9.1 Poles

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We finished last week with a classification of isolated singularitites. Now we will say more about poles, building up to Cauchy's residue theorem. All of our examples of functions on $\mathsf{Ann}(b,0,R)$ with poles were ratios $p(z)/q(z)$ where $p$ and $q$ were holomorphic on $\mathsf{B}(0,R)$ and $q(b)$ was zero. We begin by being a bit more precise about such examples.

Definition (Zero)

A function $f$ on a domain $D$ has a zero at $c\in D$ if $f(c)=0$.

We will only be interested in isolated zeros.

Definition (Isolated Zero)

Fix a domain $D\subset \mathbb{C}$. A function $f:D\to \mathbb{C}$ has an isolated zero at $c\in D$ if $f(c)=0$ and there is $r>0$ such that $f$ does not have a zero on $\mathsf{Ann}(c,0,r)$.

When $f:D\to \mathbb{C}$ is holomorphic and $b\in D$ we can write $f$ as a Taylor series $$f(z)=a_0+a_1(z-b)+a_2(z-b{)}^2+a_3(z-b{)}^3+a_4(z-b{)}^4+\cdots$$ around $b$. If $b$ is a zero of $f$ then $a_0=0$. It may be that other coefficients are zero as well. We can classify zeros by the number of consecutive coefficients in the above expansion that are themselves zero.

Definition (Simple Zero)

When $f:D\to \mathbb{C}$ is holomorphic with a zero at $b\in \mathbb{C}$ we say that $b$ is a zero of order $m\in \mathbb{N}$ if $a_0,\dots ,a_{m-1}$ in the Taylor series of $f$ around $b$ are zero, and $a_m$ is non-zero. A zero of order $1$ is called a simple zero.

Example

Here are some examples of zeroes and their orders.

1. $f(z)=z^2$ has a zero of order two at $0$.
2. $f(z)=z(z+2i{)}^3$ has a zero of order three at $-2i$ and a simple zero at $0$.
3. $f(z)=z^2+4$ has simple zeros at $2i$ and $-2i$.

We know from Taylor's theorem that $a_n=f^{(n)}(b)/n!$ so the order of a zero of $f$ is related to the number of derivatives of $f$ that have a zero at $b$. For example, if $f(b)=0$ and $f^{\prime }(b)\ne 0$ then $f$ has a simple zero at $b$.

Example

More examples of zeroes and their orders.

1. $f(z)=\mathsf{sin}(z)$ has a zero at $0$. Since $f^{\prime }(0)=\mathsf{cos}(0)=1$ the zero is a simple zero.
2. $f(z)=1-\mathsf{cos}(z)$ has a zero at $0$. Here $f^{\prime }(z)=\mathsf{sin}(z)$ which also has a zero at $0$. But $f^{″}(0)=\mathsf{cos}(0)=1$ so the zero has order two.

Lemma

If $f:D\to \mathbb{C}$ is holomorphic and has a zero of order $m$ at $b\in D$ then on every ball $\mathsf{B}(b,R)\subset D$ we have $$f(z)=(z-b{)}^{m}g(z)$$ where $g:\mathsf{B}(b,R)\to \mathbb{C}$ is holomorphic and $g(b)\ne 0$.

Proof:

Fix $\mathsf{B}(b,R)\subset D$ and let $$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ be the Taylor series of $f$ around $b$. By definition of the order of a zero the coefficients $a_0,\dots ,a_{m-1}$ are all zero and $a_m$ is non-zero. We therefore have $$\begin{array}{rl}f(z)& =\sum _{n=m}^{\mathrm{\infty }}{a}_n(z-b{)}^n\\ & =a_m(z-b{)}^m+a_{m+1}(z-b{)}^{m+1}+\cdots \\ & =(z-b{)}^m\sum _{n=0}^{\mathrm{\infty }}{a}_{n+m}(z-b{)}^n\end{array}$$ and define $$g(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n+m}(z-b{)}^m$$ on $\mathsf{B}(b,R)$. Since $a_m\ne 0$ we have $g(b)\ne 0$. Moreover, the power series defining $g$ has radius of convergence at least $R$ by the Cauchy-Hadamard theorem. $◻$

We can now describe for $f(z)=p(z)/q(z)$ a precise relationship between the zeros of $q$ and the poles of $f$.

Theorem

Fix holomorphic functions $p,q:\mathsf{B}(b,R)\to \mathbb{C}$. If $p(b)\ne 0$ and $q$ has a zero of order $m$ at $b$ then $f:\mathsf{Ann}(b,0,R)\to \mathbb{C}$ defined by $f(z)=p(z)/q(z)$ has a pole of order $m$ at $b$.

Proof:

In accordance with the previous lemma write $q(z)=(z-b{)}^{m}r(z)$ where $r:\mathsf{B}(b,R)\to \mathbb{C}$ is holomorphic and $r(b)\ne 0$. Define $g:\mathsf{B}(b,R)\to \mathbb{C}$ by $g(z)=p(z)/r(z)$. As $r(b)\ne 0$ and $b$ is an isolated zero of of $q$ the function $g$ is holomorphic on some $\mathsf{B}(b,r)$ and has a Taylor expansion $$g(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ thereon. Now for $z\in \mathsf{Ann}(b,0,r)$ we calculate that $$f(z)={\displaystyle \frac{p(z)}{q(z)}}={\displaystyle \frac{p(z)}{(z-b{)}^{m}r(z)}}={\displaystyle \frac{1}{(z-b{)}^m}}g(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^{n-m}$$ giving a Laurent series representation of $f$ on $\mathsf{Ann}(b,0,r)$. Since $a_0=g(b)\ne 0$ the function $f$ does indeed have a pole of order $m$ at $b$. $◻$

Example

Some examples of poles and their orders.

1. Define $f$ on $\mathbb{C}\setminus \{3\}$ by $f(z)=\mathsf{sin}(z)/(z-3{)}^2$. Since $\mathsf{sin}(3)\ne 0$ and $(z-3{)}^2$ has a zero of order two at $3$ we conclude from the theorem that $f$ has a pole of order two at $3$.
2. Define $f$ on $\mathbb{C}\setminus \mathbb{Z}$ by $f(z)=(z+\pi )/\mathsf{sin}(\pi z)$. Since $\mathsf{sin}(\pi z)$ has a zero at every integer and $z+\pi$ does not have a zero at any integer, it follows from the theorem that $f$ has a simple pole at every integer.