Zeros, multiplicity and the identity theorem

Try

The function z³(1+z) vanishes at zero, but nearby it behaves like z³ rather than like z. Its first nonzero Taylor coefficient occurs at degree three. That integer is the zero’s multiplicity, and it controls both local geometry and later contour counts.

Compare it with the zero function. There is no first nonzero coefficient in that case, so assigning it an ordinary finite zero order would be misleading. The zero-factorization theorem must explicitly separate these alternatives.

Predict

Predict whether a holomorphic function can vanish at 1/2, 1/3, 1/4 and all subsequent reciprocal integers without vanishing near zero. If zero belongs to its domain, the accumulating zeros force every local Taylor coefficient to vanish.

Now move the accumulation point to the boundary. The same conclusion need not hold. The location of the limit point is part of the theorem, not a technical afterthought.

Experiment

Inspect first nonzero terms at zero.

Function	First nonzero term	Zero order
sin z	z	1
1−cos z	z²/2	2
z²sin(z³)	z⁵	5

Use the known convergent series, not just a graph that appears flat. A high-order zero can look identically zero at a coarse numerical scale.

Understand

Suppose f is holomorphic near a and f(a)=0. If not all Taylor coefficients vanish, let m be the least index with nonzero coefficient. Factoring the convergent series gives f(z)=(z−a)ᵐg(z), where g is holomorphic and g(a)≠0. By continuity g remains nonzero on a smaller disk, so a is the only zero there. Thus a nonzero holomorphic germ has isolated zeros. If all coefficients vanish, the Taylor expansion makes f zero on an entire neighborhood.

For the identity theorem, let f have distinct zeros accumulating at an interior point a of a domain D. The isolated-zero alternative is impossible, so f vanishes on a neighborhood. Let S be the set of points having such a zero neighborhood. It is nonempty and open. It is relatively closed as well: if points of S approach b in D, either b already lies in S or distinct zeros accumulate at b, forcing the same local conclusion. Connectedness gives S=D.

Applying this theorem to f−g shows that two holomorphic functions agreeing on any set with an interior accumulation point agree throughout the domain. They need not agree on an open set initially; a short real interval contained in the domain is enough.

The proof explains why both holomorphy and connectedness matter. Smooth real functions can vanish on an interval and differ elsewhere, and different connected components can carry unrelated holomorphic functions.

Apply

Construct a boundary counterexample with f(z)=sin(π/(1−z)) on the unit disk. Its zeros include z=1−1/n for n≥2, approaching the excluded boundary point one. It is not identically zero: at z=1/3 its value is −1. Check both assertions explicitly.

Repair the statement “Infinitely many zeros imply identically zero.” On the whole plane, sin z has infinitely many zeros but they have no finite accumulation point. The correct condition is accumulation at a point within the domain.

Continue

Multiplicity now turns zeros into local integer data. Laurent series will supply the corresponding language for singularities and poles. Optional depth: prove that a nonzero holomorphic function has only finitely many zeros in a compact subset of its domain by combining isolated zeros with the compact accumulation-point property.