{0}, r\right)$. Then
$$\frac{1}{2 \pi i} \oint_{\gamma} \frac{1}{\zeta-z} d \zeta=1$$
Proof. To evaluate
$$\oint_{\gamma} \frac{1}{\zeta-z} d \zeta$$
we could proceed by direct computation. However, this leads to a rather messy (though calculable) integral. Instead, we reason as follows:
Consider the function
$$I(z)=\oint_{\gamma} \frac{1}{\zeta-z} d \zeta$$
defined for all $z$ with $\left|z-z_{0}\right|<r$. We shall establish two facts:

(i) $I(z)$ is independent of $z$;
(ii) $I\left(z_{0}\right)=2 \pi i$.
Then we shall have $I(z)=2 \pi i$ for all $z$ with $\left|z-z_{0}\right|<r$, as required.

## 数学代写|复变函数作业代写Complex function代考|The Cauchy Integral Formula: Some Examples

The Cauchy integral formula will play such an important role in our later work that it is worthwhile to see just how it works out in some concrete cases. In particular, in this section we are going to show how it applies to polynomials by doing some explicit calculations. These calculations will involve integrating some infinite series term by term. It is not very hard to justify this process in detail in the cases we shall be discussing. But the justification will be presented in a more general context later, so for the moment we shall treat the following calculations on just a formal basis, without worrying very much about convergence questions.

To simplify the notation and to make everything specific, we shall consider only integration around the unit circle about the origin. So define $\gamma:[0,2 \pi] \rightarrow \mathbb{C}$ by $\gamma(t)=\cos t+i \sin t$. Our first observations are the following:
(1) $\oint_{\gamma} \zeta^{k} d \zeta=0$ if $k \in \mathbb{Z}, k \neq-1$;
(2) $\oint_{\gamma} \zeta^{-1} d \zeta=2 \pi i$
Actually, we proved (2) by a calculation in the previous section, and we shall not repeat it. But we repeated the conclusion here to contrast it with formula (1).

To prove (1), set $f_{k}(\zeta)=(1+k)^{-1} \zeta^{k+1}$, for $k$ an integer unequal to $-1$. Then $f_{k}$ is holomorphic on $\mathbb{C} \backslash{0}$. By Proposition 2.1.6,
$$0=f_{k}(\gamma(2 \pi))-f_{k}(\gamma(0))=\oint_{\gamma} \frac{\partial f_{k}}{\partial \zeta} d \zeta=\oint_{\gamma} \zeta^{k} d \zeta .$$

This argument is similar to an argument used in Section 2.4. Of course, (1) can also be established by explicit computation:
\begin{aligned} \oint_{\gamma} \zeta^{k} d \zeta &=\int_{0}^{2 \pi}(\cos t+i \sin t)^{k} \cdot(-\sin t+i \cos t) d t \ &=i \int_{0}^{2 \pi}(\cos t+i \sin t)^{k+1} d t . \end{aligned}
If $k+1$ is positive, then
$$(*)=i \int_{0}^{2 \pi}[\cos (k+1) t+i \sin (k+1) t] d t=0$$
Here we have used the well-known DeMoivre formula:
$$(\cos t+i \sin t)^{n}=\cos (n t)+i \sin (n t),$$
which is easily proved by induction on $n$ and the usual angle-addition formulas for sine and cosine (Exercise 20 of Chapter 1 ).
Notice that
$$(\cos t+i \sin t)^{-1}=\cos t-i \sin t$$
since
$$(\cos t+i \sin t) \cdot(\cos t-i \sin t)=\cos ^{2} t+\sin ^{2} t=1 \text {. }$$
Hence, if $k+1<0$, then
\begin{aligned} \int_{0}^{2 \pi}(\cos t+i \sin t)^{k+1} d t &=\int_{0}^{2 \pi}(\cos t-i \sin t)^{-(k+1)} d t \ &=\int_{0}^{2 \pi}[\cos (-(k+1) t)-i \sin (-(k+1) t)] d t \ &=0 \end{aligned}

## 数学代写|复变函数作业代写Complex function代考|An Introduction to the Cauchy Integral Theorem

For many purposes, the Cauchy integral formula for integration around circles and the Cauchy integral theorem for rectangular or disc regions are adequate. But for some applications more generality is desirable. To formulate really general versions requires some considerable effort that would lead us aside (at the present moment) from the main developments that we want to pursue. We shall give such general results later, but for now we instead present a procedure which suffices in any particular case (i.e., in proofs or applications which one is likely to encounter) that may actually arise. The procedure can be easily described in intuitive terms; and, in any concrete case, it is simple to make the procedure into a rigorous proof. In particular, Proposition $2.6 .5$ will be a precise result for the region between two concentric circles; this result will be used heavily later on.

First, we want to extend slightly the class of curves over which we perform line integration.

Definition 2.6.1. A piecewise $C^{1}$ curve $\gamma:[a, b] \rightarrow \mathbb{C}, a<b, a, b \in \mathbb{R}$, is a continuous function such that there exists a finite set of numbers $a_{1} \leq a_{2} \leq$ $\cdots \leq a_{k}$ satisfying $a_{1}=a$ and $a_{k}=b$ and with the property that for every $1 \leq j \leq k-1,\left.\gamma\right|{\left[a{j}, a_{j+1}\right]}$ is a $C^{1}$ curve. As before, $\gamma$ is a piecewise $C^{1}$ curve (with image) in an open set $U$ if $\gamma([a, b]) \subseteq U$.

Intuitively, a piecewise $C^{1}$ curve is a finite number of $C^{1}$ curves attached together at their endpoints. The natural way to integrate over such a curve is to add together the integrals over each $C^{1}$ piece.

Definition 2.6.2. If $U \subseteq \mathbb{C}$ is open and $\gamma:[a, b] \rightarrow U$ is a piecewise $C^{1}$ curve in $U$ (with notation as in Definition 2.6.1) and if $f: U \rightarrow \mathbb{C}$ is a continuous, complex-valued function on $U$, then
$$\oint_{\gamma} f(z) d z \equiv \sum_{j=1}^{k} \oint_{\left.\gamma\right|{\left[a{j}, a_{j+1} \mid\right.}} f(z) d z,$$
where $a_{1}, a_{2}, \ldots, a_{k}$ are as in the definition of “piecewise $C^{1}$ curve.” $^{n}$
In order for this definition to make sense, we need to know that the sum on the right side does not depend on the choice of the $a_{j}$ or of $k$. This independence does indeed hold, and Exercise 33 asks you to verify it.

It follows from the remarks in the preceding paragraph together with the remarks in Section $2.1$ about parametrizations of curves that the complex line integral over a piecewise $C^{1}$ curve has a value which is independent of whatever parametrization for the curve is chosen, provided that the direction of traversal is kept the same. This (rather redundant) remark cannot be overemphasized. We leave as a problem for you (see Exercise 34) the proof of the following technical formulation of this assertion.

## 数学代写|复变函数作业代写Complex function代考|Cauchy Integral Theorem

C(吨)=磷+r和2圆周率一世吨.

σ(吨)=磷+r和−2圆周率一世吨

12圆周率一世∮C1G−和dG=1

∮C1G−和dG

（一世）一世(和)独立于和;
(二)一世(和0)=2圆周率一世.

## 数学代写|复变函数作业代写Complex function代考|The Cauchy Integral Formula: Some Examples

（1）∮CGķdG=0如果ķ∈从,ķ≠−1;
(2) ∮CG−1dG=2圆周率一世

0=Fķ(C(2圆周率))−Fķ(C(0))=∮C∂Fķ∂GdG=∮CGķdG.

∮CGķdG=∫02圆周率(因⁡吨+一世罪⁡吨)ķ⋅(−罪⁡吨+一世因⁡吨)d吨 =一世∫02圆周率(因⁡吨+一世罪⁡吨)ķ+1d吨.

(∗)=一世∫02圆周率[因⁡(ķ+1)吨+一世罪⁡(ķ+1)吨]d吨=0

(因⁡吨+一世罪⁡吨)n=因⁡(n吨)+一世罪⁡(n吨),

(因⁡吨+一世罪⁡吨)−1=因⁡吨−一世罪⁡吨

(因⁡吨+一世罪⁡吨)⋅(因⁡吨−一世罪⁡吨)=因2⁡吨+罪2⁡吨=1.

∫02圆周率(因⁡吨+一世罪⁡吨)ķ+1d吨=∫02圆周率(因⁡吨−一世罪⁡吨)−(ķ+1)d吨 =∫02圆周率[因⁡(−(ķ+1)吨)−一世罪⁡(−(ķ+1)吨)]d吨 =0

## 数学代写|复变函数作业代写Complex function代考|An Introduction to the Cauchy Integral Theorem

∮CF(和)d和≡∑j=1ķ∮C|[一种j,一种j+1∣F(和)d和,

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