I guess the next problem should be addressed somehow with Laurent series:
Let $0<r<s<\infty$, $A=A_{r,s}(0)$ and $f\in\mathcal{O}(A)$. Suppose that $\underset{n\rightarrow\infty}{lim}f(z_n)=0$ either for every sequence $\{z_n\}\subset A$ with $\vert z_n\vert \rightarrow r$ or for $\{z_n\}\subset A$ with $\vert z_n\vert \rightarrow s$. Show that $f(z)=0$ for all $z\in A$. Investigate weather this conclusion remains valid in either limiting case $r=0$ or $s=\infty$.
I thought that I can write f as a power series and then, from knowing it goes to 0 for all sequences in the form appears above I can say that its converging in $B_r(0)$ but I didnt succeed to get through with it
Juest extend $f$ to $\overline{A_{r,s}}$ with $f(z)=0$ when $\lvert z\rvert=r$ or $\lvert z\rvert=s$. This extension is continuous and $\overline{A_{r,s}}$ is compact. So, the absolute value of the extension has a maximum on $\overline{A_{r,s}}$. Now, apply the maximum principle.