Complex Power Series Radius of Convergence Proof
1. The problem statement, all variables and given/known data
If f(z) = [itex]\sum[/itex] a_{n}(zz_{0})^{n} has radius of convergence R > 0 and if f(z) = 0 for all z, z  z_{0} < r ≤ R, show that a_{0} = a_{1} = ... = 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 a_{i} is 0 though because they could alternate a_{1} = 1, a_{2} = 1, a_{3} = 2, a_{4} = 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  z_{0} ≠ 0 somehow then for f(z) to equal 0, the a_{i} would have to equal 0? Any help would be appreciated, thanks! 
Re: Complex Power Series Radius of Convergence Proof
Use Open Mapping Theorem.

Re: Complex Power Series Radius of Convergence Proof
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.

Re: Complex Power Series Radius of Convergence Proof
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 (a_{n}(z  z_{0})^{n}) = (na_{n}(z  z_{0})^{n1}) = (na_{n}(z  z_{0})^{n})/(z  z_{0})^{1} and then that means (z  z_{0}) can't = 0 so the a_{i} must be 0? Is that right, or sort of right, or totally wrong?

Re: Complex Power Series Radius of Convergence Proof
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Re: Complex Power Series Radius of Convergence Proof
I think I got it; f'(0) = 0 implies that a_{1} = 0 and f''(0) = 0 implies that a_{2} = 0, so f to the nth derivative of 0 = 0 implies that a_{n} = 0, so all a_{n} must be 0?

Re: Complex Power Series Radius of Convergence Proof
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Re: Complex Power Series Radius of Convergence Proof
great, thank you!

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