Let $f$ $\in$ $\mathcal{H}$$(ann(a;R_1,R_2))$ and let $(a_n)$ the sequence defined by $$a_n=\frac{1}{2\pi i}\int_{|\zeta -a|=r_2}\frac{f(\zeta )}{(\zeta-a)^{n+1}}d\zeta$$ I need to show that for every $R_1<r_2<R_2$, the Laurent serie $\sum_{n=0}^{\infty}a_nz^n$ is normally convergent inside the closed disk $\bar{D}(a,r_2)$.
Here is what I've done. Let $\varepsilon>0$ small enough such that $R_1<r_2<r_2+\varepsilon<R_2$. Without loss of generality, we may consider the case $a=0$. Let $z$ such that $R_1\leq|z|\leq r_2$, so $|z| \leq r_2<r_2+\varepsilon=|\zeta|$ for every $\zeta$ on the circle of radius $r_2+\varepsilon$. This implies that $$\frac{1}{\zeta-z}=\frac{1}{\zeta}\frac{1}{1-\frac{z}{\zeta}} = \sum_{n=0}^{\infty}\frac{z^n}{\zeta^{n+1}}$$ which is normally convergent with respect of $\zeta$. So, we get
$$\sum_{n=0}^{\infty}(\frac{1}{2\pi i}\int_{|\zeta|=r_2+\varepsilon}\frac{f(\zeta)}{\zeta^{n+1}}d\zeta)z^n$$ Now, we need to see that this series is indeed normally convergent inside the closed disk $\bar{D}(0,r_2)$. We'll use the Abel criteria to establish the normal convergence. So, we have
\begin{align} |a_nz^n| & =|a_n||z|^n \\ & \leq |\frac{1}{2\pi i}\int_{|\zeta|=r_2+\varepsilon}\frac{f(\zeta)}{\zeta^{n+1}}| r_2^n \end{align} Using the ML-inequality, we get that \begin{align} |\frac{1}{2\pi i}\int_{|\zeta|=r_2+\varepsilon}\frac{f(\zeta)}{\zeta^{n+1}}| r_2^n & \leq \frac{r_2^n}{2\pi} \sup_{t\in[0,2\pi]}\left | \frac{f((r_2+\varepsilon)e^{it})(r_2+\varepsilon)e^{it}}{(r_2+\varepsilon)^{n+1}e^{i(n+1)t}} \right |2\pi(r_2+\varepsilon) \\ & \leq \frac{r_2^n}{2\pi} \sup_{t\in[0,2\pi]}\left | f((r_2+\varepsilon)e^{it})(r_2+\varepsilon) \right |\frac{2\pi(r_2+\varepsilon)}{(r_2+\varepsilon)^{n+1}} \\ & \leq \frac{r_2^n}{2\pi} \max_{|\zeta|=r_2+\varepsilon}|f(\zeta)| \frac{2\pi(r_2+\varepsilon)}{(r_2+\varepsilon)^{n+1}} \\ & = \max_{|\zeta|=r_2+\varepsilon}|f(\zeta)| \left (\frac{r_2}{(r_2+\varepsilon)} \right )^n \end{align}
And since $\frac{r_2}{(r_2+\varepsilon)}<1$, by the Abel criteria, the series is normally convergent.
My question is the follow, is my proof correct? I don't know if during the last steps involving the sup, I can really do what I did. Since $|e^{it}|=1$ for all $t\in[0,2\pi]$, we can simplify and so on.
Thanks for the reading, hope you can help me!