Laurent series, classification of singularities, Casorati–Weierstrass

In one sentence

Count the negative powers in the Laurent series: none means removable, finitely many means a pole, infinitely many means essential — and each of those is equivalent to a statement about $|f|$ that you can check without expanding anything.

Why the exam asks it

Because "essential" gets read as "a very bad pole", and it is the opposite: a pole is the case where $|f| \to \infty$, and essential is the case where $|f|$ does not settle down at all. And because the same function has different Laurent series on different annuli, so a question can hand you the wrong one and let you classify from it.

The idea

On an annulus $r < |z - z_0| < R$ where $f$ is holomorphic,

$$f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n,$$

converging there. The principal part is the sum over $n < 0$.

The classification

$z_0$ is an isolated singularity if $f$ is holomorphic on some punctured disc around it. Then, from the Laurent series on that punctured disc:

• Removable — no negative powers. Equivalently $f$ is bounded near $z_0$ (Riemann's theorem), equivalently $\lim_{z\to z_0} f(z)$ exists. $\dfrac{\sin z}{z}$ at $0$: the series $1 - z^2/6 + \cdots$ has no negative terms at all.
• Pole of order $m$ — finitely many negative powers, the lowest being $-m$. Equivalently $|f(z)| \to \infty$. Equivalently $(z-z_0)^m f(z)$ has a removable singularity with nonzero limit.
• Essential — infinitely many negative powers. Equivalently $|f|$ has no limit, finite or infinite. $e^{1/z}$ at $0$: the series $\sum \frac{1}{n!z^n}$ never terminates.

Each of these is an equivalence, which is what makes them usable. You almost never expand a Laurent series in an exam; you take a limit and read the row.

For $e^{1/z}$: along the positive reals $e^{1/x} \to \infty$, along the negative reals $e^{1/x} \to 0$. Two different limits along two approaches, so neither "bounded" nor "$\to \infty$" holds, and the singularity is essential without expanding anything.

Zeros and poles are reciprocal

If $f$ has a zero of order $m$ at $a$ then $1/f$ has a pole of order $m$ there, and conversely. So $\dfrac{1}{\sin z}$ has simple poles exactly at $n\pi$, because $\sin$ has simple zeros there. Orders add under multiplication and subtract under division, and counting them settles most questions of the form "what kind of singularity does this have".

The annulus matters

$f(z) = \dfrac{1}{z(z-1)} = -\dfrac1z + \dfrac{1}{z-1}$.

On $0 < |z| < 1$, expand $\frac{1}{z-1} = -\frac{1}{1-z} = -(1 + z + z^2 + \cdots)$, giving

$$-\frac1z - 1 - z - z^2 - \cdots$$

— one negative power, so a simple pole at $0$, which is correct.

On $|z| > 1$, expand $\frac{1}{z-1} = \frac{1}{z}\cdot\frac{1}{1 - 1/z} = \frac1z + \frac1{z^2} + \cdots$, giving

$$\frac{1}{z^2} + \frac{1}{z^3} + \cdots$$

— infinitely many negative powers, and it would be a mistake to call the singularity essential. That series does not describe behaviour near $0$ at all; it describes behaviour near infinity. Only the expansion on the punctured disc closest to the point classifies the singularity there.

Casorati–Weierstrass and Picard

Near an essential singularity, the image of every punctured neighbourhood is dense in $\mathbb{C}$.

So $f$ comes arbitrarily close to every complex number, arbitrarily near $z_0$. Picard's great theorem sharpens this enormously: $f$ actually attains every complex value infinitely often, with at most one exception. For $e^{1/z}$ the exception is $0$, which it never takes.

Both are stated for essential singularities specifically, and both are false at poles, where $|f| \to \infty$ and the image eventually avoids every bounded set.

Not every singularity is isolated

$f(z) = \dfrac{1}{\sin(1/z)}$ has poles wherever $\sin(1/z) = 0$, that is at $z = \frac{1}{n\pi}$. These accumulate at $0$, so no punctured disc around $0$ is free of singularities, and $0$ is not an isolated singularity at all. It has no Laurent series there and the classification simply does not apply.

Branch points are the other non-isolated case: $\log z$ and $\sqrt z$ at $0$ are not isolated singularities either, because no single-valued branch exists on a punctured disc.

Where intuition breaks

"Essential means $|f| \to \infty$ faster." That describes a pole. Essential means $|f|$ has no limit — $e^{1/z}$ tends to $\infty$ one way and $0$ the other. This is the topic's central confusion.

"$f$ is unbounded near $z_0$, so there is a pole." Unbounded is not the same as tending to infinity. $e^{1/z}$ is unbounded near $0$ and has an essential singularity. The pole condition is $|f| \to \infty$, meaning every approach.

"The Laurent series has infinitely many negative powers, so the singularity is essential." Only if it is the expansion on a punctured disc around the point. The $|z|>1$ expansion of $1/(z(z-1))$ has infinitely many and the singularity at $0$ is a simple pole.

"Every singularity is isolated." $1/\sin(1/z)$ at $0$. Options quietly assume isolation so that the classification applies; it does not always.

"$f$ is bounded near $z_0$, so we cannot say much." Riemann's theorem says the singularity is removable — $f$ extends holomorphically. Bounded is a very strong hypothesis here, and recognising it as sufficient is often the whole question.

"An essential singularity means $f$ misses lots of values." The reverse. By Picard it takes every value infinitely often, with at most one exception.

"$\sqrt z$ has a pole at $0$." It has a branch point, which is not an isolated singularity. No Laurent series exists.

The exam's angle

1. Take a limit before expanding. Finite limit means removable; $\infty$ means a pole; neither means essential. That is the whole classification and it needs no series.
2. For pole order, multiply and check. $(z-z_0)^m f(z)$ having a finite nonzero limit pins $m$ exactly.
3. Count zeros of the denominator. Order of pole equals order of zero, so $1/\sin z$ has simple poles and $1/\sin^2 z$ double ones.
4. Check isolation before classifying. If the singularities accumulate — anything with $1/z$ inside a periodic function — the question is about non-isolation.
5. Read which annulus a given series belongs to. A series with infinitely many negative powers proves nothing until you know where it converges.

The night before

• No negative powers: removable $\iff$ bounded near $z_0$ $\iff$ the limit exists. $\sin z/z$.
• Finitely many, lowest $-m$: pole of order $m$ $\iff$ $|f| \to \infty$. $1/z^3$.
• Infinitely many: essential $\iff$ no limit at all. $e^{1/z}$.
• Zero of order $m$ for $f$ $\iff$ pole of order $m$ for $1/f$.
• One function, many annuli, many Laurent series. Only the innermost punctured disc classifies. $1/(z(z-1))$ has a simple pole at $0$ and an infinite principal part on $|z|>1$.
• Casorati–Weierstrass: image of every punctured neighbourhood is dense. Picard: every value infinitely often, at most one exception — $e^{1/z}$ omits $0$.
• $1/\sin(1/z)$ at $0$: poles accumulate, so not isolated, so unclassifiable.
• Branch points ($\log z$, $\sqrt z$) are not isolated singularities.