Suppose \(f\) is analytic, defined on all of \(\CC\), and for each \(z_0 \in \CC\) there is at least one coefficient in the expansion \(f(z) = \sum_{n=0}^\infty c_n(z-z_0)^n\) is zero. Prove that \(f\) is a polynomial.
Hint: use the fact that \(c_n n! = f^{(n)}(z_0)\) and use a countability argument.
Solution
The claim is that there is some \(N\) such that \(\dd{^N f}{z^N} \equiv 0\), making \(f\) a polynomial. If no derivative is identically zero, the following set is countable: \[ \Union_{n\in \NN} \inverseof{\dd{^n f}{z^n}}(0) .\] This is because each derivative is analytic and the zeros of an analytic function are countable. But by hypothesis this set is uncountable.