So I have some questions about singularities that should be rather simple to clear up. First of all, is a zero of a function a removable singularity? So if I have f(z)=$e^{z}-1$ is z=0 a removable singularity?

Can a function have multiple singularities? (I assume yes but since we're here we might as well be safe)

Now say I have something like $sin(1/z)$. When $z=0$ what type of singularity do we have? Do we have a pole (1/z clearly goes to $\infty$)? Do we have a removable singularity (the sin is bounded between 1 and -1)? Or is it essential (if it were to be neither since sin can be rewritten with the Euler formulas and isn't bounded by Liouville's theorem)

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    $\begingroup$ $z = 0$ is not a singularity of $e^z-1$... $\endgroup$ – Ivo Terek Feb 18 '15 at 0:19


A singularity is a point where things go bad at an isolated point in the domain of an analytic function. For example, $\frac1z$ has a pole at $z=0$. The function is not defined at $z=0$ and the function grows arbitrarily large as $z$ gets close to $0$. A pole at $z=a$ can be removed with multiplication by $(z-a)^n$ for a large enough integer $n$.

Removable Singularities

Certain kinds of bad can be remedied. For example, at $z=1$ in $$ f(z)=\frac{z^2-1}{z-1} $$ Technically, $f(1)$ is not defined since its evaluation requires division by $0$. However, everywhere but at $z=1$ we have $f(z)=z+1$. This singularity is removable since we can define $f(1)=2$ and then $f(z)=z+1$ everywhere.

Essential Singularities

Essential Singularities are those that are neither removable nor poles. If $f$ has an essential singularity at $z=a$, then so does $1/f$. The Great Picard Theorem describes the behavior of functions near an essential singularity. That is, the function takes on every complex value except at most one in all neighborhoods of the singularity.

Points from the question:

Unless there is a problem in the definition of an analytic function, I would not call a zero a singularity. For example, $z^2/z$ has a removable singularity at $z=0$, which happens to be a zero of the function.

Functions can have an infinite number of singularities, for example, $\cot(\pi z)$ has a pole at every integer.

The property that $|\sin(x)|\le1$ is only valid for $x\in\mathbb{R}$. In $\mathbb{C}$, $\sin(x)$ can take on any value. In fact, $\sin(1/z)$ has an essential singularity at $z=0$.


By definition a singularity at $p$ of an holomorphic map $f: U\backslash \{p\} \to \mathbb C$.

If you can define an holomorphic map $f : U \to \mathbb C$ the singularity is said removable. Here $e^z - 1$ is already defined in $0$, so it's not a singularity at all !

After, it's not clear what do you mean by "multiple singularities". The function could have one singularity with multiplicity $d > 1$ for example the map $z \mapsto z^{-d}$ or several singularities, like $z \mapsto \frac{1}{(z-1)(z-2)}$.

Finally a function is said meromorphic on $U$ if $f$ is holomorphic on $U \backslash A$ where $A$ is discrete, and for each $a \in A$ there is a neighborhood $U_a$ of $a$ and an integer $k_a$ such that $U \cap A = a$ and $z^{k_a}f$ is holomorphic on $U_a$.

If you have a singularity, you have 3 options :

1) the singularity is removable

2) we have $\lim\limits_{z \to p}|f(z)| = + \infty$ ($\Rightarrow$ the function is meromorphic)

3) the singularity is essential.

Now, since $\sin(1/z) = \sum_{n=0}^{+\infty}\frac{z^{2n+1}}{(2n+1)!}$ we have $$\sin(1/z) = \sum_{n=0}^{+\infty}\frac{1}{z^{2n+1}(2n+1)!}$$ which is clearly a non-removable singularity, and non holomorphic if multiplyed by any $z^k$. Then, $\sin(\frac{1}{z})$ has an essential singularity at the origin.


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