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

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## 数学代写|复变函数作业代写Complex function代考|Differentiability Properties of Holomorphic Functions

The first property that we shall deduce from the Cauchy integral formula is that holomorphic functions, which by definition have (continuous) first partial derivatives with respect to $x$ and $y$, have in fact continuous derivatives of all orders. This property again contrasts strongly with the situation for real functions and for general complex-valued functions on $\mathbb{C}$, where a function can have continuous derivatives up to and including some order $k$, but not have a $(k+1)^{\text {st }}$ derivative at any point (Exercises 58 and 59$)$.
Theorem 3.1.1. Let $U \subseteq \mathbb{C}$ be an open set and let $f$ be holomorphic on $U$. Then $f \in C^{\infty}(U)$. Moreover, if $\bar{D}(P, r) \subseteq U$ and $z \in D(P, r)$, then
$$\left(\frac{\partial}{\partial z}\right)^{k} f(z)=\frac{k !}{2 \pi i} \oint_{|\zeta-P|=r} \frac{f(\zeta)}{(\zeta-z)^{k+1}} d \zeta, \quad k=0,1,2, \ldots$$
Proof. Notice that, for $z \in D(P, r)$, the function
$$\zeta \longmapsto \frac{f(\zeta)}{\zeta-z}$$
is continuous on $\partial D(P, r)$. Also, $|\zeta-z| \geq r-|z-P|>0$ for all $\zeta \in \partial D(P, r)$. It follows that
$$\frac{f(\zeta)}{\zeta-w} \rightarrow \frac{f(\zeta)}{\zeta-z}$$

as $w \rightarrow z$ uniformly over $\zeta \in \partial D(P, r)$. From this assertion, and elementary algebra, it follows that
$$\frac{1}{h}\left(\frac{f(\zeta)}{\zeta-(z+h)}-\frac{f(\zeta)}{\zeta-z}\right) \rightarrow \frac{f(\zeta)}{(\zeta-z)^{2}}$$
uniformly over $\zeta \in \partial D(P, r)$ as $h \rightarrow 0$. Therefore
\begin{aligned} \lim {h \rightarrow 0} \frac{f(z+h)-f(z)}{h} &=\lim {h \rightarrow 0} \frac{1}{2 \pi i} \frac{1}{h} \oint_{|\zeta-P|=r} \frac{f(\zeta)}{\zeta-(z+h)}-\frac{f(\zeta)}{\zeta-z} d \zeta \ &=\frac{1}{2 \pi i} \oint_{|\zeta-P|=r} \lim {h \rightarrow 0} \frac{1}{h}\left(\frac{f(\zeta)}{\zeta-(z+h)}-\frac{f(\zeta)}{\zeta-z}\right) d \zeta . \end{aligned} [Because the limit occurs uniformly, it was legitimate to interchange the order of the integral and the limit-see Appendix A.] Thus $$\lim {h \rightarrow 0} \frac{f(z+h)-f(z)}{h}=\frac{1}{2 \pi i} \oint_{|\zeta-P|=r} \frac{f(\zeta)}{(\zeta-z)^{2}} d \zeta$$
Therefore $f^{\prime}(z)$ equals
$$\frac{1}{2 \pi i} \oint_{|\zeta-P|=r} \frac{f(\zeta)}{(\zeta-z)^{2}} d \zeta$$
This same argument applied to $f^{\prime}$ shows that $f^{\prime}$ itself has a complex derivative at each point of $D(P, r)$, given by the formula
$$\left(f^{\prime}(z)\right)^{\prime}=\frac{2}{2 \pi i} \oint_{|\zeta-P|=r} \frac{f(\zeta)}{(\zeta-z)^{3}} d \zeta$$

## 数学代写|复变函数作业代写Complex function代考|Complex Power Series

The theory of Taylor series in real-variable calculus associates to each infinitely differentiable function $f$ from $\mathbb{R}$ to $\mathbb{R}$ a formal power series expansion at each point of $\mathbb{R}$, namely
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(p)}{n !}(x-p)^{n}, \quad p \in \mathbb{R} .$$
There is no general guarantee that this series converges for any $x$ other than $x=p$. Moreover, there is also no general guarantee that, even if it does converge at some $x \neq p$, its sum is actually equal to $f(x)$. An instance of this latter phenomenon is the function
$$f(x)=\left{\begin{array}{lll} e^{-1 / x^{2}} & \text { if } & x \neq 0 \ 0 & \text { if } & x=0 \end{array}\right.$$
This function can be easily checked to be $C^{\infty}$ on $\mathbb{R}$ with $0=f(0)=$ $f^{\prime}(0)=f^{\prime \prime}(0)=\ldots$ (use l’Hôpital’s rule to verify this assertion). So the

Taylor expansion of $f$ at 0 is
$$0+0 x+0 x^{2}+0 x^{3}+\cdots$$
which obviously converges for all $x$ with sum $\equiv 0$. But $f(x)$ is 0 only if $x=0$. (An example of the phenomenon that the Taylor series need not even converge except at $x=p$ is given in Exercise 64.) The familiar functions of calculus- $\sin , \cos , e^{x}$, and so forth-all have convergent power series. But most $C^{\infty}$ functions on $\mathbb{R}$ do not. Real functions $f$ that have, at each point $p \in \mathbb{R}$, a Taylor expansion that converges to $f$ for all $x$ near enough to $p$ are called real analytic on $\mathbb{R}$.

In holomorphic function theory, it is natural to attempt to expand functions in powers of $z$. We would expect to need $z$ powers only and not $\bar{z}$ powers since our original definition of holomorphic functions was designed to rule out any $\bar{z}$ ‘s (in polynomial functions in particular). Compared to the real-variable case just discussed ( $C^{\infty}$ functions from $\mathbb{R}$ to $\mathbb{R}$ ), the attempt to expand a holomorphic function in powers of $z$ works remarkably well. In fact, we shall see that if $f: U \rightarrow \mathbb{C}$ is a holomorphic function on an open set $U$ and if $P \in U$, then the formal $z$ expansion at $P$,
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(P)}{n !}(z-P)^{n}$$
converges for all $z$ in some neighborhood of $P$ and it converges to $f(z)$ for all $z$ near $P$. Specifically, if $r>0$ is such that $D(P, r) \subset U$, then the series converges to $f(z)$ for all $z$ in $D(P, r)$. This expansion shows a sense in which holomorphic functions are a natural generalization of polynomials in $z$.
We prove these assertions in the next section. First, we need to introduce complex power series formally and establish a few of their properties. We assume the reader to be familiar with real power series at the level of freshman calculus.

A sequence of complex numbers is a function from ${1,2, \ldots}$ to $\mathbb{C}$ (sometimes it is convenient to renumber and think of a sequence as a function from ${0,1,2, \ldots}$ to $\mathbb{C})$. We usually write a sequence as $\left{a_{1}, a_{2}, \ldots\right}$ or $\left{a_{k}\right}_{k=1}^{\infty}$ (resp. $\left{a_{0}, a_{1}, \ldots\right}$ or $\left.\left{a_{k}\right}_{k=0}^{\infty}\right)$. The sequence is said to converge to a limit $\ell \in \mathbb{C}$ if for any $\epsilon>0$ there is an $N_{0}$ such that $k \geq N_{0}$ implies $\left|a_{k}-\ell\right|<\epsilon$. It is frequently useful to test convergence by means of the Cauchy criterion:

Lemma 3.2.1. Let $\left{a_{k}\right}_{k=1}^{\infty}$ be a sequence of complex numbers. Then $\left{a_{k}\right}$ converges to a limit if and only if for each $\epsilon>0$ there is an $N_{0}$ such that $j, k \geq N_{0}$ implies $\left|a_{j}-a_{k}\right|<\epsilon$.

## 数学代写|复变函数作业代写Complex function代考|The Power Series Expansion for a Holomorphic Function

As previously discussed, we first demonstrate that a holomorphic function has a convergent complex power series expansion (locally) about any point in its domain. Note that since a holomorphic function is defined on an arbitrary open set $U$ while a power series converges on a disc, we cannot expect a single power series expanded about a fixed point $P$ to converge to $f$ on all of $U$.

Theorem 3.3.1. Let $U \subseteq \mathbb{C}$ be an open set and let $f$ be holomorphic on $U$. Let $P \in U$ and suppose that $D(P, r) \subseteq U$. Then the complex power series
$$\sum_{k=0}^{\infty} \frac{\left(\partial^{k} f / \partial z^{k}\right)(P)}{k !}(z-P)^{k}$$
has radius of convergence at least $r$. It converges to $f(z)$ on $D(P, r)$.
Proof. Recall that from Theorem 3.1.1 we know that $f$ is $C^{\infty}$. So the coefficients of the power series expansion make sense. Given an arbitrary $z \in D(P, r)$, we shall now prove convergence of the series at this $z$. Let $r^{\prime}$ be a positive number greater than $|z-P|$ but less than $r$ so that
$$z \in D\left(P, r^{\prime}\right) \subseteq \bar{D}\left(P, r^{\prime}\right) \subseteq D(P, r) .$$
Assume without loss of generality that $P=0$ (this simplifies the notation considerably, but does not change the mathematics) and apply the Cauchy integral formula to $f$ on $D\left(P, r^{\prime}\right)$. Thus for $z \in D\left(P, r^{\prime}\right)=D\left(0, r^{\prime}\right)$ we have
$$f(z)=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta-z} d \zeta$$

\begin{aligned} &=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \frac{1}{1-z \cdot \zeta^{-1}} d \zeta \ &=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \sum_{k=0}^{\infty}\left(z \cdot \zeta^{-1}\right)^{k} d \zeta \end{aligned}
Notice that the last equality is true because $|z|<r^{\prime},|\zeta|=r^{\prime}$, hence
$$\left|z \cdot \zeta^{-1}\right|<1$$
so $\sum\left(z \cdot \zeta^{-1}\right)^{k}$ is a convergent geometric series expansion which converges to $1 /\left(1-z \cdot \zeta^{-1}\right)$ (see Exercise 7 of Chapter 2). Moreover, the series converges uniformly on $\left{\zeta:|\zeta|=r^{\prime}\right}$. This fact allows us to switch the sum and the integral. This step is so crucial that we write it out rather carefully (see Appendix A). Set $S_{N}(z, \zeta)=\sum_{k=0}^{N}\left(z \cdot \zeta^{-1}\right)^{k}$. Then we have
\begin{aligned} (*) &=\frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} \lim {N \rightarrow \infty} S{N}(z, \zeta) d \zeta \ &=\frac{1}{2 \pi i} \lim {N \rightarrow \infty} \oint{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta} S_{N}(z, \zeta) d \zeta \end{aligned}
by uniform convergence. Now this last expression equals
$$=\frac{1}{2 \pi i} \lim {N \rightarrow \infty} \sum{k=0}^{N} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta}\left(z \cdot \zeta^{-1}\right)^{k} d \zeta$$
since finite sums always commute with integration. The last equation equals
\begin{aligned} &=\lim {N \rightarrow \infty} \sum{k=0}^{N} z^{k} \cdot \frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta^{k+1}} d \zeta \ &=\sum_{k=0}^{\infty} z^{k} \frac{1}{2 \pi i} \oint_{|\zeta|=r^{\prime}} \frac{f(\zeta)}{\zeta^{k+1}} d \zeta \ &=\sum_{k=0}^{\infty} z^{k} \frac{1}{k !} \frac{\partial^{k} f}{\partial z^{k}}(0) \end{aligned}
by Theorem 3.1.1.

## 数学代写|复变函数作业代写Complex function代考|Differentiability Properties of Holomorphic Functions

(∂∂和)ķF(和)=ķ!2圆周率一世∮|G−磷|=rF(G)(G−和)ķ+1dG,ķ=0,1,2,…

G⟼F(G)G−和

F(G)G−在→F(G)G−和

1H(F(G)G−(和+H)−F(G)G−和)→F(G)(G−和)2

12圆周率一世∮|G−磷|=rF(G)(G−和)2dG

(F′(和))′=22圆周率一世∮|G−磷|=rF(G)(G−和)3dG

## 数学代写|复变函数作业代写Complex function代考|Complex Power Series

∑n=0∞F(n)(p)n!(X−p)n,p∈R.

$$f(x)=\left{和−1/X2 如果 X≠0 0 如果 X=0\对。$$

0+0X+0X2+0X3+⋯

∑n=0∞F(n)(磷)n!(和−磷)n

## 数学代写|复变函数作业代写Complex function代考|The Power Series Expansion for a Holomorphic Function

∑ķ=0∞(∂ķF/∂和ķ)(磷)ķ!(和−磷)ķ

F(和)=12圆周率一世∮|G|=r′F(G)G−和dG=12圆周率一世∮|G|=r′F(G)G11−和⋅G−1dG =12圆周率一世∮|G|=r′F(G)G∑ķ=0∞(和⋅G−1)ķdG

|和⋅G−1|<1

(∗)=12圆周率一世∮|G|=r′F(G)G林ñ→∞小号ñ(和,G)dG =12圆周率一世林ñ→∞∮|G|=r′F(G)G小号ñ(和,G)dG

=12圆周率一世林ñ→∞∑ķ=0ñ∮|G|=r′F(G)G(和⋅G−1)ķdG

=林ñ→∞∑ķ=0ñ和ķ⋅12圆周率一世∮|G|=r′F(G)Gķ+1dG =∑ķ=0∞和ķ12圆周率一世∮|G|=r′F(G)Gķ+1dG =∑ķ=0∞和ķ1ķ!∂ķF∂和ķ(0)

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