Classifying Singularities

Zeroes

If \(f(z)\) is holomorphic at \(z_0\) and the first $n$-terms of its Taylor series are 0. I.e. \(f(z) = a_n(z-z_0)^n + a_{n+1}(z-z_0)^{n+1}+\cdots = (z-z_0)^n(a_n+a_{n+1}(z-z_0)+\cdots) = (z-z_0)^ng(z)\). Then \(f(z)\) has 0 of order \(n\) at \(z_0\) and \(g(z_0)=a_n\) is non-zero and is holomorphic.

Zeros of holomorphic functions are isolated or the function is constant.

Singularities

There is an isolated singularity at \(z_0\) if \(f(z)\) is holomorphic in a neighborhood of \(z_0\) but not at \(z_0\).

If Laurent series has 0 principle part, \((b_n=0\forall n)\) but the function is not holomorphic at \(z_0\), then \(z_0\) is called a removable singularity.

Example: Rational function, piecewise function defined with discontinuity at \(z_0\), \(g(z)=\frac{\exp(z)-1}{z}\) (we can patch this by defining a new function with \(g(0)=1\)).

If the highest power of \(\frac{1}{z-z_0}\) with \(b_n\neq 0\) is \(n\), we say \(f(z) = \sum_m a_m(z-z_0)^m + \sum_{m=1}^n \frac{b_m}{(z-z_0)^m}\) has a pole of order \(n\) at \(z_0\). If \(n=1\) we call this a simple pole.

If the principal part of the Laurent expansion goes to \(\infty\), then the singularity at \(z_0\) is called an essential singularity (E.g. Picard’s theorem).

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:15

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