Undergrad Why are analyticity and convergence related in complex analysis?

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Analyticity and convergence in complex analysis are closely linked, particularly through the behavior of power series. The Laurent Series' coefficients, denoted as a_n, are shown to be zero due to Cauchy's Theorem, highlighting their relationship with analyticity. The convergence of the power series is uniform within a closed ball inside its radius of convergence, which ensures differentiability. This uniform convergence allows the derivative of the series to pass through the limit, reinforcing the connection between analyticity and convergence. Understanding these concepts is crucial for grasping the implications of the Cauchy-Riemann conditions in complex analysis.
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From Proof of Residue Theorem in Mary Boas' Text
Hello,

I am currently reading about the Residue Theorem in complex analysis. As a part of the proof, Mary Boas' text states how the a_n series of the Laurent Series is zero by Cauchy's Theorem, since this part of the Series is analytic. This appears to then be related to convergence of the a_n series.

I suppose convergence and analyticity seem to go together on an intuitive level, but I am having a hard time with the details of why this is so, especially as to how it would relate back to the Cauchy-Riemann conditions for analyticity.

Can anyone offer further insights here for me?

Thanks!
dm4b
 
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Check out the epsilon-delta definitions of limit and convergence.
 
I'm not sure what you mean by the a_n series but if you mean the part with positive powers then the proof that the power series is differentiable starts with
- notice that on any closed ball strictly inside the radius of conference, the power series converges uniformly
- if ##f_n## converges to ##f## uniformly in a region, then the derivative passes through the limit, i.e. ##f_n'## converges to ##f'##, and in particular ##f## is differentiable.
- the partial sums are all polynomials so are differentiable.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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