Help to understand this step of the proof of holomorphic functions are analytic

Question

I'm trying to understand the proof of the analyticity of holomorphic functions.

The step I don't understand is when one interchanges the series and the integral. In every source I have read, it says because the series converges uniformly.

My question is: uniformly where?

1. Over the compact closed curve (the circunference)? (the book Analyse complexe pour la Licence 3 by Patrice Tauvel pag 77 seems to say this)
2. Or over the disk that is the interior of this curve? (the book Basic Complex Analysis of One Variable by Anant Shastri pag 177 seems to say this)
3. Maybe it has to be uniform on the whole closed disk?

What is necesary and what is sufficient?

Thanks!

Edit: I'll add some info to clarify my confusion. The step I habe doubts is when one change this

$ \frac{1}{2 \pi i} \int_{|w-z_0| = r} \sum_{n=0}^{\infty} \underbrace{ \left( \frac{z-z_0}{w-z_0} \right)^n \frac{f(w)}{w-z_0} }_{g(w,z,z_0,n)} dw$

into this

$ \frac{1}{2 \pi i} \sum_{n=0}^{\infty} \int_{|w-z_0| = r} \left( \frac{z-z_0}{w-z_0} \right)^n \frac{f(w)}{w-z_0} dw$

I'm supposed to look that the series $\sum_n g(w,z,z_0,n)$ converge uniformly. But uniform convergene over which set? Even over which variable? Over all $w \in C$? Over all $z \in B(z_0,r)$?

Answer 1

Your question is why we can interchange the integral and the sum in this step? $$ \sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,\mathrm{d}w $$

This follows from the dominated convergence theorem

In the proof later you have your majorant independent from $n$, as required.