Series Expansion of Holomorphic Functions

Taylor Expansion

Proposition

Let and suppose there is a power series converges on to , then .

Proof By Abel’s theorem, the power series converges absolutely and normally on , so we can differentiate term by term to get Evaluating at yields Therefore .

The above proposition shows that the power series representation of a holomorphic function is unique if it exists. Now let’s consider the existence:

Proposition

Let . Then there exists a power series centered at converging to on .

Proof First consider the case . Pick , then . Fix some arbitrary . By Cauchy integral formula we have For , we have It follows that Since is arbitrary, the power series converges to on . As is also arbitrary, the power series converges to on . Now for the general case , we can consider the function , which is holomorphic on , and apply the above argument to get the desired result.

Proposition

Suppose the Taylor series of centered at has radius of convergence , then is the largest disk to which has a holomorphic extension.

e.g.

• What is the radius of convergence of the Taylor series of centered at ? The disk of convergence is the largest disk centered at such that defined on some has a holomorphic extension to. As the function blows up at , the radius of convergence is . Now consider the real function , which is smooth and at every point its Taylor series converges to . However, it has a radius of convergence of . From the real number perspective, there do not appear to be any “problems” with the functions, but the radius of convergence knows about the distance to the nearest “problem point” for the function in the complex plane.
• Extend the above function to a more general case , where is a polynomial with real coefficients, such that all roots lie in . Then the function is smooth on , and at every point its Taylor series converges to . However, the radius of convergence of the Taylor series is the distance to the nearest root of in the complex plane, i.e. .

The Uniqueness Theorem

Uniqueness Theorem

Let be a domain, and let . Suppose there exists a sequence of distinct points that converges to a limit such that for all . Then .

Proof Without loss of generality, we may assume that (i.e. replace by ). Let . We will show that . Since is connected, by theorem, it suffices to show that is non-empty, open, and closed in .

1. is non-empty: Suppose . Assume that . Then we denote the smallest such that as . Choose some such that . For we have a normally convergent power series representation: The power series converges pointwise on as for any fixed , is constant. Therefore it converges normally to a holomorphic function with , and we have Since are distinct and converge to , infinitely many are not equal to , and lie in , we denote them as . It follows that for any It implies that for all , so yielding a contradiction. Therefore , is non-empty.
2. is open: Fix , then there exists a disk . Since is holomorphic, the Taylor series of centered at converges to on : Therefore for all , so for all . Hence , is open.
3. is closed: Observe that Since is continuous, the preimage of a closed set is closed, so is closed.

We conclude that , so .

e.g.

• In fact, the complex exponential and trigonometric functions are the unique holomorphic extension of the corresponding real functions.
• If , then by definition there exists an open set containing and a holomorphic function such that . By the uniqueness theorem, one can pick a convergent sequence in , therefore is unique for a given .

Liouville’s Theorem

If is bounded, then it is constant. In general, if there exists positive such that for all , then is a polynomial of degree at most .