Cauchy estimates, Liouville and the fundamental theorem of algebra

Try

Suppose |f| is at most twelve on the radius-two circle and f is holomorphic on a neighborhood of the closed disk. Cauchy’s formula bounds |f′(0)| by six. A boundary amplitude bound controls an interior rate of change.

The same calculation for the third derivative gives 3!·12/2³=9. The factorial grows with derivative order; it is not optional decoration. Before using a bound, identify its center, radius, derivative order and the circle on which the modulus bound holds.

Predict

Predict why a bounded entire function must be constant but a bounded function on the unit disk need not be. The function z is already a counterexample on the disk.

The difference is access to arbitrarily large circles. For an entire function bounded by one fixed M, the first derivative estimate M/R can be made as small as desired. A bounded domain does not permit that limiting operation.

Experiment

Keep the same circle bound M while comparing derivatives.

Order n	Cauchy bound on ∣f^(n)(a)∣
0	M
1	M/R
2	2M/R²
3	6M/R³

These are upper bounds, not claims that every function attains them simultaneously. Changing the radius can also change M, so the large-radius limit must track both quantities.

Understand

The derivative formula has denominator modulus R^(n+1) on a radius-R circle, length 2πR, and numerator bounded by M. The ML estimate yields |f^(n)(a)|≤n!M/Rⁿ. This derivation explains every factor.

Liouville’s theorem follows immediately. Let f be entire, meaning holomorphic on all of C, and bounded there by a single constant M. For any point a and every R>0, |f′(a)|≤M/R. Letting R increase without bound forces f′(a)=0. Since the plane is connected, f is constant.

To prove the fundamental theorem of algebra, suppose a nonconstant polynomial p has no complex zero. Its reciprocal is then entire. If p has degree d and leading coefficient c≠0, write p(z)=czᵈ times a factor tending to one as |z| tends to infinity. Hence |p(z)| tends to infinity, so 1/p is bounded outside a large disk. On the closed disk it is continuous and bounded because p never vanishes. Liouville makes 1/p constant, a contradiction. Thus p has at least one zero. Repeated polynomial division then gives a complete linear factorization, counting multiplicities.

The proof establishes root existence before using factorization. Invoking a factorization to prove the leading-term growth would be unnecessary and could make the argument circular.

Apply

Test the stronger-looking claim “Every entire function with polynomial growth is a polynomial.” Suppose |f(z)|≤C(1+|z|ᵐ) for a nonnegative integer m. At zero, the Cauchy estimate forces every derivative of order greater than m to vanish as R grows. The Taylor theorem developed shortly then identifies f with a polynomial of degree at most m.

Explain the dependency: derivative vanishing alone is not a real-variable Taylor theorem with zero remainder. Complex analyticity is what makes the global coefficient conclusion valid.

Continue

Use Liouville as a global rigidity tool, always checking both entire holomorphy and a uniform whole-plane bound. Optional depth: prove an entire function whose real part is bounded above must be constant by applying Liouville to exp f, then differentiating exp f. Nonvanishing of exp f is needed in the last step.