Laurent Series, Residues, and the Annulus

“Beauty is so many things… and you are in all of them,” I said. “Wow! I thought that you were as romantic as a brick to the head or a nuclear bomb’s guts”, Susanne replied, Apocalypse, Anawim, #justothepoint.

The Mystery of the Missing Antiderivative

In calculus, we are used to the idea that if you integrate a function around a closed loop and return to the start, the total “displacement” (the integral) should be zero. However, complex analysis reveals a strange exception.

Consider the function $f(z) = \frac{1}{(z-a)^n}$ for an integer $n$.

1. When $n \geq 2$: The function has a valid, single-valued antiderivative (primitive) in the punctured plane $\mathbb{C} \setminus \{a\}$. For example, if $n=2$: $F(z) = \int \frac{1}{(z-a)^2} dz = -\frac{1}{z-a}$
   Using the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$), we get: $F(z) = \frac{(z-a)^{-1}}{-1} = -\frac{1}{z-a}$, and this is a valid function in the punctured domain $\mathbb{C} \setminus \{a\}$. It gives a unique complex number and it does’t jump if you walk around “a”.
   This function $F(z)$ is “well-behaved.” If you walk in any simple closed curve $\gamma$ around a (typically a circle), $F(z_{start}) = F(z_{end})$.
   Now, we could apply the Fundamental Theorem of Calculus. If a function f has an antiderivative F everywhere on the curve $\gamma$, then: $\int_\gamma f(z) dz = F(end_point) - F(start_point)$. Since $\gamma$ is a closed curve, the start and end points are the same ($z_{start_point} = z_{end_point}$), hence: $\int_\gamma \frac{1}{(z-a)^2} dz = F(z_{start}) - F(z_{start}) = 0$
2. When $n = 1$: The power rule gives us $\int \frac{1}{z-a} dz = \ln(z-a)$. But the complex logarithm is multi-valued. Every time you circle the point $a$, the angle (argument) increases by $2\pi$.
   $\int_\gamma \frac{1}{z-a} dz = [\ln|z-a| + i(\theta + 2\pi)] - [\ln|z-a| + i\theta] = 2\pi i$
   The result is not zero. This tells us that $1/(z-a)$ has no single-valued global antiderivative on the punctured disk. It leaves a “residue” behind.

Path Independence

Let $f(z) = \frac{1}{(z-a)^n}, n \ge 2$, consider two different paths, $\alpha$ and $\beta$, that both start at point P and end at point Q. $\gamma = \beta - \alpha$ represents the closed loop formed by going forward along $\beta$ (from P to Q) and then backward along $\alpha$ (from Q back to P).

Since the integral over the closed loop $\gamma$ is zero, then: $\int_\gamma f(z) dz = 0 \implies \int_\beta f(z) dz - \int_\alpha f(z) dz = 0 \implies \int_\beta f(z) dz = \int_\alpha f(z) dz$. Conclusion: Path independence: Since the integral around the closed loop is zero, the integral from P to Q doesn't depend on which path you take. It depends only on the endpoints.

However, $g(z) = \frac{1}{z-a}$. Its antiderivative G(z) = ln(z -a). The complex logarithm is multi-valued, $\int_\gamma \frac{1}{z-a} dz = G(\text{end}) - G(\text{start}) = 2\pi i \neq 0$. Because the integral is not zero, no single-valued global antiderivative exists on the punctured disk.

Introducing the Laurent Series

A Taylor series is a “local scanner” that describes a function using only positive powers: $a_0 + a_1(z-a) + a_2(z-a)^2 \dots$. However, Taylor series fail the moment they hit a singularity (a point where the function blows up).

To describe a function near a singularity, we need Laurent Series, which use both positive and negative powers: $f(z) = \underbrace{\dots + \frac{a_{-2}}{(z-a)^2} + \frac{a_{-1}}{z-a}}_{\text{Principal Part (The "Singular" Part)}} + \underbrace{a_0 + a_1(z-a) + a_2(z-a)^2 + \dots}_{\text{Analytic Part (The "Nice" Part)}}$

Any function analytic in an annulus (ring) around a can be written as: $f(z) = \dots + \frac{a_{-2}}{(z-a)^2} + \frac{\mathbf{a_{-1}}}{(z-a)} + a_0 + a_1(z-a) + \dots$

When we integrate this series around a closed curve surrounding a, we integrate term by term.

• The Power Rule Terms ($n \neq -1$): For any integer $n \neq -1$, the antiderivative is $\frac{(z-a)^{n+1}}{n+1}$. This is single-valued, hence $\int_{\gamma} (z-a)^n dz = 0 \quad (\text{for } n \neq -1)$ (this includes our previous example).
• The Logarithmic Term ($n = -1$): This is the only term that survives the integration. $\int_{\gamma} \frac{a_{-1}}{z-a} dz = a_{-1} (2\pi i)$

Conclusion: The integral of the entire function depends only on that one specific coefficient, $a_{-1}$. $\boxed{\int_{\gamma} f(z) dz = 2\pi i \cdot a_{-1}}$ This coefficient $a_{-1}$ is called the Residue (aka the “king”) because the 1/z term creates a “residue” (non-zero integral), while terms like $1/x^2$ vanish. Furthermore, a function $f(z)$ has an antiderivative in the punctured neighborhood if and only if its residue $a_{-1}$ is zero. In other words, the residue is the only part of the function that prevents the function from having a local antiderivative.

How to Build a Laurent Series

Let f be analytic in a punctured neighborhood of a, $\mathbb{B'}(a; r)$. Suppose f has a pole of order m at a. This means f(a) is infinite (it “blows up” like $1/(z-a)^m$). This prevent us from directly using Taylor series to expand f, as Taylor series require the function to be nice (smooth, finite, and differentiable) at a.

We can find its expansion using a three-step “neutralization” process:

1. Neutralize the Pole: Multiply $f(z)$ by $(z-a)^m$. This creates a helper function $h(z) = (z-a)^m f(z)$ that is now “nice” (analytic) at $a$ (it cancels out the infinity).
   Recall that if f has a pole of order m at a, then by definition there exists a holomorphic function g near a such that $f(z)=\frac{g(z)}{(z-a)^m}, g(a) \ne 0$, so h(z) = g(z). Since g is holomorphic at a, the limit exists and equals the value of g at a.
   In other words, if $f$ has a pole of order m, it behaves roughly like $\frac{1}{(z-a)^m}$. Multiplying by $(z-a)^m$ neutralizes or cancels the pole. $\lim_{z \to a} h(z) = g(a) = \text{Finite Non-Zero Number}$
   Because this limit exists, $h(z) = (z-a)^m f(z)$ has a removable singularity at $a$. We “fill the hole” by defining $h(a)$ as that limit. Now, $h(z)$ is analytic everywhere in the disk $\mathbb{B}(a; r)$.
   We define $h(z) = \begin{cases} (z-a)^mf(z), &z \in \mathbb{B'}(a; r)\\\\ \lim_{z \to a}(z-a)^{m}f(z) = g(a), &z = a \end{cases}$

2. Apply Taylor: Since h(z) is now analytic and well-behaved at a, we can safely use Taylor’s Theorem to expand it into a power series using only non-negative powers: $h(z) = c_0 + c_1(z -a) + c_2(z -a)^2 + \cdots + c_{m-1}(z -a)^{m-1}+ c_m(z -a)^m + \cdots, \forall z \in \mathbb{B}(a; r)$ where $c_0 = h(a) \ne 0$ (This confirms the pole was exactly order $m$) and the coefficients $c_n$ are just standard Taylor coefficients for h: $c_n = \frac{h^{(n)}(a)}{n!}$.
   In particular for $z \ne a, (z-a)^mf(z) = c_0 + c_1(z -a) + c_2(z -a)^2 + \cdots + c_{m-1}(z -a)^{m-1}+ c_m(z -a)^m + \cdots, \forall z \in \mathbb{B'}(a; r)$.

3. Restore the Pole:
   Obviously, we want $f(z)$, not our helper function $h(z)$, so we divide the entire Taylor series by our neutralizing factor, $(z-a)^m$.
   We have split f(z) into two distinct parts: $f(z) = \underbrace{\left[ \frac{c_0}{(z-a)^m} + \dots + \frac{c_{m-1}}{z-a} \right]}_{\text{Principal (Singular) Part}} + \underbrace{\left[ c_m + c_{m+1}(z-a) + \dots \right]}_{\text{Analytic (Regular) Part } f_1(z)}$

This is the Laurent Expansion. We have successfully split f(z) into two distinct parts: the first bracket contains all the negative powers (the “bad” and “naughty” part that causes the explosion/singularity), and the second bracket is a normal power series is a normal power series (the “nice” part).

Integrating the Series: The Birth of the Residue

Power series (like our nice $f_1(z)$) and Laurent series converge uniformly on any closed subset of their domain (like the contour $\gamma$). In analysis, uniform convergence is the “safety pass” that allows us to swap the limit (the infinite sum $\Sigma$) and the integral ($\int$), $\int_\gamma \left( \sum c_n (z-a)^n \right) dz = \sum c_n \left( \int_\gamma (z-a)^n dz \right)$

If $\gamma$ is a simple closed contour in $\mathbb{B'}(a; r)$ oriented positively (counter-clockwise) such that $a \in Int(\gamma)$.

If we integrate our Laurent series term by term: $\int_{\gamma} f(z) dz =[\text{Sum of integrals of individual terms}] \int_{\gamma} \frac{c_0}{(z-a)^m} + \int_{\gamma}\frac{c_1}{(z-a)^{m-1}} + \cdots + \int_{\gamma} \frac{c_{m-1}}{z-a} + \int_{\gamma} f_1(z) = \int_{\gamma} \frac{c_{m-1}}{z-a} = c_{m-1}\cdot (2\pi i)$

Let’s look at the integral of a general term $\int_\gamma C (z-a)^k dz$. For almost every integer $k \ne 1$, the function $(z-a)^k$ has a valid, single-valued antiderivative: $F(z) = \frac{(z-a)^{k+1}}{k+1}$. We can apply the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus for Contours: If a function has a single-valued antiderivative F everywhere on a curve $\gamma$, then: $\int_\gamma f(z)dz=F(\text{end point})-F(\text{start point})$. Since the path $\gamma$ is a closed loop, the start and end points are the exact same. Therefore,: $F(\text{end point}) - F(\text{start point}) = 0$. This means every term in the series vanishes to zero, except one, k = 1.

Instead, evaluating that single surviving term gives us c_m-1⋅(2πi).

In conclusion, when one integrates the infinite series for f(z), the majority of the terms evaluate to zero, and only one single term survives. $\int_\gamma f(z) dz = 2\pi i \cdot c_{m-1}$. This coefficient $c_{m-1}$ is called the Residue of f at a, often denoted as $\text{Res}(f, a)$. It is the “residue” left behind after the fires of integration burn away the rest of the series.

Lemma. Let f be analytic on and inside a positively oriented simple closed curve $\gamma$ except at a point a inside $\gamma$ where f has a pole of order m. Then, $\int_{\gamma}f(z)dz = c_{m-1}\cdot (2\pi i)$ where $c_{m-1}$ is the coefficient of $\frac{1}{z-a}$ in the singular part of the expansion of f as powers of (z -a) in the neighborhood of a.

• If $c_{m-1} \ne 0$: The integral is non-zero ($2\pi i \cdot c_{m-1}$). The function does not have an antiderivative on the punctured domain.
• If $c_{m-1} = 0$: The integral is zero. The “bad” logarithmic term is missing from the expansion. Therefore, $f(z)$ does have an antiderivative locally.

Behind-the-scenes Laurent’s series

Taylor series are like balloons. If you center a Taylor series for $\frac{1}{1-z}$ at $z=0$, the balloon expands until it hits the “pin” at $z=1$. It cannot describe the function for $|z| > 1$. To see “outside” the pin, we use the Annulus (Donut) Region.

Consider the Taylor series expansion of $\frac{1}{1-z}$ around 0, $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$ and this geometric series only converges if the ratio |z| < 1. f(z) = $\frac{1}{1-z}$ has a singularity (has a pole) at z = 1. In words, the function explodes at z = 1.

Taylor Series are like local scanners. They sit at a center point (say, z = 0) and describe the function perfectly as you move outward. However, they have a major weakness. They cannot see past the first singularity. It is like expanding a ballon centered at 0. It grows until it hits the “sharp pin” (a wall or point of resistance) at z = 1. The moment it touches the pin, the balloon pops. The Taylor series is valid only inside that balloon (the disk $|z| < 1$).

What if we want to describe the function outside the balloon and “break the wall”, where $|z| > 1$? The standard formula $\sum z^n$ blows up to infinity there. We need a different approach.

We rewrite the function to force a geometric series that converges for large z. We do this by factoring out the “dominant” term (if |z| > 1, z is dominant): $\frac{1}{1-z} = \frac{-1}{z}\bigr( \frac{1}{1-\frac{1}{z}} \bigl)$. Since we are in the region |z| > 1, the reciprocal (our new ratio) is small, $|\frac{1}{z}| \lt 1$ and we can expand $\frac{1}{1 - (1/z)}$. This is obviously valid for |z| > 1: $\frac{1}{1-z} = \frac{-1}{z}\bigr( \sum_{n=0}^\infty (\frac{1}{z})^n \bigl) = -\sum_{n=0}^\infty z^{-n-1} =[\text{Reindexing}] -\sum_{n=-\infty}^{-1} z^n = - \left( \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \dots \right) = -z^{-1} - z^{-2} - z^{-3} - \dots$ for |z| > 1.

We have effectively characterized the behavior beyond the singularity! The cost is that we now use negative powers of z.

The Annulus (The “Donut” Region)

We often deal with functions having multiple singularities, e.g., $\frac{3}{(1-z)(4-z)}$. It has two singularities, specifically z = 1 and z = 4.

This splits the complex plane into three distinct regions: (i) the Core, $|z| < 1$ (inside both); (ii) the Annulus or donut region, $1 < |z| < 4$ (outside the first, inside the second); and (iii) the Exterior: $|z| > 4$ (outside both).

We want to find the expansion for the Annulus ($1 < |z| < 4$). We decompose the function using Partial Fractions: $f(z) = \underbrace{\frac{1}{1-z}}_{\text{Term 1}} - \underbrace{\frac{1}{4-z}}_{\text{Term 2}}$

Analyze Term 1. When |z| > 1, we have previously demonstrated that $\frac{1}{1-z} = -\sum_{n=-\infty}^{-1} z^m$. Analyze term 2. Likewise, when |z| < 4, $\frac{1}{4-z} =[\text{Factor out 4 to get the ratio z/4 < 1}] \frac{1}{4}\bigr( \frac{1}{1-\frac{z}{4}} \bigl) =[\text{Use standard Taylor series}] \frac{1}{4} \bigr( \sum_{n=0}^\infty (\frac{z}{4})^n \bigl)$.

So combining both results, when 1 < |z| < 4, $f(z) = \underbrace{-\sum_{n=-\infty}^{-1} z^n}_{\text{Negative Powers}} - \underbrace{\frac{1}{4}\sum_{n=0}^{\infty} \left(\frac{z}{4}\right)^n}_{\text{Positive Powers}}$ and both individually converge, so combine them together they converge in the common “annular” region 1 < |z| < 4. A Laurent Series is simply the combination of these two parts, a Principal Part (negative powers) to handle the inner hole, and an Analytic Part (positive powers) to handle the outer boundary.

The Mystery of the “Invisible” Obstacle

Consider $g(x) = \frac{1}{1+x^2}$ on the real number line. The function looks perfectly normal, it is smooth everywhere. It’s a nice “hill” shape. It never goes to infinity for any real number x, and yet, if you try to build a Taylor series at x = 0, it fails to converge if $|x| \ge 1$. Why does it stop at 1 if there is no visible “bump” in the road?

The radius of convergence of a Taylor series about a equals the distance from a to the nearest singularity in the complex plane, even if you are only working with real numbers.

Complex analysis reveals the “invisible” obstacles. $1 + z^2 = 0 \implies z^2 = -1 \implies z = i, -i$. There are singularities (simple poles) at $i$ and $-i$. Even though these points are not on the real line, they are exactly distance 1 away from the origin, $|i-0|=1, |-i-0|=1.$ So the nearest singularity is exactly 1 unit away. Therefore, the Taylor series at 0 must have radius of convergence 1.

This is the Taylor series expansion of $\frac{1}{1+z^2}$ around 0, $\frac{1}{1+z^2} = \sum_{n=0}^\infty (-1)^nz^{2n}$ and this is valid for |z| < 1. The Taylor series “balloon” centered at 0 expands until it hits these imaginary obstacles at radius R = 1.

Double edged series (also called a two‑sided infinite series) is a formal sum indexed over all integers: $\sum_{n=-\infty}^\infty a_n$. Unlike ordinary series, this one extends infinitely in both directions.

Because this series extends infinitely in both directions, there is no natural notion of partial sums like in a one‑sided series. To give it a well‑defined meaning, we interpret it by splitting it into its positive and negative parts. We say that the two-sided series $\sum_{n=-\infty}^\infty a_n$ converges if and only if both of the following one-sided series converge in the usual sense, $\sum_{n=0}^\infty a_n, \sum_{n=1}^\infty a_{-n}$. When this happens, we define the value of the two-sided series to be $\sum_{n=-\infty}^\infty a_n = s_1 + s_2$ where $s_1 = \sum_{n=0}^\infty a_n$ and $s_2 = \sum_{n=1}^\infty a_{-n}$

Laurent’s theorem. Let A = {$z \in \mathbb{C}: R \lt |z -a| \lt S$} be an annular region ($0 \le R \lt S \le \infty$) and let f be analytic on A. Then, $f(z) = \sum_{n=-\infty}^\infty c_n(z -a)^n, \forall z \in A$ where $c_n = \frac{1}{2\pi i}\int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$ and $\gamma$ is a positively oriented circle of radius r centered at a and R < r < S ($\gamma$ is inside the annular region).