11

From what I could find, a singularity is a point at which an equation, surface, etc., blows up or becomes degenerate.

And a pole of a function is an isolated singular point a of single-valued character of an analytic function $f(z)$ of the complex variable $z$ for which $|f(z)|$ increases without bound when $z$ approaches $a$: $\lim_{z\rightarrow a}f(z) = \infty.$

I really don't fully understand this definition of a pole, like (what is an isolated singular point) and the limit says for $\lim_{z\rightarrow a}f(z) = \infty.$ What is $a$ that $z$ should approach?

$f = 1/(z-1) \ e^{z}$

Can I say $f$ has a singularity at $z = 1$ because we get $1/0$ at that point i.e. blows up and gives $\infty$?

quid
  • 42,135
HappyFeet
  • 706
  • 3
    "isolated" means that there is a neighborhood of that point such that no other singularities are in that neighborhood. $a$ is the point which is a (potential) singularity. All poles are singularities, but not vice versa (there are things called "essential singularities" which are "worse" and "removable singularities" which are "better") – BaronVT Dec 11 '14 at 20:47
  • 1
    We call a point $a$ a pole of order $n$ of a function f if $\lim_{z\to a}|f(z)|=\infty$ and $\lim_{z\to a}z^nf(z)$ exist and is finite. A pole is a special case of the singularity. – Frank Lu Dec 11 '14 at 20:48
  • 1
    @Frank you surely meant $(z-a)^nf(z)$. – quid Dec 11 '14 at 20:52
  • @Frank per your definition,since $\lim_{z\rightarrow 1}$ (1$/$(z-1)$) = \infty$. and $\lim_{z\rightarrow 1} ($(z-1)$ $f(z)$$ is finite does this mean $z = 1$ is a pole?. – HappyFeet Dec 11 '14 at 21:12
  • @quid Of course...thank you for pointing out. – Frank Lu Dec 13 '14 at 00:45
  • @Ozwurld Exactly, in this case $z=1$ is a pole of order 1 – Frank Lu Dec 13 '14 at 00:46

2 Answers2

5

A pole is a (usually defined as a) special case of a singularity. See http://en.wikipedia.org/wiki/Pole_%28complex_analysis%29

Thomas
  • 22,361
  • 1
    Or maybe http://en.wikipedia.org/wiki/Singularity_(mathematics)#Complex_analysis – mrp Dec 11 '14 at 20:53
5

One says that $z_0$ is an isolated singularity of $f$ if $f$ is defined in a punctured neighborhood $D\setminus\{z_0\}$ of $z_0$.

One says $z_0$ is a removable singularity of $f$ if there exists a holomorphic function $F(z)$ defined on $D$ which extends $f$.

Suppose $f$ is nonvanishing in a punctured neighborhood $D\setminus\{z_0\}$ of $z_0$. Define $F(z)$ on $D$ by $F(z) = 1/f(z)$ if $z \neq z_0$ and $F(z_0)=0$. Then $z_0$ is a pole of $f$ if $F$ is holomorphic at $z_0$.

A singularity $z_0$ is an essential singularity of $f$ if $z_0$ is neither a pole nor a removable singularity.

quid
  • 42,135
Frost Boss
  • 558