Isolated singularity

This article covers singularities of complex functions. For singularities of real functions, see definition gap .

Isolated singularities are considered in the mathematical sub-area of function theory. Isolated singularities are special isolated points in the source set of a holomorphic function . In the case of isolated singularities, a distinction is made between liftable singularities , poles and essential singularities .

definition

It is an open subset , . Furthermore, let it be a holomorphic complex-valued function . Then is called the isolated singularity of . ${\displaystyle \Omega \subseteq \mathbb {C} }$ ${\displaystyle z_{0}\in \Omega }$ ${\displaystyle f\colon \Omega \setminus \{z_{0}\}\to \mathbb {C} }$ ${\displaystyle z_{0}}$ ${\displaystyle f}$

Classification

Each isolated singularity belongs to one of the following three classes:

• The point is called liftable singularity if it can be continued on holomorphic. According to Riemann's law of lifting capacity , this is z. B. the case when in an environment of is restricted. ${\displaystyle z_{0}}$ ${\displaystyle f}$ ${\displaystyle \Omega }$ ${\displaystyle f}$ ${\displaystyle z_{0}}$
• The point is called a pole or pole if there is no liftable singularity and there is a natural number such that a liftable singularity has at . If this is chosen to be minimal, then one says that I have a pole in order. ${\displaystyle z_{0}}$ ${\displaystyle z_{0}}$ ${\displaystyle k}$ ${\displaystyle (z-z_{0})^{k}\cdot f(z)}$ ${\displaystyle z_{0}}$ ${\displaystyle k}$ ${\displaystyle f}$ ${\displaystyle z_{0}}$ ${\displaystyle k}$
• Otherwise, is one essential singularity of . ${\displaystyle z_{0}}$ ${\displaystyle f}$

Detachable singularities and poles are also summarized under the term non-essential singularity .

Isolated singularities and the Laurent series

The type of singularity can also be found in the Laurent series

${\displaystyle \sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}}$

read from in : ${\displaystyle f}$ ${\displaystyle z_{0}}$

• A liftable singularity is present if and only if the main part vanishes, i.e. H. for all negative integers . ${\displaystyle a_{n}=0}$ ${\displaystyle n}$
• A pole -th order is present if and only if the main part breaks off into terms, i.e. H. and for everyone . ${\displaystyle k}$ ${\displaystyle k}$ ${\displaystyle a_{-k}\neq 0}$ ${\displaystyle a_{n}=0}$ ${\displaystyle n<-k}$
• An essential singularity is present if and only if an infinite number of terms with negative exponents do not vanish.

Statements about the properties of holomorphic functions at essential singularities are made by Picard's Great Theorem and, as a simpler special case, the Casorati-Weierstrass theorem .

Examples

Plot of the function . It has an essential singularity at the zero point (center of the picture). The hue corresponds to the complex argument of the function value, while the lightness represents its amount. Here you can see that the essential singularity behaves differently depending on how you approach it (in contrast, a pole would be uniformly white). ${\displaystyle \exp(1/z)}$

It be and ${\displaystyle \Omega =\mathbb {C} }$ ${\displaystyle z_{0}=0.}$

• ${\displaystyle f\colon \Omega \setminus \{0\}\to \mathbb {C} ,\,z\mapsto {\tfrac {\sin(z)}{z}}}$, by steadily to be continued, so has at one liftable singularity. ${\displaystyle f(0)=1}$ ${\displaystyle \Omega }$ ${\displaystyle f}$ ${\displaystyle 0}$
• ${\displaystyle f\colon \Omega \setminus \{0\}\to \mathbb {C} ,\,z\mapsto {\tfrac {1}{z}}}$has at a pole of the first order, because through can be continued continuously on . ${\displaystyle 0}$ ${\displaystyle g(z)=z^{1}\cdot f(z)}$ ${\displaystyle g(0)=1}$ ${\displaystyle \Omega }$
• ${\displaystyle f\colon \Omega \setminus \{0\}\to \mathbb {C} ,\,z\mapsto \exp \left({\tfrac {1}{z}}\right)}$has an essential singularity because for is always unlimited for solid , or because in the Laurent series around an infinite number of terms of the main part do not disappear, because it is valid ${\displaystyle 0}$ ${\displaystyle z^{k}\exp \left({\tfrac {1}{z}}\right)}$ ${\displaystyle z\to 0}$ ${\displaystyle k\in \mathbb {N} }$ ${\displaystyle z_{0}}$

${\displaystyle f(z)=\sum _{n=0}^{\infty }{\frac {1}{n!\,z^{n}}}}$.

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• Eberhard Freitag , Rolf Busam: Theory of functions 1. Springer-Verlag, Berlin, ISBN 3-540-67641-4 .