Isolated Zeros Theorem
General statement, general proof — the drag above instantiates it for the toy f(z) = z² + z³.
Theorem
If f is holomorphic near z0, f(z0) = 0, and f ≢ 0 near z0, then there exists δ > 0 such that f(z) ≠ 0 for all z with 0 < |z − z0| < δ.
Setup: f holomorphic near
z0,
f(
z0) = 0,
f ≢ 0 near
z0. Then
f(
z) = (
z −
z0)
mg(
z), where
m is the order of the zero,
g is itself holomorphic near
z0 (a convergent power series), and
g(
z0) ≠ 0.
Toy instance used above: f(z) = z² + z³ = z²(1 + z), so z0 = 0, m = 2, g(z) = 1 + z, g(0) = 1.
Claim: there exists δ > 0 such that g(z) ≠ 0 — in fact |g(z)| > |g(z0)|/2 — for every z with |z − z0| < δ.
1.g is holomorphic at z0 ⇒ g is continuous at z0 (holomorphic ⇒ differentiable ⇒ continuous).
2.Continuity means: for every ε > 0, there exists δ > 0 such that |z − z0| < δ ⇒ |g(z) − g(z0)| < ε.
3.Choose ε = |
g(
z0)|/2 — legal because
g(
z0) ≠ 0 makes this a genuine positive number.
Why ε = |g(z₀)|/2, not the full distance |g(z₀)|: using the toy, g's only zero is at z = −1, at distance exactly |g(0)| = 1 from g(0). Choosing ε = 1 would give a matching δ = 1 — technically g(z) ≠ 0 everywhere in that disk (the zero sits exactly on the boundary, excluded only by strict inequality), but |g(z)| → 0 as z → −1 from inside the disk, so there's no positive floor — a knife's edge. Halving the distance instead (ε = 1/2) gives a genuine, robust quantitative floor of 1/2.
4.Reverse triangle inequality, in the form that bounds |g(z)| from below: |g(z0)| − |g(z)| ≤ |g(z) − g(z0)|, i.e. |g(z)| ≥ |g(z0)| − |g(z) − g(z0)|.
5.Since |
g(
z) −
g(
z0)| < ε = |
g(
z0)|/2 (from step 3), substitute:
|g(z)| ≥ |g(z0)| − |g(z) − g(z0)| > |g(z0)| − |g(z0)|/2 = |g(z0)|/2 > 0
—
6.So g has a hard positive floor of |g(z0)|/2 on the whole disk |z − z0| < δ (including z0 itself) — g is never 0 there. Since (z − z0)m is zero only at z = z0 (and nonzero everywhere else on the disk), f(z) = (z − z0)mg(z) vanishes only at z = z0 on this disk — the zero is isolated.
—
Because δ = ε = 0.5 for this toy (g(z) − g(0) = z exactly), the δ-disk on the left and the ε-disk on the right are the same size — drag z inside the left disk and g(z) is guaranteed to stay inside the right disk, floor intact.