06.01.27 · riemann-surfaces / complex-analysis

Power series and Laurent series

shipped3 tiersLean: none

Anchor (Master): Cauchy 1831 Turin memoir (radius of convergence); Weierstrass 1842 lectures (M-test, published 1894 Werke Vol. 1); Laurent 1843 J. Math. Pures Appl. 8; Ahlfors Complex Analysis Ch. 5

Intuition [Beginner]

A power series is an infinite polynomial: a sum of terms $c_{0} + c_{1} z + c_{2} z^{2} + c_{3} z^{3} + \dots$ where each term multiplies a coefficient $c_{n}$ by $z$ raised to the $n$ th power. Unlike a polynomial, which stops after finitely many terms, a power series keeps going. But it only produces finite answers when $z$ is small enough — within a disk whose size depends on the coefficients.

The radius of this disk is called the radius of convergence. Inside the disk the series adds up to a definite complex number. Outside the disk the terms grow without bound and the series diverges. On the boundary circle, convergence varies point by point.

A Laurent series extends the idea by allowing negative powers: $c_{- 2} z^{- 2} + c_{- 1} z^{- 1} + c_{0} + c_{1} z + c_{2} z^{2} + \dots$ . The negative-power terms capture the behavior of a function near a singular point — a place where the function blows up. Power series describe smooth regions; Laurent series describe what happens near the rough spots.

Holomorphic functions are determined by their derivatives at a single point. The power series packages these derivative values into a single expression that reconstructs the function in an entire disk. Laurent series extend this picture to functions with isolated singularities, providing a complete local description.

Visual [Beginner]

A disk in the complex plane centred at the origin, with radius $R$ marked. Inside the disk, several points are labelled with their series values. Outside the disk, a few points show divergent partial sums.

Power series are the microscope of complex analysis: they describe a function locally as an infinite polynomial, and the radius of convergence tells you how far you can zoom out before the picture breaks.

Worked example [Beginner]

Consider the series $1 + z + z^{2} + z^{3} + \dots$ . Every coefficient is $1$ and the powers go up from $0$ .

Step 1. When $∣ z ∣ < 1$ , the terms shrink because $∣ z^{n} ∣ = ∣ z ∣^{n}$ gets smaller. The series converges to $1/ (1 - z)$ . For $z = 1/2$ : the series gives $1 + 1/2 + 1/4 + 1/8 + \dots = 1/ (1 - 1/2) = 2$ .

Step 2. When $∣ z ∣ > 1$ , the terms grow: $∣ z^{n} ∣ = ∣ z ∣^{n}$ blows up. The series diverges. At $z = 2$ , the partial sums are $1, 3, 7, 15, 31, \dots$ , increasing without bound.

Step 3. The radius of convergence is $1$ . The series converges inside the disk $∣ z ∣ < 1$ and diverges outside. On the boundary $∣ z ∣ = 1$ , the series diverges at every point because $∣ z^{n} ∣ = 1$ for all $n$ , so the terms never approach zero.

What this tells us: a power series has a sharp boundary between its convergence region and its divergence region, separated by a circle whose radius depends on the growth rate of the coefficients.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Power series. A power series centred at $z_{0} \in C$ is a series of the form

n = 0 \sum \infty c_{n} (z - z_{0})^{n}

where $c_{n} \in C$ are the coefficients. The radius of convergence $R \in [0, + \infty]$ is defined by

\frac{1}{R} = \limsup_{n \to \infty}} |c_n|^{1/n},

with the conventions $1/0 = + \infty$ and $1/ + \infty = 0$ . The series converges absolutely for $∣ z - z_{0} ∣ < R$ and diverges for $∣ z - z_{0} ∣ > R$ . On the boundary $∣ z - z_{0} ∣ = R$ , convergence depends on the specific series. ^{[Stein-Shakarchi Ch. 1]}

Laurent series. A Laurent series centred at $z_{0}$ is a doubly infinite series

n = - \infty \sum \infty c_{n} (z - z_{0})^{n},

decomposing into the analytic part $\sum_{n = 0}^{\infty} c_{n} (z - z_{0})^{n}$ (non-negative powers) and the principal part $\sum_{n = 1}^{\infty} c_{- n} (z - z_{0})^{- n}$ (negative powers). A Laurent series converges on an annulus $r < ∣ z - z_{0} ∣ < R$ where $0 \leq r < R \leq + \infty$ . ^{[Ahlfors Ch. 5]}

Laurent decomposition theorem. If $f$ is holomorphic on the annulus $A = {z : r < ∣ z - z_{0} ∣ < R}$ , then $f$ has a unique Laurent expansion $f (z) = \sum_{n = - \infty}^{\infty} c_{n} (z - z_{0})^{n}$ converging absolutely and uniformly on compact subsets of $A$ . The coefficients are given by

c_{n} = \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = ρ} \frac{f ( w )}{( w - z _{0} ) ^{n + 1}} d w

for any $ρ$ with $r < ρ < R$ .

Counterexamples to common slips

The radius of convergence can be zero or infinity. The series $\sum n! z^{n}$ has $R = 0$ (converges only at the centre). The series $\sum z^{n} / n!$ has $R = \infty$ (converges everywhere). The formula $1/ R = lim sup ∣ c_{n} ∣^{1/ n}$ handles both extremes via the conventions above.
A Laurent series is not a power series with negative indices. The negative-power terms are structurally different: they encode singularity information. The function $f (z) = 1/ z$ has the Laurent series $c_{- 1} = 1$ with all other coefficients zero — no power series centred at $0$ can represent it.
Convergence on the boundary is not determined by $R$ alone. The geometric series $\sum z^{n}$ has $R = 1$ and diverges everywhere on $∣ z ∣ = 1$ because the terms do not tend to zero. The series $\sum z^{n} / n^{2}$ also has $R = 1$ but converges everywhere on $∣ z ∣ = 1$ by comparison with $\sum 1/ n^{2}$ .

Key theorem with proof [Intermediate+]

Theorem (Convergence of power series). Let $\sum_{n = 0}^{\infty} c_{n} (z - z_{0})^{n}$ be a power series with radius of convergence $R = 1/ lim sup_{n \to \infty} ∣ c_{n} ∣^{1/ n}$ . Then:

(i) For every $z$ with $∣ z - z_{0} ∣ < R$ , the series converges absolutely.

(ii) For every $z$ with $∣ z - z_{0} ∣ > R$ , the series diverges.

(iii) For every $0 < ρ < R$ , the series converges uniformly on the closed disk ${z : ∣ z - z_{0} ∣ \leq ρ}$ .

Proof. Without loss of generality set $z_{0} = 0$ (otherwise replace $z$ by $z - z_{0}$ throughout).

(i) Let $∣ z ∣ < R$ . Choose $σ$ with $∣ z ∣ < σ < R$ . Then $1/ σ > 1/ R = lim sup ∣ c_{n} ∣^{1/ n}$ . By the definition of limsup, there exists $N$ such that $∣ c_{n} ∣^{1/ n} < 1/ σ$ for all $n \geq N$ , hence $∣ c_{n} ∣ < σ^{- n}$ for $n \geq N$ . For $n \geq N$ ,

∣ c_{n} z^{n} ∣ < σ^{- n} ∣ z ∣^{n} = (\frac{∣ z ∣}{σ})^{n} .

Since $∣ z ∣/ σ < 1$ , the geometric series $\sum (∣ z ∣/ σ)^{n}$ converges, and by comparison $\sum ∣ c_{n} z^{n} ∣$ converges. Hence $\sum c_{n} z^{n}$ converges absolutely.

(ii) Let $∣ z ∣ > R$ . Then $1/∣ z ∣ < 1/ R = lim sup ∣ c_{n} ∣^{1/ n}$ . By the definition of limsup as the supremum of subsequential limits, there exists a subsequence $n_{k} \to \infty$ such that $∣ c_{n_{k}} ∣^{1/ n_{k}} \to 1/ R$ . For sufficiently large $k$ , $∣ c_{n_{k}} ∣^{1/ n_{k}} > 1/∣ z ∣$ , giving $∣ c_{n_{k}} z^{n_{k}} ∣ > 1$ . The general term $c_{n} z^{n}$ does not tend to zero, so the series diverges.

(iii) Fix $0 < ρ < R$ . Choose $σ$ with $ρ < σ < R$ . As in part (i), there exists $N$ such that $∣ c_{n} ∣ < σ^{- n}$ for all $n \geq N$ . For all $z$ with $∣ z ∣ \leq ρ$ and all $n \geq N$ ,

∣ c_{n} z^{n} ∣ \leq ∣ c_{n} ∣ ρ^{n} < σ^{- n} ρ^{n} = (\frac{ρ}{σ})^{n} .

The bound $(ρ / σ)^{n}$ is independent of $z$ and summable (since $ρ / σ < 1$ ). By the Weierstrass $M$ -test (Theorem 1 in Advanced results below), $\sum c_{n} z^{n}$ converges uniformly on ${∣ z ∣ \leq ρ}$ . $□$

Bridge. This theorem builds toward 06.01.01 holomorphic functions, where it appears again as the result that every holomorphic function equals its own Taylor series locally. The foundational reason is that the radius of convergence encodes the distance to the nearest singularity, and this is exactly the bridge between the local series representation and the global analytic structure of the function. The theorem generalises to Laurent series on annuli, and putting these together identifies the coefficients $c_{n}$ as contour integrals via the Cauchy formula 06.01.02. The central insight is that a single number $R$ controls convergence, absolute convergence, and uniform convergence on compact subsets simultaneously — a rigidity with no analogue in real-variable analysis.

Exercises [Intermediate+]

Exercise 3 (medium, symbolic).

Prove that if $\sum_{n = 0}^{\infty} c_{n} z^{n}$ has radius of convergence $R$ , then the term-by-term derivative $\sum_{n = 1}^{\infty} n c_{n} z^{n - 1}$ also has radius of convergence $R$ .

Hint

Use the ratio test. For the original series, $R = lim ∣ c_{n} / c_{n + 1} ∣$ . Compare with $R^{'} = lim ∣ n c_{n} / (n + 1) c_{n + 1} ∣$ .

Answer

Write $d_{n} = (n + 1) c_{n + 1}$ for $n \geq 0$ , so the derivative series is $\sum d_{n} z^{n}$ . By the ratio test,

\frac{d _{n + 1}}{d _{n}} = \frac{( n + 2 ) ∣ c _{n + 2} ∣}{( n + 1 ) ∣ c _{n + 1} ∣} = \frac{n + 2}{n + 1} \cdot \frac{∣ c _{n + 2} ∣}{∣ c _{n + 1} ∣} .

Since $(n + 2) / (n + 1) \to 1$ and $∣ c_{n + 2} / c_{n + 1} ∣ \to 1/ R$ (assuming the ratio test applies; otherwise use the root test below), the radius of convergence of the derivative series is also $R$ .

By the root test: $∣ d_{n} ∣^{1/ n} = ((n + 1) ∣ c_{n + 1} ∣)^{1/ n} = (n + 1)^{1/ n} \cdot ∣ c_{n + 1} ∣^{1/ n}$ . Since $(n + 1)^{1/ n} \to 1$ and $∣ c_{n + 1} ∣^{1/ n} = (∣ c_{n + 1} ∣^{1/ (n + 1)})^{(n + 1) / n} \to (1/ R)^{1} = 1/ R$ , we get $lim sup ∣ d_{n} ∣^{1/ n} = 1/ R$ , giving the same radius $R$ . $□$

Exercise 6 (medium, symbolic).

Prove the uniqueness of the Laurent expansion: if $\sum_{n = - \infty}^{\infty} c_{n} (z - z_{0})^{n} = 0$ for all $z$ in an annulus $r < ∣ z - z_{0} ∣ < R$ , then $c_{n} = 0$ for every $n$ .

Hint

Multiply by $(z - z_{0})^{- k - 1}$ and integrate around a circle centred at $z_{0}$ . Use uniform convergence to exchange the series and the integral.

Answer

Fix an integer $k$ . Multiply the identity $\sum c_{n} (z - z_{0})^{n} = 0$ by $(z - z_{0})^{- k - 1}$ and integrate over the circle $γ$ given by $∣ z - z_{0} ∣ = ρ$ for any $ρ$ with $r < ρ < R$ :

0 = \frac{1}{2 π i} \oint_{γ} (n = - \infty \sum \infty c_{n} (z - z_{0})^{n - k - 1}) d z .

Uniform convergence on the compact circle justifies exchanging the series and the integral:

0 = n = - \infty \sum \infty c_{n} \cdot \frac{1}{2 π i} \oint_{γ} (z - z_{0})^{n - k - 1} d z .

The integral $\oint_{γ} (z - z_{0})^{m} d z$ equals $2 π i$ when $m = - 1$ and $0$ otherwise (by direct parametrisation or the residue theorem 06.01.03). So the only surviving term is $n - k - 1 = - 1$ , i.e., $n = k$ . This gives $c_{k} = 0$ . Since $k$ was arbitrary, all coefficients vanish. $□$

Exercise 7 (hard, symbolic).

Prove Abel's theorem: if $\sum_{n = 0}^{\infty} c_{n} R^{n}$ converges (to a finite limit $S$ ), then $lim_{r \to 1^{-}} \sum_{n = 0}^{\infty} c_{n} (r R)^{n} = S$ .

Outline. Without loss of generality take $R = 1$ . Set $s_{n} = \sum_{k = 0}^{n} c_{k}$ and use Abel's summation formula $\sum_{n = 0}^{N} c_{n} r^{n} = s_{N} r^{N} + (1 - r) \sum_{n = 0}^{N - 1} s_{n} r^{n}$ .

Hint

Let $N \to \infty$ first (noting $s_{N} r^{N} \to 0$ for $0 < r < 1$ when $s_{N}$ is bounded), then analyse $(1 - r) \sum s_{n} r^{n}$ as $r \to 1^{-}$ by splitting at a large index $M$ .

Answer

Assume $R = 1$ and $\sum c_{n} = S$ converges. Set $s_{n} = \sum_{k = 0}^{n} c_{k}$ ; then $s_{n} \to S$ .

Abel's summation formula gives, for $0 < r < 1$ :

n = 0 \sum N c_{n} r^{n} = s_{N} r^{N} + (1 - r) n = 0 \sum N - 1 s_{n} r^{n} .

Let $N \to \infty$ . Since $s_{N}$ is bounded (it converges to $S$ ) and $r^{N} \to 0$ , we have $s_{N} r^{N} \to 0$ . On the right side, the partial sums converge, giving

n = 0 \sum \infty c_{n} r^{n} = (1 - r) n = 0 \sum \infty s_{n} r^{n} .

Write $s_{n} = S + (s_{n} - S)$ and split:

(1 - r) n = 0 \sum \infty s_{n} r^{n} = S (1 - r) n = 0 \sum \infty r^{n} + (1 - r) n = 0 \sum \infty (s_{n} - S) r^{n} .

The first term is $S (1 - r) \cdot 1/ (1 - r) = S$ . For the second term, fix $ϵ > 0$ and choose $M$ such that $∣ s_{n} - S ∣ < ϵ$ for $n \geq M$ . Split the series at $M$ :

(1 - r) n = 0 \sum \infty (s_{n} - S) r^{n} \leq (1 - r) n = 0 \sum M - 1 ∣ s_{n} - S ∣ r^{n} + (1 - r) ϵ n = M \sum \infty r^{n} .

The finite sum $(1 - r) \sum_{n = 0}^{M - 1} ∣ s_{n} - S ∣ r^{n} \to 0$ as $r \to 1^{-}$ (it is a polynomial in $r$ multiplied by $(1 - r)$ ). The tail satisfies $(1 - r) ϵ \sum_{n = M}^{\infty} r^{n} = ϵ r^{M} < ϵ$ . Taking $lim sup_{r \to 1^{-}}$ gives an error at most $ϵ$ . Since $ϵ > 0$ was arbitrary, $lim_{r \to 1^{-}} \sum c_{n} r^{n} = S$ . $□$

Exercise 8 (hard, symbolic).

Prove the identity theorem for power series: if two power series $\sum a_{n} z^{n}$ and $\sum b_{n} z^{n}$ both converge in a disk $∣ z ∣ < R$ and agree on a set $E \subset {∣ z ∣ < R}$ that has a limit point in ${∣ z ∣ < R}$ , then $a_{n} = b_{n}$ for all $n$ .

Hint

Consider the difference $f (z) = \sum (a_{n} - b_{n}) z^{n}$ , which vanishes on $E$ . Show $f \equiv 0$ by proving all derivatives of $f$ vanish at the limit point.

Answer

Set $f (z) = \sum_{n = 0}^{\infty} d_{n} z^{n}$ with $d_{n} = a_{n} - b_{n}$ . Then $f$ converges for $∣ z ∣ < R$ and $f (z) = 0$ for all $z \in E$ . Let $z_{0}$ be a limit point of $E$ with $∣ z_{0} ∣ < R$ .

If $f$ is not identically zero near $z_{0}$ , then $f (z) = (z - z_{0})^{m} g (z)$ where $m \geq 1$ , $g$ is holomorphic, and $g (z_{0}) \neq = 0$ (this is the factorisation of a holomorphic function vanishing to finite order at $z_{0}$ ). Since $g$ is continuous and $g (z_{0}) \neq = 0$ , there exists a neighbourhood of $z_{0}$ where $g (z) \neq = 0$ , hence $f (z) \neq = 0$ for $z \neq = z_{0}$ in that neighbourhood. But $z_{0}$ is a limit point of $E$ , so there exist points $z_{k} \in E$ with $z_{k} \neq = z_{0}$ and $z_{k} \to z_{0}$ . At each $z_{k}$ , $f (z_{k}) = 0$ , contradicting $f (z) \neq = 0$ for $z \neq = z_{0}$ near $z_{0}$ .

Therefore $f$ vanishes identically in a neighbourhood of $z_{0}$ , and all its derivatives vanish at $z_{0}$ . Since $f^{(n)} (0) = n! \cdot d_{n}$ (term-by-term differentiation, justified by uniform convergence on compact subsets), we get $d_{n} = 0$ for all $n$ , hence $a_{n} = b_{n}$ . $□$

Advanced results [Master]

Theorem 1 (Weierstrass $M$ -test). Let $f_{n}$ be holomorphic on an open set $Ω$ and let $M_{n} \geq 0$ with $\sum M_{n} < \infty$ . If $∣ f_{n} (z) ∣ \leq M_{n}$ for all $z \in K$ , where $K \subset Ω$ is compact, then $\sum f_{n}$ converges uniformly on $K$ to a holomorphic function. The derivative satisfies $(\sum f_{n})^{'} = \sum f_{n}^{'}$ on $Ω$ . ^{[Weierstrass 1842]}

The $M$ -test is the workhorse for establishing uniform convergence: it reduces the question to a numerical series $\sum M_{n}$ independent of $z$ . The holomorphicity of the limit follows from Morera's theorem (the uniform limit of holomorphic functions is holomorphic).

Theorem 2 (Term-by-term differentiation and integration). If $\sum_{n = 0}^{\infty} c_{n} (z - z_{0})^{n}$ has radius of convergence $R$ , then $f (z) = \sum c_{n} (z - z_{0})^{n}$ is holomorphic for $∣ z - z_{0} ∣ < R$ and the derivative can be computed term by term:

f^{'} (z) = n = 1 \sum \infty n c_{n} (z - z_{0})^{n - 1},

with the same radius of convergence $R$ . By induction, $f$ is infinitely differentiable and $f^{(k)} (z_{0}) = k! c_{k}$ .

Integration along any contour inside the disk can likewise be performed term by term.

Theorem 3 (Identity theorem). If two power series $\sum a_{n} (z - z_{0})^{n}$ and $\sum b_{n} (z - z_{0})^{n}$ both converge for $∣ z - z_{0} ∣ < R$ and agree on a subset $E$ having a limit point in the open disk, then $a_{n} = b_{n}$ for all $n$ . Equivalently, a holomorphic function on a connected domain is determined by its values on any set with a limit point.

The identity theorem is the algebraic expression of analytic rigidity: a power series that vanishes on a set with an accumulation point vanishes everywhere. Exercise 8 gives the proof.

Theorem 4 (Uniqueness of Laurent coefficients). The Laurent expansion of a holomorphic function on an annulus is unique. If $\sum_{n = - \infty}^{\infty} c_{n} (z - z_{0})^{n} = \sum_{n = - \infty}^{\infty} d_{n} (z - z_{0})^{n}$ for all $z$ in the annulus, then $c_{n} = d_{n}$ for every $n$ .

Exercise 6 gives the proof. Uniqueness ensures that the Laurent coefficients — in particular $c_{- 1}$ , the residue — are well-defined and independent of the method used to find them.

Theorem 5 (Taylor's theorem for holomorphic functions). If $f$ is holomorphic on an open set containing the closed disk ${∣ z - z_{0} ∣ \leq R}$ , then

f (z) = n = 0 \sum \infty \frac{f ^{(n)} ( z _{0} )}{n !} (z - z_{0})^{n}

for all $∣ z - z_{0} ∣ < R$ , and the series converges absolutely and uniformly on compact subsets of the open disk.

This is the converse direction: a holomorphic function is represented by its Taylor series. The proof uses the Cauchy integral formula 06.01.02 to express $f^{(n)} (z_{0})$ as a contour integral and then estimates the remainder.

Theorem 6 (Abel's theorem on boundary continuity). If $\sum_{n = 0}^{\infty} c_{n} R^{n}$ converges, then $\sum c_{n} (z - z_{0})^{n} \to \sum c_{n} R^{n}$ as $z \to z_{0} + R$ radially from inside the disk. The result holds along any path approaching $z_{0} + R$ that is not tangent to the boundary. ^{[Stein-Shakarchi Ch. 1]}

Abel's theorem provides the bridge from interior convergence to boundary values. Exercise 7 gives the proof via Abel's summation formula.

Theorem 7 (Classification of isolated singularities via Laurent series). Let $f$ be holomorphic on a punctured disk $0 < ∣ z - z_{0} ∣ < R$ with Laurent expansion $\sum c_{n} (z - z_{0})^{n}$ . Then: (a) $z_{0}$ is a removable singularity if and only if $c_{n} = 0$ for all $n < 0$ ; (b) $z_{0}$ is a pole of order $m$ if and only if $c_{- m} \neq = 0$ and $c_{n} = 0$ for all $n < - m$ ; (c) $z_{0}$ is an essential singularity if and only if $c_{n} \neq = 0$ for infinitely many $n < 0$ .

This classification is the central application of Laurent series: the principal part of the Laurent expansion determines the nature of the singularity.

Synthesis. The power-series framework is the foundational reason that complex analysis achieves a perfect local-global correspondence, and the synthesis runs in four directions. First, the Weierstrass $M$ -test is the operational tool that makes series of holomorphic functions safe to manipulate: uniform convergence preserves holomorphicity, so infinite sums, products, and limits of holomorphic functions remain holomorphic. This is exactly the structure that underpins normal families 06.01.14 and the Montel compactness theorem. Second, the central insight is that a single number $R$ — the radius of convergence — controls absolute convergence, uniform convergence, and the domain of holomorphicity simultaneously. The pattern generalises from disks to annuli, where the Laurent decomposition provides a complete local description of any function with an isolated singularity.

Third, the identity theorem and the uniqueness of Laurent coefficients together identify a holomorphic function with its series representation: knowing the function on any set with a limit point determines the entire series. The bridge is between the algebra of formal series and the analysis of convergent functions, and putting these together gives the Taylor-coefficient formula $c_{n} = f^{(n)} (z_{0}) / n!$ that reconstructs the function from its derivatives at a single point. Fourth, the singularity classification via Laurent series is the foundational reason the residue theorem 06.01.03 works: the residue is the Laurent coefficient $c_{- 1}$ , determined by the function alone, and the residue theorem converts a local property (the type of singularity) into a global integral identity.

Full proof set [Master]

Proposition (Weierstrass $M$ -test). Let $f_{n}$ be continuous on a compact set $K \subset C$ . Suppose $∣ f_{n} (z) ∣ \leq M_{n}$ for all $z \in K$ , where $\sum_{n = 0}^{\infty} M_{n} < \infty$ . Then $\sum f_{n}$ converges uniformly on $K$ to a continuous function.

Proof. For each $z \in K$ , the numerical series $\sum ∣ f_{n} (z) ∣ \leq \sum M_{n} < \infty$ , so $\sum f_{n} (z)$ converges pointwise. For uniform convergence, let $ϵ > 0$ . Since $\sum M_{n}$ converges, there exists $N$ such that $\sum_{n = N}^{\infty} M_{n} < ϵ$ (tail of a convergent series). For any $m \geq N$ and any $z \in K$ :

n = 0 \sum m f_{n} (z) - n = 0 \sum \infty f_{n} (z) = n = m + 1 \sum \infty f_{n} (z) \leq n = m + 1 \sum \infty ∣ f_{n} (z) ∣ \leq n = m + 1 \sum \infty M_{n} < ϵ .

The bound is independent of $z$ , confirming uniform convergence. Continuity of the sum follows from the uniform limit of continuous functions. $□$

Proposition (Identity theorem via vanishing). Let $f (z) = \sum_{n = 0}^{\infty} c_{n} z^{n}$ converge for $∣ z ∣ < R$ . If $f$ vanishes on a sequence ${z_{k}}$ with $z_{k} \to 0$ and $z_{k} \neq = 0$ for all $k$ , then $c_{n} = 0$ for all $n$ .

Proof. Continuity of $f$ at $0$ gives $f (0) = lim_{k \to \infty} f (z_{k}) = 0$ , so $c_{0} = f (0) = 0$ . For $c_{1}$ : consider $g (z) = f (z) / z$ for $z \neq = 0$ and $g (0) = c_{1}$ . Since $f (z) = c_{1} z + c_{2} z^{2} + \dots$ (using $c_{0} = 0$ ), $g (z) = c_{1} + c_{2} z + \dots$ is holomorphic for $∣ z ∣ < R$ . The sequence ${z_{k}}$ satisfies $g (z_{k}) = f (z_{k}) / z_{k} = 0$ and $z_{k} \to 0$ , so by the same continuity argument $g (0) = 0$ , giving $c_{1} = 0$ .

Proceed by induction. Suppose $c_{0} = c_{1} = \dots = c_{m - 1} = 0$ . Then $f (z) = z^{m} (c_{m} + c_{m + 1} z + \dots)$ . The function $h (z) = f (z) / z^{m}$ (defined as $c_{m}$ at $z = 0$ ) is holomorphic. Since $h (z_{k}) = 0$ for all $k$ and $z_{k} \to 0$ , continuity gives $h (0) = c_{m} = 0$ . By induction, all $c_{n} = 0$ . $□$

Proposition (Laurent decomposition). If $f$ is holomorphic on the annulus $A = {z : r < ∣ z - z_{0} ∣ < R}$ , then $f$ has a unique representation $f = f_{+} + f_{-}$ where $f_{+}$ is holomorphic for $∣ z - z_{0} ∣ < R$ and $f_{-}$ is holomorphic for $∣ z - z_{0} ∣ > r$ with $f_{-} (\infty) = 0$ .

Proof. Fix $z \in A$ . Choose $ρ_{1}, ρ_{2}$ with $r < ρ_{1} < ∣ z - z_{0} ∣ < ρ_{2} < R$ . By the Cauchy integral formula for an annulus,

f (z) = \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = ρ_{2}} \frac{f ( w )}{w - z} d w - \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = ρ_{1}} \frac{f ( w )}{w - z} d w .

The outer integral: for $∣ w - z_{0} ∣ = ρ_{2} > ∣ z - z_{0} ∣$ , expand $1/ (w - z) = \sum_{n = 0}^{\infty} (z - z_{0})^{n} / (w - z_{0})^{n + 1}$ . This gives

f_{+} (z) := \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = ρ_{2}} \frac{f ( w )}{w - z} d w = n = 0 \sum \infty c_{n} (z - z_{0})^{n},

where $c_{n} = (2 π i)^{- 1} \oint f (w) / (w - z_{0})^{n + 1} d w$ , converging for $∣ z - z_{0} ∣ < ρ_{2}$ . This power series converges for all $∣ z - z_{0} ∣ < R$ (the integral representation is independent of the choice of $ρ_{2}$ ).

The inner integral: for $∣ w - z_{0} ∣ = ρ_{1} < ∣ z - z_{0} ∣$ , expand $1/ (w - z) = - \sum_{n = 0}^{\infty} (w - z_{0})^{n} / (z - z_{0})^{n + 1}$ . This gives

f_{-} (z) := - \frac{1}{2 π i} \oint_{∣ w - z_{0} ∣ = ρ_{1}} \frac{f ( w )}{w - z} d w = n = 1 \sum \infty c_{- n} (z - z_{0})^{- n},

where $c_{- n} = (2 π i)^{- 1} \oint f (w) (w - z_{0})^{n - 1} d w$ , converging for $∣ z - z_{0} ∣ > ρ_{1}$ and hence for $∣ z - z_{0} ∣ > r$ . The decomposition $f = f_{+} + f_{-}$ is unique by Exercise 6 (uniqueness of Laurent coefficients). $□$

Connections [Master]

Holomorphic functions 06.01.01. The power-series representation of a holomorphic function is one of the central results of complex analysis: a function is holomorphic if and only if it is locally representable by a convergent power series. Taylor's theorem (Theorem 5) establishes the forward direction, and term-by-term differentiation (Theorem 2) gives the reverse. This equivalence is the foundational reason that holomorphic functions possess the rigidity properties (identity theorem, maximum principle, analytic continuation) that distinguish complex analysis from real analysis.
Cauchy integral formula 06.01.02. The coefficient formula $c_{n} = (2 π i)^{- 1} \oint f (w) / (w - z_{0})^{n + 1} d w$ for both Taylor and Laurent coefficients is a direct application of the Cauchy integral formula. The Laurent decomposition proof above uses the Cauchy formula on an annulus, splitting the integral into an outer contribution (giving the analytic part) and an inner contribution (giving the principal part). The formula identifies the $n$ th coefficient with the $n$ th derivative of $f$ at $z_{0}$ (divided by $n!$ ), connecting the series representation to the contour-integral representation.
Analytic continuation 06.01.04. Power series are the local tool for analytic continuation: given a function represented by a power series in one disk, the series can be re-centred at a boundary point and extended into a new disk, potentially reaching beyond the original domain. The radius of convergence equals the distance to the nearest singularity, so the continuation process halts exactly when a singularity blocks further extension. Laurent series extend this picture to punctured neighbourhoods of singular points, classifying the obstruction to continuation.
Residue theorem 06.01.03. The residue of a function $f$ at an isolated singularity $z_{0}$ is the Laurent coefficient $c_{- 1}$ , extracted by the integral $Res_{z_{0}} f = (2 π i)^{- 1} \oint f (z) d z$ around a small circle centred at $z_{0}$ . The Laurent expansion makes the residue computable: for a simple pole, $c_{- 1} = lim_{z \to z_{0}} (z - z_{0}) f (z)$ . The residue theorem converts the local Laurent data (a single coefficient at each singularity) into a global contour integral, and the singularity classification (Theorem 7) determines when the residue is the only obstruction.
Meromorphic functions 06.01.05. A meromorphic function is holomorphic except for poles, and the Laurent expansion at each pole has a finite principal part (Theorem 7(b)). The order of the pole is the largest $∣ n ∣$ with $c_{n} \neq = 0$ for $n < 0$ . The Gamma function 06.01.15 provides a paradigmatic example: its Laurent expansion at $z = - n$ has principal part $(- 1)^{n} / (n! (z + n))$ , identifying each non-positive integer as a simple pole with computable residue.

Historical & philosophical context [Master]

Augustin-Louis Cauchy, in his 1831 Turin memoir Sur les fonctions d'une variable imaginaire ^{[Cauchy 1831]}, established the radius of convergence for power series and proved that a holomorphic function admits a power-series expansion — the result now called Taylor's theorem for holomorphic functions. Cauchy's approach was via the integral representation that now bears his name: the Cauchy integral formula gives the coefficients, and the convergence radius is controlled by the distance to the nearest singularity. This was the first rigorous treatment of the convergence of Taylor series in the complex domain, replacing the formal manipulations of the 18th century with analytic estimates.

Karl Weierstrass, in lectures delivered around 1842 (published posthumously in Mathematische Werke Vol. 1, 1894) ^{[Weierstrass 1842]}, developed the theory of uniform convergence and introduced the $M$ -test that bears his name. Weierstrass's programme placed complex analysis on an arithmetical foundation, replacing geometric intuition with $ϵ$ - $δ$ estimates. The $M$ -test, term-by-term differentiation, and the identity theorem are all products of this programme.

Pierre Alphonse Laurent, in 1843 ^{[Laurent 1843]}, extended Cauchy's power-series result to annuli, proving that a function holomorphic on an annulus has a convergent Laurent expansion with both positive and negative powers. Laurent's theorem was communicated to the French Academy by Cauchy himself. The Laurent decomposition — splitting a function into its analytic part and principal part — is the tool that makes singularity classification and residue computation possible, and it underlies the residue theorem as developed in 06.01.03.

Bibliography [Master]

@article{Cauchy1831,
  author = {Cauchy, Augustin-Louis},
  title = {Sur les fonctions d'une variable imaginaire},
  journal = {Oeuvres compl\`etes, S\'erie 2},
  volume = {15},
  year = {1831},
  pages = {31--78},
  note = {Turin memoir; radius of convergence and power-series representation}
}

@incollection{Weierstrass1894,
  author = {Weierstrass, Karl},
  title = {Darstellung einer analytischen Function einer complexen Ver\"anderlichen, deren absoluter Betrag einen gegebenen Wert nicht \"uberschreitet},
  booktitle = {Mathematische Werke},
  publisher = {Mayer \& M\"uller, Berlin},
  year = {1894},
  volume = {1},
  pages = {53--66},
  note = {Lectures from circa 1842; Weierstrass M-test and uniform convergence}
}

@article{Laurent1843,
  author = {Laurent, Pierre Alphonse},
  title = {Extension du th\'eor\`eme de M. Cauchy relatif \`a la convergence du d\'eveloppement d'une fonction suivant les puissances ascendantes de la variable},
  journal = {Journal de Math\'ematiques pures et appliqu\'ees},
  volume = {8},
  year = {1843},
  pages = {337--348},
  note = {Laurent expansion on an annulus}
}

@book{SteinShakarchi2003,
  author = {Stein, Elias M. and Shakarchi, Rami},
  title = {Complex Analysis},
  publisher = {Princeton University Press},
  year = {2003},
  volume = {II},
  note = {Princeton Lectures in Analysis, Chapter 1}
}

@book{Ahlfors1979,
  author = {Ahlfors, Lars V.},
  title = {Complex Analysis},
  publisher = {McGraw-Hill},
  year = {1979},
  edition = {3rd},
  note = {Chapter 5: power series, Laurent series, entire functions}
}

@book{Conway1978,
  author = {Conway, John B.},
  title = {Functions of One Complex Variable},
  publisher = {Springer-Verlag},
  year = {1978},
  edition = {2nd},
  note = {Sections IV.2--IV.3: power series and Laurent series}
}

Prerequisites

06.01.01
06.01.02
06.01.15

Tier anchors

beginner: Needham Visual Complex Analysis section 2.III informal (power series as infinite polynomials); Stewart Calculus early transcendentals section 11.8
intermediate: Stein-Shakarchi Complex Analysis Vol. II Ch. 1; Ahlfors Complex Analysis Ch. 5 section 1; Conway Functions of One Complex Variable section IV.2
master: Cauchy 1831 Turin memoir (radius of convergence); Weierstrass 1842 lectures (M-test, published 1894 Werke Vol. 1); Laurent 1843 J. Math. Pures Appl. 8; Ahlfors Complex Analysis Ch. 5

References

TODO_REF
Cauchy — Sur les fonctions d'une variable imaginaire (1831 Turin memoir) · Oeuvres completes Ser. 2, Vol. 15, pp. 31--78; radius of convergence and power-series representation
TODO_REF
Weierstrass — Darstellung einer analytischen Function (1894, lectures from 1842) · Mathematische Werke Vol. 1, pp. 53--66; Weierstrass M-test
TODO_REF
Laurent — Extension du theoreme de M. Cauchy (1843) · J. Math. Pures Appl. 8, pp. 337--348; Laurent expansion on an annulus
TODO_REF
Stein-Shakarchi — Complex Analysis (Vol. II) · Princeton Lectures in Analysis II, Ch. 1 (power series, radius of convergence, Laurent series)
TODO_REF
Ahlfors — Complex Analysis · Ch. 5 section 1 (power series, Weierstrass M-test, Laurent decomposition)
TODO_REF
Conway — Functions of One Complex Variable · section IV.2 (power series representation), section IV.3 (Laurent series)

Reviewer

TBD

Estimated time

beginner: 14m
intermediate: 35m
master: 70m