Assume $f$ is analytic on the annulus $R_1<|z-a|<R_2$. Assume $R_1<r<|z-a|$. Define $f_2$ by
$$f_2(z)=\frac1{2\pi i}\int_{|x-a|=r}\frac{f(x)dx}{x-z}$$
$f_2$ is analytic on $|z-a|>r$. To find Taylor’s expansion consider the change of variables $x=a+\frac1{x'}$, $z=a+\frac1{z'}$. This transformation reduces $|x-a|=r$ to $|x'|=\frac1r$ with a negative orientation. We can easily compute
$$f_2(a+\frac1{z'})=\frac1{2\pi i}\int_{|x'|=\frac1r}\frac{z'}{x'}\frac{f(a+\frac1{x'})dx'}{x'-z'}=\sum_{n=1}^\infty B_nz'^n$$
Here $B_n$ are obviously the coefficients in Maclaurin's expansion of $f_2(a+\frac1{z'})$, are they not? How does the above expression justify this from of the expansion?
Why is $B_0=0$? How can we see that $$B_n=\frac1{2\pi i}\int_{|x'|=\frac1r}\frac{f(a+\frac1{x'})dx'}{x'^{n+1}}$$
EDIT: On the way to Taylor's formula, I saw a lemma that allows us to differentiate functions like $f_2$ under the sign of the integral, but I do not see that it helps.