Complex Analysis 30 | Identity Theorem [dark version]

TL;DR

If f and g are holomorphic on the same connected open domain D and agree on a set with an accumulation point inside D, then f and g agree everywhere on D.

Briefing Cornell Notes

Briefing

A single accumulation point of agreement between two holomorphic functions forces them to be identical everywhere on a connected domain. That’s the identity theorem’s core punchline—and it’s why holomorphic functions are so rigid compared with ordinary differentiable functions.

Start with two holomorphic functions f and g on the same open, path-connected domain D. Consider the set M = {z in D : f(z) = g(z)}. Even if M is “small” (not necessarily a curve or region), the theorem says that if M has at least one accumulation point in D, then f and g must coincide on all of D. In other words, knowing that the functions match on a subset that clusters somewhere inside the domain determines the entire holomorphic function.

The theorem also has a second, equivalent formulation in terms of derivatives at a point. If there exists a point c in D such that all derivatives match—meaning f^(n)(c) = g^(n)(c) for every natural number n (including n = 0)—then f and g are again forced to be equal on all of D. This derivative condition is essentially the statement that their Taylor (power series) expansions at c are identical, so the functions agree not just at c but throughout a neighborhood; holomorphicity then propagates that agreement across the whole connected domain.

A key concept behind both versions is the accumulation point of a set: a point p is an accumulation point of M if every neighborhood around p contains other points of M (formally, for every open neighborhood U of p, (U \ {p}) ∩ M is non-empty). The transcript illustrates why this matters by contrasting discrete sets like the natural numbers (which have no accumulation points in C) with sets like {1/n}, which accumulate at 0 even though 0 is not itself in the set.

For the proof sketch, the argument reduces the problem to a single holomorphic function by setting h = f − g. Then the question becomes: when does h vanish identically? The equivalences translate into statements about the zero set of h, and about whether all derivatives of h vanish at some point. The proof uses power series expansions: if h has a zero of minimal order m at a point c (so the first nonzero derivative occurs at order m), then near c the leading term a_m (z − c)^m dominates, implying h(z) ≠ 0 for z close to but not equal to c. That blocks c from being an accumulation point of the zeros.

Another step defines closed sets A_k = {z in D : h^(k)(z) = 0} and intersects them to form A = ⋂_k A_k, the set where all derivatives vanish. Using continuity and the local power series behavior, A is shown to be both open and closed. Since D is connected, that forces A to be either empty or all of D; under the derivative-vanishing assumption it must be all of D, so h is identically zero and thus f = g.

The practical implication is extension uniqueness: once a holomorphic function is fixed on a subset with an accumulation point, there is no alternative holomorphic continuation. The transcript highlights the familiar example that the real sine function has a unique holomorphic extension to the complex plane consistent with complex differentiability.

Cornell Notes

The identity theorem says two holomorphic functions on a connected open set D that agree “often enough” must agree everywhere. If M = {z in D : f(z)=g(z)} has an accumulation point in D, then f and g coincide on all of D. An equivalent criterion uses derivatives: if for some c in D, f^(n)(c)=g^(n)(c) for every n ≥ 0, then f=g on all of D. The proof idea reduces to h=f−g and uses power series expansions: a holomorphic function with a zero of finite order cannot have that zero as an accumulation point of its zero set. It then uses sets where derivatives vanish, showing they become both open and closed, which—by connectedness—forces h to be identically zero.

What does it mean for a point p to be an accumulation point of a set M inside a domain D?

p is an accumulation point of M (relative to D) if every open neighborhood U of p contains points of M other than p itself. Formally, for all open sets U with p ∈ U, the intersection (U \ {p}) ∩ M is non-empty. This captures the idea that M “clusters” at p even if p might not belong to M.

Why does agreement on a subset with an accumulation point force two holomorphic functions to be identical?

Let h = f − g, which is holomorphic on D. The set where f and g agree is exactly the zero set of h. If that zero set has an accumulation point in D, then h must be the zero function. The reason is that holomorphic functions admit power series expansions: if h had a zero of finite order at a point, the leading nonzero term would prevent zeros from accumulating there. So accumulation of zeros forces all derivatives to vanish at some point, which then forces h ≡ 0 on the connected domain.

How does the derivative version of the identity theorem relate to Taylor series?

If f^(n)(c)=g^(n)(c) for all n ≥ 0 at some c, then their Taylor (power series) expansions at c match term-by-term. That means f and g coincide on a neighborhood of c. Holomorphicity then extends that local equality across the entire connected domain D, yielding f ≡ g on all of D.

What role does connectedness (path-connectedness) of D play in the proof?

In the proof sketch, the set A where all derivatives of h vanish is shown to be both open and closed in D. In a connected domain, the only subsets that are simultaneously open and closed are either empty or the whole domain. Since the derivative-vanishing assumption makes A non-empty, connectedness forces A = D, so h ≡ 0.

Why can’t a holomorphic function have a zero that is an accumulation point unless it is identically zero?

Near any point c, h has a power series expansion. If the first nonzero term occurs at order m, then h(z) behaves like a_m (z − c)^m plus higher-order terms. For z sufficiently close to c but not equal to c, the leading term dominates and keeps h(z) ≠ 0. That means c cannot be an accumulation point of the zeros unless there is no such finite m—i.e., unless all derivatives vanish and h is identically zero.

Review Questions

Given two holomorphic functions f and g on a connected domain D, how would you check the identity theorem using only the zero set of h=f−g?
Explain the difference between a discrete set like {1,2,3,...} and a set like {1/n} in terms of accumulation points in C.
Why does showing a set is both open and closed imply it must be all of D (or empty) when D is connected?

Key Points

1
If f and g are holomorphic on the same connected open domain D and agree on a set with an accumulation point inside D, then f and g agree everywhere on D.
2
The identity theorem can be phrased using zeros: for h=f−g, an accumulation point of the zero set forces h to be identically zero.
3
A derivative-based version is equivalent: if all derivatives of f and g match at a point c, then the functions are identical on all of D.
4
Accumulation points are defined via neighborhoods: every neighborhood of p contains other points of the set, even if p itself is not in the set.
5
Power series expansions provide the mechanism: a holomorphic function with a zero of finite order cannot have that zero as an accumulation point of its zeros.
6
Connectedness of D is essential because it prevents a nontrivial set from being both open and closed.
7
Holomorphic continuation is unique: fixing a holomorphic function on a subset with an accumulation point determines its extension to the whole domain.

Highlights

Agreement at even one accumulation point is enough to determine a holomorphic function on an entire connected domain.

Matching all derivatives at a single point forces equality everywhere, because Taylor expansions coincide term-by-term.

The proof reduces to h=f−g and uses the fact that power series control how zeros can behave near a point.

A set where all derivatives vanish becomes both open and closed; connectedness then forces it to be the whole domain.

The identity theorem makes holomorphic extensions unique, illustrated by the sine function’s unique complex-analytic extension.