Complex Power Series Radius of Convergence Proof

In summary, if a power series f(z) = ∑ an(z-z0)^n has a radius of convergence R > 0 and if f(z) = 0 for all z, |z - z0| < r ≤ R, then it can be shown that a0 = a1 = ... = 0 by using the Open Mapping Theorem and the fact that all derivatives of f(z) are zero at z0. This implies that all coefficients an must be zero for the series to equal zero.
  • #1
jsi
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Homework Statement



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.

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 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!
 
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  • #2
Use Open Mapping Theorem.
 
  • #3
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
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
jsi said:
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?

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
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
jsi said:
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?

Yes, that's it. Except you want to evaluate all of the derivatives at z=z0. Not z=0.
 
  • #8
great, thank you!
 

1. What is the radius of convergence for a complex power series?

The radius of convergence for a complex power series is the distance from the center of the series to the nearest point where the series diverges. It is represented by the letter R and can be calculated using the ratio test or the root test.

2. How do you prove the radius of convergence for a complex power series?

The radius of convergence can be proven using the ratio test or the root test. These tests involve taking the limit of the absolute value of the terms in the series and comparing it to a known value, such as 1 or 0. If the limit is less than 1, the series converges, and the radius of convergence can be calculated. If the limit is greater than 1, the series diverges, and the radius of convergence is 0. If the limit is equal to 1, the test is inconclusive and another method may need to be used.

3. What is the significance of the radius of convergence in a complex power series?

The radius of convergence is important because it determines the interval of values for which the series will converge. If a value falls within the radius of convergence, the series will converge at that point. If a value falls outside the radius of convergence, the series will diverge at that point. This information is crucial in determining the validity and behavior of a complex power series.

4. Can the radius of convergence for a complex power series be negative?

No, the radius of convergence for a complex power series cannot be negative. The radius of convergence is a distance, and therefore it must be a positive value. However, it is possible for the radius of convergence to be infinite, meaning the series converges for all complex numbers.

5. How does the radius of convergence change when the center of the series is shifted?

When the center of the series is shifted, the radius of convergence remains the same. This is because the radius of convergence is determined by the behavior of the terms in the series, not the location of the center. However, the interval of convergence may change depending on the new center and the behavior of the series at that point.

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