Why Agreement at Many Points Can Determine a Holomorphic Function
The Maximum Modulus Principle showed that a holomorphic function cannot have a nonconstant interior maximum of its modulus. The Identity Theorem gives another strong restriction: once two holomorphic functions agree on a sufficiently large set, their agreement may be forced throughout the domain. The key condition is not simply that there are many points of agreement. Those points must accumulate at a point that lies inside the domain.
The reason comes from the local Taylor expansion of a holomorphic function. Near a zero, either every Taylor coefficient vanishes, so the function vanishes on a neighborhood, or there is a first nonzero coefficient, which makes the zero isolated. The Identity Theorem turns this local fact into a conclusion on a whole connected domain.
Zeros and Their Orders
Because \(h(a)=0\), the constant coefficient in this Taylor expansion is zero. If the first nonzero coefficient is \(c_m\), the series can be factored locally:
where \(g\) is holomorphic near \(a\). Continuity of \(g\) implies that \(g(z)\neq0\) for all \(z\) sufficiently close to \(a\). Thus \(h\) has no zeros near \(a\) other than \(a\) itself. This gives the basic local result.
Proof. The Taylor expansion of \(h\) at \(a\) cannot have all coefficients equal to zero. If they were all zero, analyticity would make \(h\) identically zero on a neighborhood of \(a\), contrary to the hypothesis. Let \(m\) be the least index with \(c_m\neq0\). Since \(h(a)=0\), \(m\geq1\), and the Taylor expansion factors as \(h(z)=(z-a)^m g(z)\), where \(g\) is holomorphic near \(a\) and \(g(a)=c_m\neq0\). By continuity, there is a \(\delta>0\) such that \(g(z)\neq0\) whenever \(|z-a|<\delta\). For \(0<|z-a|<\delta\), both factors in \(h(z)=(z-a)^m g(z)\) are nonzero, so \(h(z)\neq0\). \(\square\)
Worked Example: Finding the Orders of Polynomial Zeros
Consider \(h(z)=(z-1)^2(z+2i)\). At \(a=1\), the factorization already has the form of a zero of order two, provided the remaining factor is nonzero there. Indeed, \(1+2i\neq0\), so \(z=1\) is a zero of order two. At \(a=-2i\), the factor \(z+2i\) occurs once, and the other factor has value
Thus \(z=-2i\) is a simple zero. The factorization also verifies that these are the only zeros: a product of complex numbers is zero only if at least one factor is zero. Each zero is isolated, in agreement with the theorem.
The Identity Theorem
Proof. Define \(h=f-g\). It is holomorphic on \(U\), and \(h\) vanishes at every point where \(f\) and \(g\) agree. Let \(a\in U\) be an accumulation point of those points. There is a sequence of distinct zeros of \(h\) converging to \(a\).
Since \(a\) is a limit of zeros of \(h\), continuity gives \(h(a)=0\). If \(h\) were not identically zero on any neighborhood of \(a\), the theorem that zeros are isolated would give a neighborhood of \(a\) containing no zero of \(h\) other than possibly \(a\). That contradicts the sequence of distinct zeros converging to \(a\). Therefore \(h\) vanishes on some neighborhood of \(a\).
To extend this local equality across \(U\), let \(A\) be the set of points \(z\in U\) at which \(h\) is identically zero on some neighborhood. The neighborhood just found shows that \(A\) is nonempty, and the definition shows that \(A\) is open in \(U\).
The set \(A\) is also closed relative to \(U\). Suppose \(z_n\in A\) and \(z_n\to z\in U\). For each integer \(k\geq0\), \(h^{(k)}(z_n)=0\), since \(h\) vanishes on a neighborhood of each \(z_n\). Holomorphic functions have continuous derivatives of every order, so \(h^{(k)}(z)=0\) for every \(k\geq0\). The Taylor expansion of \(h\) at \(z\) therefore has all coefficients zero. Analyticity implies that \(h\) vanishes on a neighborhood of \(z\), so \(z\in A\). This proves relative closedness. Since \(U\) is connected and \(A\) is nonempty, open, and closed in \(U\), \(A=U\). Hence \(h=0\) throughout \(U\), or \(f=g\) throughout \(U\). \(\square\)
Applying the theorem to \(h=f\) gives the equivalent zero-set form: if a holomorphic function on a connected open set has zeros accumulating at a point of that set, it vanishes everywhere on the set. The conclusion depends on connectedness. Without it, equality is forced only on the connected component containing the accumulation point.
Worked Example: Agreement on a Nondegenerate Real Interval
Suppose \(f\) and \(g\) are holomorphic on a connected open set \(U\) containing the real interval \([0,1]\), and suppose \(f(x)=g(x)\) for every \(x\in[0,1]\). The points \(x_n=1/(n+1)\), for positive integers \(n\), are distinct points of the interval, and \(x_n\to0\). The point \(0\) belongs to \(U\), so it is an accumulation point inside the domain of the agreement set. The Identity Theorem therefore gives \(f(z)=g(z)\) for every \(z\in U\). In particular, their values and all their derivatives agree at every point of \(U\).
The interval must be nondegenerate: it must contain more than one point, and in this example it supplies a sequence of distinct agreement points with an interior limit. A singleton does not supply such a sequence.
Uniqueness from Derivatives and from a Sequence
The Identity Theorem is often useful even when equality is not initially given on a whole interval. For example, a sequence of distinct points of agreement is enough if it converges to a point in the domain. Likewise, if two holomorphic functions have the same derivatives of every order at one point, their Taylor expansions agree near that point. The Identity Theorem then extends that local equality throughout the connected domain.
Worked Example: Equality from a Sequence of Agreement Points
Let \(f\) and \(g\) be holomorphic on a connected open set \(U\), and suppose there are distinct points \(z_n\in U\) such that \(z_n\to a\in U\) and \(f(z_n)=g(z_n)\) for every \(n\). Set \(h=f-g\). Then \(h(z_n)=0\) for all \(n\). If \(h\) did not vanish on any neighborhood of \(a\), its zero at \(a\), if it has one, would be isolated; if \(h(a)\neq0\), continuity would instead give a neighborhood with no zeros. Both possibilities contradict the distinct zeros \(z_n\) approaching \(a\). The Identity Theorem gives \(h=0\) throughout \(U\), so \(f=g\) everywhere on \(U\).
For instance, this reasoning applies whenever two holomorphic functions agree at \(z_n=a+1/n\) for every sufficiently large positive integer \(n\), provided those points lie in \(U\). Their convergence to the interior point \(a\), rather than the mere number of agreement points, is decisive.
Worked Example: Why a Single Agreement Point Is Not Enough
On the unit disk \(U=\{z\in\mathbb{C}:|z|<1\}\), let \(f(z)=0\) and \(g(z)=z\). Both functions are holomorphic, and \(f(0)=g(0)=0\). However, \(f(1/2)=0\) while \(g(1/2)=1/2\), so they do not agree throughout \(U\). The agreement set is just \(\{0\}\), which has no accumulation point. This shows why the hypothesis of the Identity Theorem cannot be replaced by agreement at one point.
Why the Accumulation Point Must Be Inside the Domain
The condition that the accumulation point belongs to the domain is also essential. A sequence of zeros can approach a boundary point without forcing a holomorphic function to vanish identically. For example, on the open unit disk define
This function is holomorphic on the disk because \(1-z\neq0\) there. For every sufficiently large positive integer \(n\), let \(z_n=1-1/(n\pi)\). These points lie in the unit disk, are distinct, and converge to \(1\), which is on the boundary rather than in the domain. Direct substitution gives \(1/(1-z_n)=n\pi\), and therefore \(h(z_n)=\sin(n\pi)=0\). But \(h(0)=\sin(1)\neq0\). Thus the zeros accumulate at the boundary without making \(h\) identically zero.
A related consequence of the isolated-zeros theorem is that a nonzero holomorphic function cannot have infinitely many zeros in a compact subset of its domain. Otherwise, compactness would give an accumulation point in that subset, and the Identity Theorem would force the function to vanish identically on the connected domain.
Using the Theorem Carefully
When applying the Identity Theorem, check three features: both functions must be holomorphic on the domain in question; the domain must be connected; and the set of agreement must have an accumulation point inside that domain. Agreement on a nondegenerate real interval contained in the domain is sufficient, because every point of the interval is approached by other points of the interval. Agreement on a singleton is not sufficient, and accumulation only at the boundary is not sufficient.
The theorem explains why holomorphic functions are rigid: local information can determine a function globally. This rigidity is useful for proving formulas, identifying analytic continuations, and establishing uniqueness. It does not say that arbitrary values at finitely many points determine a holomorphic function; the accumulation condition is what makes the conclusion possible.
Check Your Understanding
Use the zero-set criterion and the Identity Theorem to answer the following questions.
- Why is a zero of finite order isolated?
- If two holomorphic functions agree at distinct points converging to an interior point, what can be concluded when the domain is connected?
- Why does agreement on a nondegenerate real interval contained in the domain imply equality throughout the domain?
- Give an example showing that agreement at a single point does not imply equality.
- Why does the function \(\sin(1/(1-z))\) on the unit disk not contradict the Identity Theorem?