A point where $f$ is not analytic, but analytic on a small punctured disk around it, is an isolated singularity. Its Laurent series on that punctured disk sorts it into one of three kinds, and the kind decides how integrals around it behave.
Idea clave
Let $f(z) = \sum c_n(z - a)^n$ on $0 < |z - a| < r$.
• Removable: no negative powers. $f$ has a finite limit at $a$ and becomes analytic once $f(a)$ is defined as that limit. Example: $\frac{\sin z}{z}$ at $0$.
• Pole of order $m$: the negative powers stop at $(z - a)^{-m}$. Equivalently $f = \frac{g(z)}{(z - a)^m}$ with $g$ analytic and $g(a) \ne 0$; then $|f| \to \infty$. Example: $\frac{1}{z^2(z - 1)}$ has a pole of order 2 at $0$.
• Essential: infinitely many negative powers. Example: $e^{1/z}$ at $0$, which comes arbitrarily close to every complex value near $0$.
A zero of order $m$ means $f = (z - a)^mg(z)$ with $g(a) \ne 0$; a zero of order $m$ on the bottom of a quotient gives a pole of order $m$.
• Removable: no negative powers. $f$ has a finite limit at $a$ and becomes analytic once $f(a)$ is defined as that limit. Example: $\frac{\sin z}{z}$ at $0$.
• Pole of order $m$: the negative powers stop at $(z - a)^{-m}$. Equivalently $f = \frac{g(z)}{(z - a)^m}$ with $g$ analytic and $g(a) \ne 0$; then $|f| \to \infty$. Example: $\frac{1}{z^2(z - 1)}$ has a pole of order 2 at $0$.
• Essential: infinitely many negative powers. Example: $e^{1/z}$ at $0$, which comes arbitrarily close to every complex value near $0$.
A zero of order $m$ means $f = (z - a)^mg(z)$ with $g(a) \ne 0$; a zero of order $m$ on the bottom of a quotient gives a pole of order $m$.
Por qué funciona
The three cases cover everything: the negative part of the Laurent series either is empty, stops, or goes on forever. If it stops at $(z - a)^{-m}$, multiplying by $(z - a)^m$ leaves an ordinary power series $g$ with $g(a) = c_{-m} \ne 0$. For a quotient $\frac{h}{k}$ where $h(a) \ne 0$ and $k$ has a zero of order $m$, write $k = (z - a)^m q(z)$ with $q(a) \ne 0$; then $\frac{h}{k} = \frac{h/q}{(z - a)^m}$, a pole of order $m$. If $h$ also vanishes at $a$, subtract its order from $m$.
Worked examples
Ejemplo resuelto: Order of a pole
Classify the singularity of $\dfrac{z + 2}{(z - 1)^3(z + 4)}$ at $z = 1$.
- Write it as $\frac{g(z)}{(z - 1)^3}$ with $g(z) = \frac{z + 2}{z + 4}$.
- $g$ is analytic near $1$ and $g(1) = \frac{3}{5} \ne 0$.
- A pole of order $3$.
Respuesta: pole of order 3
Ejemplo resuelto: Cancelling zeros
Classify the singularity at $0$ of $\dfrac{1 - \cos z}{z^4}$.
- $1 - \cos z = \frac{z^2}{2} - \frac{z^4}{24} + \cdots$, a zero of order $2$.
- So $\frac{1 - \cos z}{z^4} = \frac{1}{2z^2} - \frac{1}{24} + \cdots$.
- A pole of order $4 - 2 = 2$.
Respuesta: pole of order 2
Ejemplo resuelto: Removable
Classify the singularity of $\dfrac{\sin z}{z}$ at $0$.
- $\frac{\sin z}{z} = 1 - \frac{z^2}{6} + \frac{z^4}{120} - \cdots$: no negative powers.
- Removable, with limit $1$; defining the value at $0$ to be $1$ makes the function entire.
Respuesta: removable
Error común
Misreading the kind. $\frac{\sin z}{z}$ looks like it has a pole at $0$ because of the $z$ in the denominator, but the top vanishes there too: it is removable. And $e^{1/z}$ is not a pole: along the positive real axis it blows up, but along the negative real axis it tends to $0$, so it is not true that $|f| \to \infty$. Check with the Laurent series: count the negative powers.