Cauchy Integral Formula For Derivatives

Formula for the Derivative:

Let U \subseteq \!\, C be an open set and let f be holomorphic on U. Then f \in \!\, C\infty \!\,(U). Moreover, if D(P, r) \subseteq \!\, U and z \in \!\, D(P, r), then

This formula is obtained simply by differentiating the standard Cauchy formula under the integral sign.


The Cauchy Estimates:

If f is a holomorphic on a region containing the closed disc and if  | f | ≤ M on then,

This is proved by direct estimation of the Cauchy Formula.


Entire Functions and Liouville’s Theorem:

A function f is said to be entire if it is defined and holomorphic on all of C, i.e., f : C → C is holomorphic. For instance, any holomorphic polynomial is entire, ez is entire, and sin z, cos z are entire. The function f(z) = 1/z is not entire because it is undefined at z = 0. [In a sense that we shall make precise later, this last function has a “singularity” at 0.] The question we wish to consider is: “Which entire functions are bounded?” This question has a very elegant and complete answer as follows:

Liouville’s Theorem A bounded entire function is constant.

Proof: Let f be entire and assume that | f(z) | ≤ M for all z \in \!\, C. Fix a P \in \!\, C and let r > 0. We apply the Cauchy estimate for k = 1 on . So

Since this inequality is true for every r > 0, we conclude that

Since P was arbitrary, we conclude that


The end of the last proof bears some commentary. We prove that df/dz ≡ 0. But we know, since f is holomorphic, that df/dz ≡ 0. It follows from linear algebra that df/dx  ≡ 0 and df/dy ≡ 0. Then calculus tells us that f is constant. The reasoning that establishes Liouville’s theorem can also be used to prove this more general fact: If f : C → C is an entire function and if for some real number C and some positive integer k, it holds that,

for all z, then f is a polynomial in z of degree at most k.