Cauchy estimates
Theorem. If \(|f|\le M\) on \(|\zeta-a|=R\), then \(|f^{(n)}(a)|\le n!M/R^n\). This follows directly from the derivative form of Cauchy's integral formula and the length estimate.
Taylor series radius
Corollary. If \(f\) is holomorphic on \(D(a,R)\), then its Taylor series at \(a\) converges to \(f\) on every smaller disc. Coefficients are controlled by Cauchy's estimates, and the integral formula identifies the sum.
Liouville from estimates
Example. If \(f\) is entire and bounded by \(M\), then \(|f'(a)|\le M/R\) for every \(R>0\), hence \(f'(a)=0\). Since \(a\) is arbitrary, \(f\) is constant.
Derivative bounds
Proposition. Bounds on a holomorphic function automatically give bounds on all derivatives in smaller discs. This is a major contrast with real differentiable functions, where a function can be small while its derivative is large.
Removable singularity criterion
Corollary. If \(f\) is holomorphic on a punctured disc and bounded near the puncture, then the singularity is removable. Cauchy's estimates force the negative Laurent coefficients to vanish.
Analytic rigidity
Remark. Cauchy estimates are the quantitative form of rigidity: holomorphic functions cannot oscillate arbitrarily fast without becoming large nearby.
Derivative estimates
Theorem (Cauchy estimates). If \(f\) is holomorphic on a neighborhood of \(\overline{D(z_0,R)}\) and \(M_R=\max_{|z-z_0|=R}|f(z)|\), then \[ |f^{(n)}(z_0)|\le\frac{n!M_R}{R^n}. \]
Proof. Apply the derivative form of Cauchy's formula and the ML estimate: \[ |f^{(n)}(z_0)| \le\frac{n!}{2\pi}(2\pi R)\frac{M_R}{R^{n+1}}. \]
Liouville and polynomial growth
Theorem. If an entire function satisfies \[ |f(z)|\le C(1+|z|^m) \] for some integer \(m\ge0\), then \(f\) is a polynomial of degree at most \(m\).
Proof. For \(n>m\), Cauchy's estimate on \(|z|=R\) gives \[ |f^{(n)}(0)| \le n!C(1+R^m)R^{-n}\to0 \quad(R\to\infty). \] Thus all Taylor coefficients above degree \(m\) vanish. The case \(m=0\) is Liouville's theorem.
Quantitative Taylor remainders
Proposition. If \(|f|\le M_R\) on \(|\zeta-z_0|=R\), then for \(|z-z_0|=r<R\), \[ \left|f(z)-\sum_{n=0}^{N} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n\right| \le \frac{M_R(r/R)^{N+1}}{1-r/R}. \]
Proof. Cauchy's coefficient estimate gives \(|a_n|\le M_R/R^n\). Sum the geometric tail \(\sum_{n>N}M_R(r/R)^n\).
Removable singularities from growth
Theorem. If \(f\) is holomorphic on \(0<|z-a|<R\) and bounded near \(a\), then \(f\) extends holomorphically to \(a\).
Proof. Put \(g(z)=(z-a)^2f(z)\) for \(z\ne a\) and \(g(a)=0\). Boundedness makes \(g\) complex differentiable at \(a\) with \(g'(a)=0\), so \(g\) is holomorphic. Its Taylor series has zero constant and linear terms: \(g(z)=(z-a)^2h(z)\). Therefore \(f=h\) away from \(a\), giving the extension.
A sharper criterion is \(|f(z)|=o(|z-a|^{-1})\), which also forces removability by applying the same argument to \((z-a)f(z)\) and checking its derivative.