# Complex Analysis Proof Review

1. Dec 12, 2011

### Poopsilon

Let D ⊂ ℂ be a domain and let f be analytic on D. Show that if there is an a ∈ D such that the kth derivative of f at a is zero for k=n, n+1, n+2,..., then f is a polynomial with degree at most n.

So I believe I have a proof, but the theorems are so powerful I feel like I might be overlooking something, here it is:

Basically f being analytic at a means it can be expanded as a convergent power-series in some open neighborhood around a. Now since the kth derivative and beyond is zero at a, we know this power-series is just at most an nth degree polynomial, which coincides with the values of f for all z in our neighborhood.

Now there is a theorem in my book which says that if g:D→ℂ is an analytic function on a domain D which is not identically zero, then the set of zeros of g is discrete in D. By discrete we mean that the set of zeros does not contain an accumulation point.

Thus since the kth derivative and beyond of f is analytic on D and are zero on an open neighborhood of D, and since this neighborhood is open in ℂ, then it clearly contains accumulation points, and thus the kth derivative and beyond of f must be zero on all of D in order not to contradict the theorem given above.

Therefore f is a polynomial with degree at most n.

Thanks =].