Removable singularity

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A graph of a parabola with a removable singularity at x = 2

In complex analysis, a removable singularity (sometimes called a cosmetic singularity) of a holomorphic function is a point at which the function is undefined, but it is possible to define the function at that point in such a way that the function is regular in a neighbourhood of that point.

For instance, the function

f(z)={\frac  {\sin z}{z}}

has a singularity at z = 0. This singularity can be removed by defining f(0) := 1, which is the limit of f as z tends to 0. The resulting function is holomorphic. In this case the problem was caused by f being given an indeterminate form. Taking a power series expansion for {\frac  {\sin(z)}{z}} shows that

f(z)={\frac  {1}{z}}\left(\sum _{{k=0}}^{{\infty }}{\frac  {(-1)^{k}z^{{2k+1}}}{(2k+1)!}}\right)=\sum _{{k=0}}^{{\infty }}{\frac  {(-1)^{k}z^{{2k}}}{(2k+1)!}}=1-{\frac  {z^{2}}{3!}}+{\frac  {z^{4}}{5!}}-{\frac  {z^{6}}{7!}}+\cdots .

Formally, if U\subset {\mathbb  C} is an open subset of the complex plane {\mathbb  C}, a\in U a point of U, and f:U\setminus \{a\}\rightarrow {\mathbb  C} is a holomorphic function, then a is called a removable singularity for f if there exists a holomorphic function g:U\rightarrow {\mathbb  C} which coincides with f on U\setminus \{a\}. We say f is holomorphically extendable over U if such a g exists.

Riemann's theorem

Riemann's theorem on removable singularities states when a singularity is removable:

Theorem. Let D\subset C be an open subset of the complex plane, a\in D a point of D and f a holomorphic function defined on the set D\setminus \{a\}. The following are equivalent:

  1. f is holomorphically extendable over a.
  2. f is continuously extendable over a.
  3. There exists a neighborhood of a on which f is bounded.
  4. \lim _{{z\to a}}(z-a)f(z)=0.

The implications 1 ⇒ 2 ⇒ 3 ⇒ 4 are trivial. To prove 4 ⇒ 1, we first recall that the holomorphy of a function at a is equivalent to it being analytic at a (proof), i.e. having a power series representation. Define

h(z)={\begin{cases}(z-a)^{2}f(z)&z\neq a,\\0&z=a.\end{cases}}

Clearly, h is holomorphic on D \ {a}, and there exists

h'(a)=\lim _{{z\to a}}{\frac  {(z-a)^{2}f(z)-0}{z-a}}=\lim _{{z\to a}}(z-a)f(z)=0

by 4, hence h is holomorphic on D and has a Taylor series about a:

h(z)=c_{0}+c_{1}(z-a)+c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.

We have c0 = h(a) = 0 and c1 = h'(a) = 0; therefore

h(z)=c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.

Hence, where z≠a, we have:

f(z)=h(z)/(z-a)^{2}=c_{2}+c_{3}(z-a)+\cdots \,.

However,

g(z)=c_{2}+c_{3}(z-a)+\cdots \,.

is holomorphic on D, thus an extension of f.

Other kinds of singularities

Unlike functions of a real variable, holomorphic functions are sufficiently rigid that their isolated singularities can be completely classified. A holomorphic function's singularity is either not really a singularity at all, i.e. a removable singularity, or one of the following two types:

  1. In light of Riemann's theorem, given a non-removable singularity, one might ask whether there exists a natural number m such that \lim _{{z\rightarrow a}}(z-a)^{{m+1}}f(z)=0. If so, a is called a pole of f and the smallest such m is the order of a. So removable singularities are precisely the poles of order 0. A holomorphic function blows up uniformly near its poles.
  2. If an isolated singularity a of f is neither removable nor a pole, it is called an essential singularity. It can be shown that such an f maps every punctured open neighborhood U\setminus \{a\} to the entire complex plane, with the possible exception of at most one point.

See also

External links

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