[복소해석학] 3.1. Zeros and poles

Definition. A point singularity of a function $f$ is a complex number $z_0$ such that $f$ is defined in a neighborhood of $z_0$ but not at the point itself. We also call such point isolated singularity, since that one is the only point satisfying such properties in its neighborhood. By its definition, the singularity is a local property.

A holomorphic function can have three different types of singularities. In order of increasing severity, these are removable singularities, poles and essential singularities. Soon we will get to know that Riemann's work on singularities categorize the types of the singularities in connection with the value $|f(z)|$.

Since singularities often appear because the denominator of a fraction vanishes, we begin with a local study of the zeros of a holomorphic function. A complex number $z_0$ is a zero for the holomorphic function $f$ if $f(z_0)=0$. The identity theorem([1], Chapter 2, Theorem 4.8.) shows that the zeros of a holomorphic function $f$ are isolated unless $f$ is identically zero. That is, if $f$ is holomorphic function which is not identically zero, for $z_0$, the zero of $f$, we may take a neighborhood $U$ of $z_0$ such that $z\in U\setminus\{z_0\} \Rightarrow f(z)\ne 0$.
 
Theorem 1. Suppose that $f$ is holomorphic in a connected open set $\Omega$ having a zero at $z_0\in\Omega$ and is not a identical zero in $\Omega$. Then we may take a neighborhood $U\subset\Omega$ of $z_0$, a non-vanishing holomorphic function $g$ on $U$ and a unique positive integer $k$ such that
 
$$f(z)=(z-z_0)^kg(z)\;\;\;\;\;\;\operatorname{for\;all}\;z\in U$$

proof. Suppose $\Omega$ is not connected then we have a separation $\Omega=A\cup B$ of open sets. If $z_0\in A$ then we may define $f$ by identically zero in $A$ and not identically zero in $B$, which makes $f$ vanishes in the neighborhood of $z_0$. But since $\Omega$ is connected and $f$ is not identically zero, $f$ is not identically zero in any neighborhood of $z_0$. Then by identity theorem, we may take a small disc $D$ centered at $z_0$ such that $z\in D\setminus\{z_0\}\Rightarrow f(z)\ne 0$. Since $f$ is holomorphic in $\Omega$, containing $\overline{D}$ so we have a power series expansion
 
$$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$$

for $z\in D$ of $f$ and there exists the smallest integer $k\in\mathbb{N}$ such that $a_k\ne 0$. Then we may rewrite the power series as
    \begin{align*}
        f(z)&=\sum_{n=0}^{\infty}a_n(z-z_0)^n\\
        &=a_k(z-z_0)^k+a_{k+1}(z-z_0)^{k+1}+\cdots\\
        &=(z-z_0)^k\left(a_k+a_{k+1}(z-z_0)+\cdots\right)=(z-z_0)^k\cdot\sum_{n=k+1}^{\infty}a_n(z-z_0)^{n-k}
    \end{align*}
Now define $g(z)=\sum_{n=k+1}^{\infty}a_n(z-z_0)^{n-k}$ for $z\in D$ then $g$ is absolutely holomorphic in $D$ and $g(z)\ne 0$ for $z\in D$ since the only zero of $f(z)=(z-z_0)^k g(z)$ in $D$ is $z_0$. For the uniqueness of $k$, suppose we have
 
$$f(z)=(z-z_0)^k g(z)=(z-z_0)^m h(z)$$

and assume $k>m$. Then $h(z)=(z-z_0)^{k-m}g(z)$ so $h(z)\to 0$ as $z\to z_0$ which is a contradiction. Similarly we can easily show that $k<m$ yields contradiction so $k=m$ as desired.

We say that the multiplicity of $z_0$ in $f$ is $k$ in this case. If the multiplicity of zero is $1$, we say that it is simple.
 
Definition. A deleted neighborhood of $z_0$ is an open disc centered at $z_0$ minus the point $z_0$. That is, the set
 
$$\{z:0<|z-z_0|<r\}$$
 
for some $r>0$. We say a function $f$ defined in a deleted neighborhood of $z_0$ has a \textbf{pole} at $z_0$ if the function $1/f$, defined to be zero at $z_0$, is holomorphic in a full neighborhood of $z_0$.

Example) Let $f(z)=1/(z-\alpha)(z-\beta)$ for $\alpha\in\mathbb{C}$ be defined in the deleted neighborhood $D_\alpha^{*}$ centered at $\alpha$ of radius $r=|\alpha-\beta|/2$. Then $1/f$ in $D_\alpha$ defined by
 
$$\left(\frac{1}{f}\right)(z)=\begin{cases}
                        (z-\alpha)(z-\beta) & \text{if}\;\; z\in D_\alpha^*\\
                        0 & \text{if}\;\; z=\alpha
                   \end{cases}$$

is holomorphic in $D_\alpha$ so $f$ defined in $D_\alpha^*$ has a pole at $\alpha$. Similarly if the same $f$ is defined in $\mathbb{C}\setminus\{\alpha, \beta\}$ then we say that $f$ has a pole at $\alpha, \beta$.

Theorem 2. If $f$ has a pole at $z_0\in\Omega$, then in a neighborhood of that point there exists a non-vanishing holomorphic function $h$ and a unique positive integer $k$ such that
 
$$f(z)=(z-z_0)^{-k}h(z)$$

proof. We simply apply \textbf{Theorem 2.2.} Since $1/f$ has a zero at $z_0\in\Omega$, holomorphic in $\Omega$ and not identically zero in $\Omega$ so we may write $1/f$ as
 
$$1/f(z)=(z-z_0)^k g(z)$$

for unique $k\in\mathbb{N}$. Then we have a sufficiently small open disc, say $U$ of $z_0$ where $g$ does not vanish - see the argument of the previous theorem. So $h=1/g$ is well defined for all $z\in U$ and we conclude that
 
$$f(z)=(z-z_0)^{-k}h(z)$$
 
as desired.
 
Now since $h$ is holomorphic in $U$ we may take a smaller disc $U'$ such that $\overline{U'}\subset U$. Then $h$ is also holomorphic in $\overline{U'}$ and thus for $z\in U'$ we may write $h(z)=A_0+A_1(z-z_0)+\cdots$, obtaining
\begin{align*}
    f(z)&=(z-z_0)^{-k}h(z)=(z-z_0)^{-k}(A_0+A_1(z-z_0)+\cdots)\\
    &=\frac{a_{-k}}{(z-z_0)^k}+\frac{a_{-k+1}}{(z-z_0)^{k-1}}+\cdots+\frac{a_{-1}}{(z-z_0)}+a_0+a_1(z-z_0)+\cdots\\
\end{align*}
 
Theorem 3. If $f$ has a pole of order $k$ at $z_0$, then
 
$$f(z)==\frac{a_{-k}}{(z-z_0)^k}+\frac{a_{-k+1}}{(z-z_0)^{k-1}}+\cdots+\frac{a_{-1}}{(z-z_0)}+G(z)$$

where $G(z)$ is a holomorphic function in a neighborhood of $z_0$.

$$\frac{a_{-k}}{(z-z_0)^k}+\frac{a_{-k+1}}{(z-z_0)^{k-1}}+\cdots+\frac{a_{-1}}{(z-z_0)}$$
is called the principal part of $f$ at the pole $z_0$ and denoted by $P(z)$. For any closed curve $\gamma\subset U'$ having $z_0$ inside of $\gamma$ we have
\begin{align*}
    \int_\gamma f(z)dz&=\int_\gamma P(z)+G(z) dz\\
    &=\int_\gamma \frac{a_{-k}}{(z-z_0)^k}+\frac{a_{-k+1}}{(z-z_0)^{k-1}}+\cdots+\frac{a_{-1}}{(z-z_0)}dz\\
    &=\int_\gamma \frac{a_{-1}}{(z-z_0)}dz\\
    &=\int_{|z-z_0|=\epsilon} \frac{a_{-1}}{(z-z_0)}dz=2\pi i\cdot a_{-1}
\end{align*}
for sufficiently small $\epsilon>0$ by the generalized Cauchy theorem and the fact that $1/(z-z_0)^k$ has a primitive for $k>1$. From this perspective, as a remainder after an integration, we call $a_{-1}$ of the principal part the \textbf{residue} of $f$ at the pole and denote it as $\operatorname{res}_{z_0} f$ or $\operatorname{res}(f,z_0)$. In the next section, we generalize this result as the residue formula.

We easily obtain that

$$\operatorname{res}(f,z_0)=\lim_{z\to z_0}(z-z_0) f(z)$$
 
by simple calculation. Even if the pole has a higher order, a similar formula holds.
 
Theorem 4. If $f$ has a pole of order $k$ at $z_0$ then
 
$$\operatorname{res}(f,z_0)=\lim_{z\to z_0}\frac{1}{(k-1)!}\left(\frac{d}{dz}\right)^{k-1}(z-z_0)^k f(z)$$

proof. Simply calculate the $k^{th}$ derivative of $(z-z_0)^k f(z)$ then we obtain desired result from the fact that holomorphic function is infinitely complex differentiable.

지금까지 보인 정리들은 다음에 나오는 유수 공식의 증명에 사용된다. 추가로 다음 절에서 공부할 몇 가지 정리를 익히고 나면 $f$가 $z_0$에서 pole을 가질 때 $z_0$의 적당한 근방 안에서 $f$를 $f=P+G$로 쓸 수 있게 된다. 이 내용과 서문에서 증명한 일반화된 코시의 정리를 이용하면 유수 공식을 쉽게 증명할 수 있다.

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[3], 김영원, 2023. 복소해석학 강의노트. 한빛아카데미.
[4], Rudin, Walter 1976. Principles of Mathematical Analysis 3ed. McGraw-Hill Press.
[5], Munkres, J.R., 2000. Topology 2ed. Pearson Education.