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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 $\Ann(b,0,R)$ with poles were ratios $p(z) / q(z)$ where $p$ and $q$ were holomorphic on $\ball(0,R)$ and $q(b)$ was zero. We begin by being a bit more precise about such examples.
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.
Fix a domain $D \subset \C$. A function $f : D \to \C$ 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 $\Ann(c,0,r)$.
When $f : D \to \C$ 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.
When $f : D \to \C$ is holomorphic with a zero at $b \in \C$ we say that $b$ is a zero of order $m \in \N$ 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.
Here are some examples of zeroes and their orders.
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 $f'(b) \ne 0$ then $f$ has a simple zero at $b$.
More examples of zeroes and their orders.
If $f : D \to \C$ is holomorphic and has a zero of order $m$ at $b \in D$ then on every ball $\ball(b,R) \subset D$ we have \[ f(z) = (z-b)^m g(z) \] where $g : \ball(b,R) \to \C$ is holomorphic and $g(b) \ne 0$.
Fix $\ball(b,R) \subset D$ and let \[ f(z) = \sum_{n=0}^\infty a_n (z-b)^n \] 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 \[ \begin{align*} f(z) & = \sum_{n=m}^\infty a_n (z-b)^n \\ & = a_m (z-b)^m + a_{m+1} (z-b)^{m+1} + \cdots \\ & = (z-b)^m \sum_{n=0}^\infty a_{n+m} (z-b)^n \end{align*} \] and define \[ g(z) = \sum_{n=0}^\infty a_{n+m} (z-b)^m \] on $\ball(b,R)$. 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. $\square$
We can now describe for $f(z) = p(z)/q(z)$ a precise relationship between the zeros of $q$ and the poles of $f$.
Fix holomorphic functions $p,q : \ball(b,R) \to \C$. If $p(b) \ne 0$ and $q$ has a zero of order $m$ at $b$ then $f : \Ann(b,0,R) \to \C$ defined by $f(z) = p(z) / q(z)$ has a pole of order $m$ at $b$.
In accordance with the previous lemma write $q(z) = (z-b)^m r(z)$ where $r : \ball(b,R) \to \C$ is holomorphic and $r(b) \ne 0$. Define $g : \ball(b,R) \to \C$ 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 $\ball(b,r)$ and has a Taylor expansion \[ g(z) = \sum_{n=0}^\infty a_n (z-b)^n \] thereon. Now for $z \in \Ann(b,0,r)$ we calculate that \[ f(z) = \dfrac{p(z)}{q(z)} = \dfrac{p(z)}{(z-b)^m r(z)} = \dfrac{1}{(z-b)^m} g(z) = \sum_{n=0}^\infty a_n (z-b)^{n-m} \] giving a Laurent series representation of $f$ on $\Ann(b,0,r)$. Since $a_0 = g(b) \ne 0$ the function $f$ does indeed have a pole of order $m$ at $b$. $\square$
Some examples of poles and their orders.