Complex Power Series Radius of Convergence Proof

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

    If f(z) = [itex]\sum[/itex] 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. jcsd
  3. Use Open Mapping Theorem.
     
  4. Dick

    Dick 25,738
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    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.
     
  5. 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?
     
  6. Dick

    Dick 25,738
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    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.
     
  7. 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?
     
  8. Dick

    Dick 25,738
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    Yes, that's it. Except you want to evaluate all of the derivatives at z=z0. Not z=0.
     
  9. great, thank you!
     
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