KindiakMath

Infinite Series Revisited

Previously, we have seen that if f is holomorphic on a domain D, then all of its derivatives are holomorphic on D.

Theorem 1. If f is holomorphic on D, then for any z_0 \in D, there exists r > 0 such that

\displaystyle f(z_0 + w) = \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot w^k,\quad |w| < r.

Furthermore, the convergence holds uniformly, and the coefficients in the series expansion are unique. The converse holds trivially.

Proof. Denoting \Gamma := \partial B(0,r) = \{\omega \in \mathbb C : |\omega| = r\} for r to be determined, applying Cauchy’s integral formula to the function f(z_0 + \cdot), for any |w| < r, since |w/\omega| < 1,

\begin{aligned} 2\pi i \cdot f(z_0 + w) &= \oint_{\Gamma} \frac{f(z_0 + \omega)}{\omega - w}\, \mathrm d\omega \\ &= \oint_{\Gamma} \frac{f(z_0 + w)}{\omega} \cdot \frac{1}{1 - \frac{ w }{ \omega }}\, \mathrm d\omega \\ &= \oint_{\Gamma} \frac{f(z_0 + w)}{\omega} \cdot \sum_{k=0}^\infty \left( \frac{w}{\omega} \right)^k\, \mathrm d\omega \\ &= \sum_{k=0}^\infty \oint_{\Gamma} \frac{f(z_0 + w)}{\omega^{k+1}}\, \mathrm d\omega \cdot w^k \\ &= 2\pi i \cdot \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot w^k, \end{aligned}

where the interchange of the infinite sum and the integral is justified since the geometric series converges uniformly. For the uniqueness claim, assume

\displaystyle f(z_0 + w) = \sum_{k=0}^\infty c_k \cdot w^k = \sum_{k=0}^\infty d_k \cdot w^k.

We claim that c_n = 0 for any n. Setting w = 0, c_0 = d_0. Suppose c_k = d_k for k = 0,1,\dots, n. Differentiating n+1 times, uniform convergence yields

\displaystyle (n+1)! \cdot c_{n+1} + O(w) = (n+1)! \cdot d_{n+1} + O(w).

Setting w = 0, c_{n+1} = d_{n+1}, as required.

Definition 1 (Taylor Series). A function f : \mathbb C \to \mathbb C is analytic at a point z_0 \in \mathbb C if all of its derivatives are holomorphic at z_0 and there exists r > 0 such that

\displaystyle f(z_0 + w) = \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot w^k,\quad |w| < r.

The function f is analytic on a domain D if it is analytic at every point in D. We call the right-hand side the Maclaurin series of the function f(z_0 + \cdot). The Taylor series at z_0 is obtained via the substitution z := z_0 + w:

\displaystyle f(z) = \sum_{k=0}^\infty \frac{ f^{(k)}(z_0) }{ k! } \cdot (z-z_0)^k,\quad z \in B(z_0, r).

Corollary 1. A function f : \mathbb C \to \mathbb C is holomorphic on a domain D if and only if it is analytic on D.

When are two holomorphic functions identical to each other? Some sufficient conditions are surprisingly mild.

Theorem 2 (Identity Theorem). Let f, g : \mathbb C \to \mathbb C be holomorphic on a domain D. Suppose for any z_0 \in D such that there exists a sequence z_k \to z_0, f(z_k) = g(z_k) for any k. Then f|_D = g|_D.

Proof. Assume without loss of generality that g = 0, so that we apply the result to f=g \iff f-g = 0. Fix z_0 \in \mathbb C and z_n \to z_0 such that f(z_k) = 0. By continuity, f(z_0) = 0, so that z_0 is a root of f. Fix r> 0 such that

\displaystyle f(z) = \sum_{m=0}^\infty \frac{f^{(m)}(z_0)}{m!} \cdot (z-z_0)^m,\quad z \in B(z_0, r) =: B.

Denote c_m := f^{(m)}(z_0)/m! for brevity. Suppose for a contradiction that f|_B \neq 0. Then there exists M \in \mathbb N such that c_M \neq 0. By Corollary 1, there exists some g(z) \in O(z) holomorphic on D such that

\displaystyle f(z) = c_M \cdot (z - z_0)^M \cdot (1 + g(z-z_0)).

For sufficiently large k,

0 = f(w_k) = (w_k - z_0)^M \cdot (1 + g(w_k - z_0)) \neq 0,

a contradiction. Therefore, f|_B = 0.

Now we prove that f|_D = 0. Define

\displaystyle U := \bigcup_{S \in \mathcal U} S,\quad \mathcal U := \{B \subseteq D\ \text{open} : f|_B = 0 \}.

Then U is nonempty, open, and closed. This means that V := D \backslash U is open and closed as well. Since D is connected, U \sqcup V = D and U \neq \emptyset implies V = \emptyset, so that D = U. In particular, f|_D = 0, as required.

Definition 2. Define the annulus with centre z_0 and radii 0 < r_1 < r_2 by

\text{Ann}(z_0, r_1, r_2) := \{ z \in \mathbb C : r_1 < |z - z_0| < r_2 \}.

Define the closed disk by D(z_0, r) := \bar{B}(z_0, r), and the punctured disk by D(z_0, r) \backslash \{z_0\} = \text{Ann}(z_0, 0, r).

Theorem 3 (Laurent Series). Suppose f : \mathbb C \to \mathbb C is holomorphic on an open set containing \overline{\text{Ann}}(z_0, r_1, r_2). Define \Gamma_1 := \partial B(z_0, r_1) and \Gamma_2 := \partial B(z_0, r_2), both oriented anti-clockwise, so that \Gamma_1 \cup \Gamma_2 = \partial \overline{\text{Ann}}(z_0, r_1, r_2). Then for any r_1<r<r_2,

\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k (z - z_0)^k,\quad z \in \text{Ann}(z_0, r_1, r_2),

where the convergence is uniform,

\displaystyle c_k = \frac 1{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{(w-z_0)^{k+1}}\, \mathrm dw,

and \partial B(z_0, r) is oriented anti-clockwise.

Proof. Assume z_0 = 0 for simplicity. By the Cauchy-Goursat theorem, for aby z \in \text{Ann}(z_0, r_1, r_2) and anti-clockwise circle \gamma \subseteq \text{Ann}(z_0, r_1, r_2) with centre z,

\displaystyle \left( \oint_{\Gamma_2} - \oint_{\Gamma_1} - \oint_{\gamma} \right) \frac{ f(w) }{ w - z }\, \mathrm dw = \oint_{\Gamma_2 + (- \Gamma_1) + (-\gamma)} \frac{ f(w) }{ w - z }\, \mathrm dw = 0.

By Cauchy’s integral formula,

\displaystyle f(z) = \frac{1}{2\pi i} \oint_{\gamma} \frac{ f(w) }{ w - z }\, \mathrm dw = \frac{1}{2\pi i} \oint_{\Gamma_2}\frac{ f(w) }{ w - z }\, \mathrm dw - \frac{1}{2\pi i} \oint_{\Gamma_1}\frac{ f(w) }{ w - z }\, \mathrm dw.

For the first integral, for any w \in \Gamma_2, |w| > |z| \Rightarrow |z/w| < 1. By the uniform convergence of the geometric series,

\displaystyle \frac 1{w-z} = \frac 1w \cdot \frac 1{1- \frac zw} = \frac 1w \cdot \sum_{k=0}^\infty \left( \frac zw \right)^k = \sum_{k=0}^\infty \frac{1}{w^{k+1}} \cdot z^k.

Thanks to uniform convergence,

\displaystyle \frac{1}{2\pi i} \oint_{\Gamma_2}\frac{ f(w) }{ w - z }\, \mathrm dw = \sum_{k=0}^\infty \left( \frac{1}{2\pi i} \oint_{\Gamma_2} \frac{f(w)}{w^{k+1}}\, \mathrm dw \right) \cdot z^k.

Similarly for the second integral, for any w \in \Gamma_1, |w| < |z| \Rightarrow |w/z| < 1. By the uniform convergence of the geometric series,

\displaystyle \frac 1{w-z} = -\frac 1z \cdot \frac 1{1-\frac wz} = -\sum_{k=0}^\infty \frac{1}{w^{-k}} \cdot z^{-(k+1)} = -\sum_{k=-\infty}^{-1} \frac 1{w^{k+1}} \cdot z^{k}.

Thanks to uniform convergence,

\displaystyle -\frac{1}{2\pi i} \oint_{\Gamma_1}\frac{ f(w) }{ w - z }\, \mathrm dw = \sum_{k=-\infty}^{-1} \frac{1}{2\pi i} \left( \oint_{\Gamma_1} \frac {f(w)}{w^{k+1}}\, \mathrm dw \right) \cdot z^{k}.

By the Cauchy-Goursat theorem,

\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k z^k,\quad c_k = \frac 1{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{w^{k+1}}\, \mathrm dw.

For uniqueness, write

\displaystyle f(z) = \sum_{k=-\infty}^{\infty} d_k z^k.

By the definition of c_n and uniform convergence that justifies integral-summation swapping,

\begin{aligned} 2\pi i \cdot c_n &= \oint_{\partial B(z_0, r)} \frac{f(w)}{w^{k+1}}\, \mathrm dw \\ &= \oint_{\partial B(z_0, r)} \frac{1}{w^{n+1}}\, \sum_{k=-\infty}^{\infty} d_k w^k \cdot \mathrm dw \\ &= \sum_{k=-\infty}^{\infty} d_k \oint_{\partial B(z_0, r)} \frac{1}{w^{(n-k)+1} }\, \mathrm dw \\ &= \sum_{k=-\infty}^{\infty} d_k \cdot 2\pi i \cdot \mathbb I_{n}(k) = 2 \pi i \cdot d_n.\end{aligned}

What’s the goal of the Laurent series? Roughly speaking, the Laurent series of a function f tells us how “badly behaved” it is at a point z_0. If, for instance, c_k = 0 for k < 0, then we get the usual Taylor series which tells us that f is holomorphic at z_0. Otherwise, even though f may not be holomorphic z_0, we might be able to multiply sufficiently many copies of (z-z_0) so that the resulting function is holomorphic at z_0. In this case, though badly behaved, the bad behavior f is somewhat controllable. We will explore these singularities the next post, and develop the calculus of residues that help us compute many otherwise intractable integrals.

—Joel Kindiak, 21 Aug 25, 2236H

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