Complex Analysis/Global theory of holomorphic functions
Proof 1: First note that
Let and . Then Cauchy's integral formula and the triangle inequality for integrals together imply that
for a certain . The latter expression may be computed explicitly; it equals
which tends to zero as . Hence, vanishes and is a polynomial of degree .
Proof: Let be any point. Since holomorphic functions are analytic, the function posesses a power series expansion
which converges on a sufficiently small neighbourhood of .
Suppose first that is a cluster point of the set .
Let be the least natural number such that .
Proof: Since , we may pick the following subset of :
where is sufficiently small. Since the restriction of a holomorphic function is holomorphic, is holomorphic on . Moreover,
- Use Liouville's theorem to demonstrate that every non-constant polynomial has at least one root in (Hint: Consider the function ).
- In this exercise, we want to look at the simplest sufficient conditions for the possibility of extending a function given by a real power series to a function on the complex plane.
- Let be a power series with real coefficients which converges absolutely on an open neighbourhood of the origin of . Prove that may be extended to a function on an open neighbourhood of the origin of the complex plane.
- Let be a power series such that for all is real and positive. Suppose further that converges for all st. , where is a real number. Prove that may be extended to a holomorphic function on .
- Prove that the extensions considered in the first two sub-exercises are unique.
- Let be an entire function and let , and such that . Prove that is a polynomial of degree .