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Complex Analysis 30 | Identity Theorem thumbnail

Complex Analysis 30 | Identity Theorem

5 min read

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TL;DR

The identity theorem states that if two holomorphic functions agree on a set with an accumulation point inside a connected open domain, they agree everywhere on that domain.

Briefing

Complex analysis hinges on a strict “no surprises” rule for holomorphic functions: if two holomorphic functions agree on a set with an accumulation point, they must agree everywhere on the connected domain. That matters because it turns partial information—values on a tiny subset—into a complete determination of an analytic object, eliminating the possibility of different holomorphic extensions.

The identity theorem formalizes this. Start with two holomorphic functions f and g on an open, path-connected domain D. Consider the set M = {z in D : f(z) = g(z)}. If M has at least one accumulation point inside D, then f and g coincide on all of D. In other words, agreement at infinitely many nearby points forces global equality; a “small” coincidence set can’t be isolated.

A second, equivalent formulation uses derivatives at a point. If there exists c in D such that f and g have matching derivatives of every order at c—meaning f^(n)(c) = g^(n)(c) for all natural numbers n including 0—then f and g are identical on D. This equivalence reflects the power-series nature of holomorphic functions: matching all Taylor coefficients at one point forces the same Taylor expansion, hence the same function in a neighborhood, and then everywhere by connectedness.

The transcript also clarifies what an accumulation point is. For a subset M of D, a point p is an accumulation point if every punctured neighborhood of p (every open set U containing p, with p removed) still contains points of M. This captures the idea of “points of M clustering near p,” even if p itself is not in M. Examples include the discrete set {1/n} accumulating at 0.

For the proof sketch, the key move is to reduce the problem to a single holomorphic function by defining h = f − g. Then the identity theorem becomes a statement about zeros of h: if h vanishes on a set with an accumulation point, then h must be the zero function. The argument uses Taylor expansions. If all derivatives of h at a point c do not vanish at some minimal order m, then near c the leading term a_m (z − c)^m dominates, preventing h from having zeros arbitrarily close to c; this blocks accumulation points of the zero set.

Another step introduces sets A_k of points where the k-th derivative of h vanishes, and their intersection A where all derivatives vanish. Continuity makes each A_k closed, while power-series expansion shows A is also open. Since D is connected (path-connected), a set that is both open and closed must be either empty or all of D; under the theorem’s assumptions it becomes all of D, forcing h ≡ 0.

A final implication highlights the theorem’s strength: holomorphic functions are uniquely determined by their values on a subset with an accumulation point. Even the real sign function has no alternative holomorphic extension to the complex plane once its restriction is fixed on the real line, because any two holomorphic candidates would be forced to match everywhere.

Cornell Notes

The identity theorem says that holomorphic functions on a connected open set cannot agree “locally” without agreeing everywhere. If f and g are holomorphic on a path-connected domain D and the set of points where f(z)=g(z) has an accumulation point in D, then f=g on all of D. An equivalent version uses derivatives: if all derivatives of f and g match at a single point c in D, then the functions are identical on D. The proof idea reduces to h=f−g and uses Taylor expansions to show that if h has a zero set with an accumulation point, then h must be the zero function. Connectedness then spreads local equality to global equality.

Why does agreement on a set with an accumulation point force two holomorphic functions to be identical on the whole domain?

Let h=f−g. The set where f=g is exactly the zero set of h. If that zero set has an accumulation point in D, then zeros occur arbitrarily close to some point c. A Taylor expansion of h at c shows that if the first nonzero derivative occurs at order m, then near c the leading term a_m (z−c)^m keeps h from vanishing for all nearby z≠c. That contradicts the existence of an accumulation point of zeros. So h must be identically zero, meaning f=g everywhere on D.

What exactly counts as an accumulation point, and why can it lie outside the set?

A point p is an accumulation point of a subset M⊂D if every neighborhood U of p contains points of M other than p itself; formally, (Up)∩M is non-empty. p need not belong to M. For example, M={1/n} has 0 as an accumulation point even though 0 is not in M, because points 1/n get arbitrarily close to 0.

How do matching derivatives at one point relate to matching functions everywhere?

If f^(n)(c)=g^(n)(c) for all n≥0, then all Taylor coefficients of f and g at c match, so their Taylor expansions coincide. That forces f and g to agree on some neighborhood of c. The identity theorem then extends this local agreement to the entire connected domain D, because the set where f=g contains that neighborhood and therefore has accumulation points in D.

Why does connectedness (path-connectedness) matter in the proof?

The proof constructs a set A where all derivatives of h vanish. Using power-series behavior, A is shown to be both open and closed in D. In a connected domain, the only subsets that are simultaneously open and closed are the empty set and the whole domain. Since the assumptions ensure A is not empty, connectedness forces A=D, implying h≡0.

What role does the Taylor expansion play in preventing accumulation points of zeros?

Suppose not all derivatives of h vanish at c. Let m be the smallest index with h^(m)(c)≠0. Then near c, h(z) has the form a_m (z−c)^m plus higher-order terms. Because a_m≠0, the leading term controls the sign/size for z close enough to c, making h(z) nonzero for all z in a punctured neighborhood of c. That means the zero set cannot accumulate at c.

Review Questions

  1. Given a holomorphic function h on a path-connected domain D, how would you use the Taylor expansion at a point c to argue that zeros of h cannot accumulate at c if some derivative at c is nonzero?
  2. Explain why a set that is both open and closed must be either empty or all of D when D is connected. How does this connect to the identity theorem?
  3. Provide an example of a set with an accumulation point that is not contained in the set, and explain how that definition is used in the theorem.

Key Points

  1. 1

    The identity theorem states that if two holomorphic functions agree on a set with an accumulation point inside a connected open domain, they agree everywhere on that domain.

  2. 2

    The coincidence set M={z∈D: f(z)=g(z)} having one accumulation point is enough to force global equality f=g.

  3. 3

    An equivalent criterion uses derivatives: if f^(n)(c)=g^(n)(c) for all n≥0 at some c∈D, then f=g on all of D.

  4. 4

    An accumulation point p of M⊂D means every neighborhood of p contains other points of M; p itself may or may not lie in M.

  5. 5

    The proof reduces to h=f−g and studies the zero set of h, using Taylor expansions to rule out accumulation of zeros unless h is identically zero.

  6. 6

    Connectedness (path-connectedness) is essential because it turns a set that is both open and closed into either the whole domain or nothing, forcing h≡0 under the theorem’s assumptions.

  7. 7

    Holomorphic functions are uniquely determined by their values on any subset with an accumulation point, ruling out alternative holomorphic extensions from the same local data.

Highlights

Agreement on infinitely many nearby points is not just suggestive in holomorphic functions—it is decisive: one accumulation point forces equality everywhere.
Matching all derivatives at a single point is equivalent to matching the entire holomorphic function on the whole connected domain.
The accumulation-point definition allows clustering behavior even when the limiting point is not itself in the set (e.g., {1/n} accumulating at 0).
The proof’s backbone is Taylor expansion: the first nonzero derivative creates a leading term that prevents zeros from accumulating near that point.
Connectedness converts local “all derivatives vanish” information into global equality across D.

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