Complex Power Series Radius of Convergence Proof

1. jsi

24
1. The problem statement, all variables and given/known data

If f(z) = $\sum$ an(z-z0)n has radius of convergence R > 0 and if f(z) = 0 for all z, |z - z0| < r ≤ R, show that a0 = a1 = ... = 0.

2. Relevant equations

3. The attempt at a solution

I know it is a power series and because R is positive I know it converges. And if f(z) = 0 then the sum itself would be 0 which would mean that each term must add to 0. This doesn't necessarliy imply that each coefficient ai is 0 though because they could alternate a1 = 1, a2 = -1, a3 = 2, a4 = -2 etc... Am I right in thinking this? And if so, I'm not sure how to go from there... Unless f(z) = 0 for all z implies that |z - z0| ≠ 0 somehow then for f(z) to equal 0, the ai would have to equal 0? Any help would be appreciated, thanks!

2. Some Pig

38
Use Open Mapping Theorem.

3. Dick

25,853
It's pretty easy to see that all of the derivatives of f(z) are zero at z0, isn't it? Try to use derivatives to say something about the coefficients.

4. jsi

24
oh ok, so if say, since f(z) = 0, d/dz = 0, then taking the derivative of every term of the series yields d/dz (an(z - z0)n) = (nan(z - z0)n-1) = (nan(z - z0)n)/(z - z0)1 and then that means (z - z0) can't = 0 so the ai must be 0? Is that right, or sort of right, or totally wrong?

5. Dick

25,853
Take it one step at a time. The series is f(z)=a0+a1(z-z0)+a2(z-z0)^2+... Since f(z0)=0 that means a0=0. f'(z)=a1+2*a2(z-z0)+... What does f'(z0)=0 tell you? Now use the second derivative to say something about a2. Etc.

6. jsi

24
I think I got it; f'(0) = 0 implies that a1 = 0 and f''(0) = 0 implies that a2 = 0, so f to the nth derivative of 0 = 0 implies that an = 0, so all an must be 0?

7. Dick

25,853
Yes, that's it. Except you want to evaluate all of the derivatives at z=z0. Not z=0.

8. jsi

24
great, thank you!