Complex Power Series Radius of Convergence Proof

[jsi]

Homework Statement

If f(z) = \sum a_n(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.

Homework Equations

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 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 a_i would have to equal 0? Any help would be appreciated, thanks!

 

[Some Pig]

Use Open Mapping Theorem.

 

[Dick]

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.

 

[jsi]

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 - z₀)ⁿ) = (na_n(z - z₀)^n-1) = (na_n(z - z₀)ⁿ)/(z - z₀)¹ and then that means (z - z₀) can't = 0 so the a_i must be 0? Is that right, or sort of right, or totally wrong?

 

[Dick]

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 - z₀)ⁿ) = (na_n(z - z₀)^n-1) = (na_n(z - z₀)ⁿ)/(z - z₀)¹ and then that means (z - z₀) can't = 0 so the a_i must be 0? Is that right, or sort of right, or totally wrong?

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.

 

[jsi]

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 a_n = 0, so all a_n must be 0?

 

[Dick]

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 a_n = 0, so all a_n must be 0?

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

 

[jsi]

great, thank you!