Laurent Series
Isolated singular points classification
Domain of convergence
We consider a generalization of the power series known as the Laurent series. Let z_0 be a point in the complex plane and \{c_n\}_{n \in \mathbb{Z}} be a sequence of complex coefficients. A Laurent series centered at z_0 is an expression of the form:
\sum_{n=-\infty}^\infty c_n \left(z - z_0 \right)^n
To analyze the convergence properties of this series, it is advantageous to decompose it into two distinct components:
\sum_{n=-\infty}^\infty c_n \left(z - z_0 \right)^n = \sum_{n=0}^\infty c_n \left(z - z_0 \right)^n + + \sum_{n=1}^\infty c_{-n} \frac{1}{\left(z - z_0 \right)^n}
The domain where the Laurent series converges is the intersection of the convergence domains of these two individual series.
The first component, comprising the terms with non-negative indices, is designated the regular part of the series. This is a standard power series:
f_+(z) = \sum_{n=0}^\infty c_n \left(z - z_0 \right)^n
As established by Abel’s theorem, this series converges absolutely inside an open disk centered at z_0. We denote the radius of this disk by R_1. The series converges to an analytic function f_+(z) within this disk. The domain of convergence is therefore:
|z - z_0| < R_1
The second component, containing all terms with negative indices, is known as the principal part of the series. To ascertain its domain of convergence, we introduce a change of variable. Let:
\zeta = \frac{1}{z-z_0}
With this substitution, the principal part transforms into a conventional power series in the variable \zeta:
g(\zeta) = \sum_{n=1}^\infty c_{-n} \zeta^n
This power series in \zeta converges within a disk |\zeta| < \rho for some radius of convergence \rho. Let us define this radius as \rho = 1/R_2. The function g(\zeta) is analytic inside this disk:
|\zeta| < \frac{1}{R_2}
Reverting to the original variable z, the condition for convergence of the principal part becomes:
|z - z_0| > R_2
This domain is the exterior of a disk of radius R_2 centered at z_0. Within this region, the principal part converges to an analytic function f_-(z).
The full Laurent series converges only where both its regular and principal parts converge simultaneously. This requires that a point z satisfies both conditions: |z - z_0| < R_1 and |z - z_0| > R_2. Such a region is non-empty if and only if R_2 < R_1. When this condition holds, the domain of convergence is an open annulus:
R_2 < |z - z_0| < R_1
In this annulus, the Laurent series converges to an analytic function f(z), which is the sum of the functions defined by its two parts:
f(z) = f_+(z) + f_-(z) = \sum_{n=-\infty}^\infty c_n \left(z - z_0 \right)^n
The function f(z) is analytic in this annulus because both f_+(z) and f_-(z) are analytic there. Should the condition R_2 < R_1 not be met, meaning R_2 \ge R_1, the domains of convergence for the two series are disjoint (or meet only at the boundary circle if R_1 = R_2). In this scenario, there is no open set in the complex plane where the Laurent series converges.
Theorem: a function f(z), analytic in the open annulus R_2 < |z - z_0| < R_1, is represented uniquely in that annulus by a convergent Laurent series:
f(z) = \sum_{n=-\infty}^\infty c_n(z - z_0)^n
The coefficients c_n are given by the integral formula:
c_n = \frac{1}{2\pi i} \oint_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \mathrm d\zeta, \quad n = 0, \pm 1, \pm 2, \dots
where C is any simple closed counter-clockwise contour lying within the annulus and enclosing the point z_0.
Proof: Let us fix an arbitrary point z within the annulus R_2 < |z - z_0| < R_1. We can then construct two concentric circles, C_{R_1^\prime} and C_{R_2^\prime} with radii R_1^\prime and R_2^\prime respectively, also centered at z_0, such that the point z is situated between them. This requires the radii to satisfy the condition R_2 < R_2^\prime < |z - z_0| < R_1^\prime < R_1. The region bounded by these two circles is a smaller annulus that lies entirely within the domain of analyticity of f(z).
By Cauchy’s Integral Formula for a multiply connected domain, the value of the function f(z) can be expressed as:
f(z) = \frac{1}{2\pi i} \oint_{C_{R_1^\prime}} \frac{f(\zeta)}{\zeta - z} \mathrm d\zeta - \frac{1}{2\pi i} \oint_{C_{R_2^\prime}} \frac{f(\zeta)}{\zeta - z} \mathrm d\zeta
Here, both contours C_{R_1^\prime} and C_{R_2^\prime} are traversed in the standard counter-clockwise direction. We will analyze each integral separately.
For the first integral, over the outer circle C_{R_1^\prime}, the variable of integration \zeta satisfies |\zeta - z_0| = R_1^\prime. For the chosen point z, we have |z - z_0| < R_1^\prime. Consequently, the ratio |\frac{z-z_0}{\zeta - z_0}| < 1. We can expand the kernel of the integral as a geometric series:
\begin{aligned} \frac{1}{\zeta - z} &= \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \cdot \frac{1}{1 - \frac{z - z_0}{\zeta - z_0}} \\ &= \frac{1}{\zeta - z_0} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta - z_0}\right)^n = \sum_{n=0}^\infty \frac{(z-z_0)^n}{(\zeta - z_0)^{n+1}} \end{aligned}
This series converges uniformly for \zeta on C_{R_1^\prime} because the ratio of successive terms is bounded by a constant less than one. This uniform convergence permits the interchange of summation and integration. Substituting this expansion into the first integral gives:
\begin{aligned} \frac{1}{2\pi i} \oint_{C_{R_1^\prime}} \frac{f(\zeta)}{\zeta - z} \mathrm d\zeta &= \frac{1}{2\pi i} \oint_{C_{R_1^\prime}} f(\zeta) \sum_{n=0}^\infty \frac{(z-z_0)^n}{(\zeta - z_0)^{n+1}} \mathrm d\zeta \\ &= \sum_{n=0}^\infty \left( \frac{1}{2\pi i} \oint_{C_{R_1^\prime}} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \mathrm d\zeta \right) (z-z_0)^n \\ &= \sum_{n=0}^\infty c_n (z-z_0)^n \end{aligned}
This is the regular part of the Laurent series, with coefficients c_n for n \ge 0.
For the second integral, over the inner circle C_{R_2^\prime}, the variable \zeta satisfies |\zeta - z_0| = R_2^\prime. In this case, |z - z_0| > R_2^\prime, which implies that the ratio |\frac{\zeta-z_0}{z-z_0}| < 1. We now expand the kernel in a different manner:
\begin{aligned} \frac{1}{\zeta - z} &= \frac{1}{(\zeta - z_0) - (z - z_0)} = -\frac{1}{z - z_0} \cdot \frac{1}{1 - \frac{\zeta - z_0}{z - z_0}} \\ &= -\frac{1}{z - z_0} \sum_{n=0}^\infty \left(\frac{\zeta-z_0}{z - z_0}\right)^n = -\sum_{n=0}^\infty \frac{(\zeta-z_0)^n}{(z - z_0)^{n+1}} \end{aligned}
Substituting this uniformly convergent series into the second integral yields:
\begin{aligned} -\frac{1}{2\pi i} \oint_{C_{R_2^\prime}} \frac{f(\zeta)}{\zeta - z} \mathrm d\zeta &= \frac{1}{2\pi i} \oint_{C_{R_2^\prime}} f(\zeta) \sum_{n=0}^\infty \frac{(\zeta-z_0)^n}{(z - z_0)^{n+1}} \mathrm d\zeta \\ &= \sum_{n=0}^\infty \left( \frac{1}{2\pi i} \oint_{C_{R_2^\prime}} f(\zeta)(\zeta-z_0)^n \mathrm d\zeta \right) \frac{1}{(z - z_0)^{n+1}} \end{aligned}
To align this with the standard form of a Laurent series, we perform a change of index. Let m = -(n+1), so n = -m-1. As n goes from 0 to \infty, m goes from -1 to -\infty. The sum becomes:
\sum_{m=-\infty}^{-1} \left( \frac{1}{2\pi i} \oint_{C_{R_2^\prime}} \frac{f(\zeta)}{(\zeta-z_0)^{m+1}} \mathrm d\zeta \right) (z-z_0)^m = \sum_{m=-\infty}^{-1} c_m (z-z_0)^m
This constitutes the principal part of the series.
The integrands for the coefficients c_n are analytic throughout the annulus R_2 < |\zeta - z_0| < R_1. By Cauchy’s theorem for multiply connected domains, the value of the integrals does not change if the contour of integration is deformed, as long as it remains within this annulus. We can therefore replace both C_{R_1^\prime} and C_{R_2^\prime} with a single, arbitrary closed contour C that lies in the annulus and encircles z_0. This unifies the formulas for all coefficients into a single expression:
c_n = \frac{1}{2\pi i} \oint_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \mathrm d\zeta, \quad n \in \mathbb{Z}
Combining the results for the two integrals confirms that f(z) is represented by the full Laurent series.
To demonstrate uniqueness, assume there exists another Laurent series representation:
f(z) = \sum_{n=-\infty}^\infty c_n^\prime (z-z_0)^n
valid in the same annulus. Equating the two series gives:
\sum_{n=-\infty}^\infty c_n (z-z_0)^n = \sum_{n=-\infty}^\infty c_n^\prime (z-z_0)^n
Let us multiply both sides by (z-z_0)^{-m-1} for an arbitrary integer m and integrate term-by-term around a circular contour C_R with radius R such that R_2 < R < R_1. The uniform convergence of the series on C_R justifies this operation. We rely on the integration result (\zeta - z_0 = Re^{i\theta}, \mathrm d\zeta / \mathrm d\theta =iRe^{i\theta}, \mathrm d\zeta = iRe^{i\theta}\mathrm d\theta):
\oint_{C_R} (\zeta-z_0)^k \mathrm d\zeta = \int_0 ^{2\pi} R^{k+1}i e^{(k+1)\theta}\mathrm d\theta = \begin{cases} 2\pi i, & k = -1 \\ 0 & k \ne -1 \end{cases}
Applying this to our series, the integral of (\zeta-z_0)^n (\zeta-z_0)^{-m-1} = (\zeta-z_0)^{n-m-1} is non-zero only when n-m-1 = -1, which means n=m. After integration, only one term from each infinite sum survives:
c_m (2\pi i) = c_m^\prime (2\pi i)
This implies c_m = c_m^\prime. Since m was an arbitrary integer, all coefficients must be identical. The Laurent series representation is unique.
It follows from this theorem that the annulus of convergence for a Laurent series is the largest possible annulus, bounded by the singularities of the analytic function it represents.
Isolated singular points classification
A point z_0 is designated as an isolated singular point of a function f(z) if f(z) is single-valued and analytic in an annulus, defined by 0 < |z - z_0| < R for some positive radius R. Within this annulus, the function can be represented by a convergent Laurent series. The character of the singularity at z_0 is determined entirely by the structure of this series expansion. There are three mutually exclusive possibilities for the Laurent series of f(z) centered at z_0:
- the series contains no terms with negative powers of (z - z_0), so the principal part, is identically zero;
- the principal part of the series contains a finite, non-zero number of terms;
- the principal part of the series contains an infinite number of terms.
Removable singularity
The first case corresponds to a removable singularity. Here, the Laurent series for f(z) in the neighborhood of z_0 lacks a principal part and takes the form of a Taylor series:
f(z) = \sum_{n=0}^\infty c_n (z-z_0)^n
The existence of this series representation implies that the limit of f(z) as z approaches z_0 exists and is finite. This leads to a formal statement.
Theorem: if a point z_0 is a removable singularity of the function f(z), then the limit of the function as z \to z_0 exists and is equal to the constant term of its series expansion:
\lim_{z \to z_0} f(z) = c_0
If the function was not initially defined at z_0, we can extend its definition by setting f(z_0) = c_0. This action renders the function f(z) analytic in the entire disk |z - z_0| < R, effectively “removing” the singularity. Consequently, the function f(z) is bounded in any neighborhood of z_0. If \lim_{z \to z_0} f(z) = 0, then c_0=0, and the function has a zero at z_0. The order of this zero is the smallest integer m > 0 such that c_m \neq 0.
A converse statement, known as Riemann’s theorem on removable singularities, provides a criterion for identifying such points based on the function’s behavior.
Theorem: if a function f(z) is analytic and bounded in the annulus 0 < |z - z_0| < R, then the point z_0 is a removable singularity of f(z).
Proof: we examine the coefficients of the Laurent series expansion of f(z). The formula for these coefficients is given by:
c_n = \frac{1}{2\pi i} \oint_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \mathrm d\zeta
Here, C can be any circle |\zeta - z_0| = \rho with 0 < \rho < R. By hypothesis, there exists a constant M such that |f(z)| \le M within the annulus. We can bound the magnitude of the coefficients:
\begin{aligned} |c_n| & \le \frac{1}{2\pi} \oint_{C} \frac{|f(\zeta)|}{|\zeta - z_0|^{n+1}} |\mathrm d\zeta| \\ & \le \frac{1}{2\pi} \frac{M}{\rho^{n+1}} (2\pi \rho) = M \rho^{-n} \end{aligned}
Now, consider the coefficients of the principal part, for which n < 0. Let n = -k where k is a positive integer. The inequality becomes:
|c_{-k}| \le M \rho^k
Since the coefficients c_n are independent of the choice of integration contour, this inequality must hold for any \rho such that 0 < \rho < R. As we can let \rho \to 0^+, the term M\rho^k approaches zero. This forces |c_{-k}| = 0 for all k \ge 1. All coefficients of the principal part are zero, which confirms that z_0 is a removable singularity.
Pole
The second case arises when the principal part of the Laurent series contains a finite number of terms. If the highest power of (z-z_0)^{-1} is m, the series is:
f(z) = \sum_{n=-m}^{\infty} c_n (z - z_0)^n, \quad \text{where } c_{-m} \neq 0
In this scenario, the point z_0 is called a pole of order m of the function f(z).
Theorem: if the point z_0 is a pole of the function f(z), then the modulus of the function, |f(z)|, increases without bound as z \to z_0, regardless of the path of approach.
Proof: we can factor the function f(z) in a neighborhood of z_0:
\begin{aligned} f(z) & = \frac{c_{-m}}{(z-z_0)^m} + \frac{c_{-m+1}}{(z-z_0)^{m-1}} + \cdots + \sum_{n=0}^{\infty} c_n (z - z_0)^n \\ & = (z - z_0)^{-m} \left[ c_{-m} + c_{-m+1}(z-z_0) + \cdots + c_{-1}(z-z_0)^{m-1} + \sum_{n=0}^{\infty} c_n (z - z_0)^{n+m} \right] \end{aligned}
Let us define an auxiliary function \phi(z) as the expression in the brackets:
\phi(z) = c_{-m} + c_{-m+1}(z-z_0) + \cdots
This function \phi(z) is analytic in a disk around z_0 and \lim_{z\to z_0} \phi(z) = c_{-m} \neq 0. The original function can be written as:
f(z) = \frac{\phi(z)}{(z-z_0)^m}
As z \to z_0, the denominator (z-z_0)^m approaches zero, while the numerator \phi(z) approaches the non-zero constant c_{-m}. Consequently, |f(z)| \to \infty.
The converse of this theorem provides a practical test for poles.
Theorem: if a function f(z) is analytic in a punctured neighborhood of z_0 and \lim_{z\to z_0} |f(z)| = \infty, then the point z_0 is a pole of f(z).
Proof: by hypothesis, for any large number A > 0, there exists a neighborhood of z_0 where |f(z)| > A. Consider the reciprocal function g(z) = 1/f(z). In this neighborhood, g(z) is analytic and satisfies |g(z)| < 1/A. Since \lim_{z \to z_0} |f(z)| = \infty, it follows that \lim_{z \to z_0} g(z) = 0.
According to Riemann’s theorem, since g(z) is bounded (and approaches zero), the point z_0 must be a removable singularity for g(z). Because its limit is zero, z_0 is a zero of g(z) of some integer order m \ge 1. We can therefore write:
g(z) = (z-z_0)^m h(z)
where h(z) is analytic and h(z_0) \neq 0. For the original function f(z), we have:
f(z) = \frac{1}{g(z)} = \frac{1}{(z-z_0)^m h(z)} = \frac{\phi(z)}{(z-z_0)^m}
where \phi(z) = 1/h(z) is analytic and non-zero at z_0. This is the characteristic form of a function with a pole of order m at z_0.
This establishes a duality: a point z_0 is a zero of order m for an analytic function g(z) if and only if it is a pole of order m for the reciprocal function f(z) = 1/g(z).
Essential singularity
The third and final case is that of an essential singularity, where the principal part of the Laurent series contains infinitely many terms:
f(z) = \sum_{n=-\infty}^\infty c_n (z-z_0)^n
The behavior of a function near an essential singularity is described by the Sokhotski-Weierstrass Theorem.
Theorem: in any neighborhood of an essential singularity z_0, the values taken by the function f(z) are dense in the complex plane. That is, for any complex number w, any \varepsilon > 0, and any \delta > 0, there exists a point z in the disk 0 < |z-z_0| < \delta such that |f(z) - w| < \varepsilon.
Proof: the proof is by contradiction. We assume the theorem is false. Then there exists some complex number w, some \varepsilon > 0, and some \delta > 0 such that for all z in the annulus 0 < |z - z_0| < \delta, we have:
| f(z) - w | \ge \varepsilon
Now, consider the auxiliary function:
g(z) = \frac{1}{f(z)-w}
This function is analytic in the punctured disk 0 < |z - z_0| < \delta. The condition |f(z) - w| \ge \varepsilon implies that |g(z)| \le 1/\varepsilon. Since g(z) is analytic and bounded in this annulus, Riemann’s theorem asserts that z_0 is a removable singularity for g(z). This means that \lim_{z \to z_0} g(z) exists and is finite. Let this limit be L.
Two possibilities arise for the value of L:
- If L \neq 0, then \lim_{z\to z_0} (f(z)-w) = 1/L. This implies \lim_{z\to z_0} f(z) = w + 1/L, which means z_0 is a removable singularity of f(z). This contradicts the hypothesis that z_0 is an essential singularity.
- If L = 0, then z_0 is a zero of g(z) of some order m \ge 1. This implies that f(z)-w = 1/g(z) has a pole of order m at z_0. This would mean f(z) itself has a pole at z_0, which also contradicts the hypothesis.
Since both possibilities lead to a contradiction, the initial assumption must be false, and the theorem is proven.
This theorem illustrates that a function with an essential singularity does not approach any specific finite or infinite limit. Instead, its values oscillate wildly, coming arbitrarily close to every complex number.
Point at infinity
The nature of a function at the point at infinity, z=\infty, can be analyzed by considering it as an isolated singular point. This is possible if there exists a radius R such that f(z) is analytic for all |z|>R. The behavior of f(z) at infinity is defined by the behavior of an auxiliary function at the origin.
Let the transformation be z = 1/\zeta. This maps the region |z|>R to the annulus 0 < |\zeta| < 1/R. We define a new function:
g(\zeta) = f(1/\zeta) = f(z)
The point z=\infty is classified as a removable singularity, a pole, or an essential singularity for f(z) according to whether \zeta=0 is a removable singularity, a pole, or an essential singularity for g(\zeta).
This definition translates directly to the structure of the Laurent series for f(z) in the annulus R < |z| < \infty:
f(z) = \sum_{n=-\infty}^\infty c_n z^n
- Removable singularity at infinity: The point \zeta=0 is removable for g(\zeta). This occurs if the Laurent series for g(\zeta) has no principal part. Since z^n = \zeta^{-n}, this means the Laurent series for f(z) must have no terms with positive powers of z. The function approaches a finite limit c_0 as z \to \infty. If this limit is zero, z=\infty is a zero of f(z).
- Pole at infinity: The point \zeta=0 is a pole of order m for g(\zeta). This means the principal part of the series for g(\zeta) goes down to \zeta^{-m}. This corresponds to the series for f(z) having a highest positive power of z^m. In this case, |f(z)| \to \infty as z \to \infty.
- Essential Singularity at infinity: The point \zeta=0 is an essential singularity for g(\zeta), meaning its Laurent series has infinitely many negative powers of \zeta. This corresponds to the series for f(z) containing infinitely many positive powers of z. The Sokhotski-Weierstrass theorem applies, meaning the function takes values dense in the complex plane in any neighborhood of infinity.
The character of a singularity is invariant under analytic mappings.
Theorem: Let z_0 be an isolated singular point of a function f(z). Let \zeta = g(z) be an analytic function that is a one-to-one mapping from a neighborhood of z_0 to a neighborhood of \zeta_0 = g(z_0). Then \zeta_0 is an isolated singular point for the composite function F(\zeta) = f[g^{-1}(\zeta)], and its character (removable, pole, or essential) is the same as that of z_0 for f(z).
Proof: This theorem is a direct consequence of the properties of analytic functions. The composition of analytic functions is analytic, so the mapping preserves the local structure of the function. The behavior of f(z) near z_0 is directly translated to the behavior of F(\zeta) near \zeta_0, preserving the classification of the singularity.