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)^kz^{2k+1}}{(2k+1)!} \right) = \sum_{k=0}^{\infty} \frac{(-1)^kz^{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[edit]

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 \ne 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)^2f(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[edit]

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[edit]

External links[edit]