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

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## 数学代写|复分析作业代写Complex function代考|Further Results on the Zeros of Holomorphic Functions

In the previous sections of this chapter, we have developed a detailed understanding of the local behavior of holomorphic functions, that is, of their behavior in a small neighborhood of a particular point. The methods we used, and especially Proposition 5.1.2, can be applied in a wider context to the “global behavior” of a holomorphic function on its whole domain of definition. In this section we state and prove two important results of this sort.

Theorem 5.3.1 (Rouché’s theorem). Suppose that $f, g: U \rightarrow \mathbb{C}$ are holomorphic functions on an open set $U \subseteq \mathbb{C}$. Suppose also that $\bar{D}(P, r) \subseteq U$ and that, for each $\zeta \in \partial D(P, r)$,
$$|f(\zeta)-g(\zeta)|<|f(\zeta)|+|g(\zeta)|$$
Then
$$\frac{1}{2 \pi i} \oint_{\partial D(P, r)} \frac{f^{\prime}(\zeta)}{f(\zeta)} d \zeta=\frac{1}{2 \pi i} \oint_{\partial D(P, r)} \frac{g^{\prime}(\zeta)}{g(\zeta)} d \zeta$$

That is, the number of zeros of $f$ in $D(P, r)$ counting multiplicities equals the number of zeros of $g$ in $D(P, r)$ counting multiplicities.

Before beginning the proof of Rouché’s theorem, we note that the (at first strange looking) inequality (*) implies that neither $f(\zeta)$ nor $g(\zeta)$ can vanish on $\partial D(P, r)$. In particular, neither $f$ nor $g$ vanishes identically; moreover, the integrals of $f^{\prime} / f$ and of $g^{\prime} / g$ on $\partial D(P, r)$ are defined.

Also, $()$ implies that the function $f(\zeta) / g(\zeta)$ cannot take a value in ${x+i 0: x \leq 0}$ for any $\zeta \in \partial D(P, r)$. If it did, say $$\frac{f(\zeta)}{g(\zeta)}=\lambda \leq 0$$ for some $\zeta \in \partial D(P, r)$, then \begin{aligned} \left|\frac{f(\zeta)}{g(\zeta)}-1\right| &=|\lambda-1| \ &=-\lambda+1 \ &=\left|\frac{f(\zeta)}{g(\zeta)}\right|+1 \end{aligned} hence $$|f(\zeta)-g(\zeta)|=|f(\zeta)|+|g(\zeta)|$$ This equality contradicts $()$.

## 数学代写|复分析作业代写Complex function代考|The Maximum Modulus Principle

Consider the $C^{\infty}$ function $g$ on the unit disc given by $g(z)=2-|z|^{2}$. Notice that $1<|g(z)| \leq 2$ and that $g(0)=2$. The function assumes an interior maximum at $z=0$. One of the most startling features of holomorphic functions is that they cannot behave in this fashion: In stating the results about this phenomenon, the concept of a connected open set occurs so often that it is convenient to introduce a single word for it.

Definition 5.4.1. A domain in $\mathbb{C}$ is a connected open set. A bounded domain is a connected open set $U$ such that there is an $R>0$ with $|z|<R$ for all $z \in U$.

Theorem 5.4.2 (The maximum modulus principle). Let $U \subseteq \mathbb{C}$ be a domain. Let $f$ be a holomorphic function on $U$. If there is a point $P \in U$ such that $|f(P)| \geq|f(z)|$ for all $z \in U$, then $f$ is constant.

Proof. Assume that there is such a $P$. If $f$ is not constant, then $f(U)$ is open by the open mapping principle. Hence there are points $\zeta$ of $f(U)$ with $|\zeta|>|f(P)|$. This is a contradiction. Hence $f$ is a constant.

Here is a consequence of the maximum modulus principle that is often useful:

Corollary 5.4.3 (Maximum modulus theorem). Let $U \subseteq \mathbb{C}$ be a bounded domain. Let $f$ be a continuous function on $\bar{U}$ that is holomorphic on $U$. Then the maximum value of $|f|$ on $\bar{U}$ (which must occur, since $\bar{U}$ is closed and bounded) must occur on $\partial U$.

## 数学代写|复分析作业代写Complex function代考|Further Results on the Zeros of Holomorphic Functions

$$|f(\zeta)-g(\zeta)|<|f(\zeta)|+|g(\zeta)|$$

$$\frac{1}{2 \pi i} \oint_{\partial D(P, r)} \frac{f^{\prime}(\zeta)}{f(\zeta)} d \zeta=\frac{1}{2 \pi i} \oint_{\partial D(P, r)} \frac{g^{\prime}(\zeta)}{g(\zeta)} d \zeta$$

$$\frac{f(\zeta)}{g(\zeta)}=\lambda \leq 0$$

$$\left|\frac{f(\zeta)}{g(\zeta)}-1\right|=|\lambda-1| \quad=-\lambda+1=\left|\frac{f(\zeta)}{g(\zeta)}\right|+1$$

$$|f(\zeta)-g(\zeta)|=|f(\zeta)|+|g(\zeta)|$$

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