Rearranging a power series centered in $z_0$ to a power series centered in 0: is that possible?

Question

Suppose we have this power series $S(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n$ that converges $\forall z \in D(z_0,r)$, $r$ being the radius of convergence. Is that possible to transform $S(z)$ to $T(z) = \sum_{n=0}^{\infty}b_nz^n$? Under which conditions the previous question makes sense (absolute convergence of terms - made of their own power series)? What can we say about "$S(z)=T(z)$"? How does the radius of convergence change? The center of the power really changes?

EDIT: I mean something like this:

$$ S(z) = \sum_{n \geq 0} a_n (z-z_0)^n = \sum_{n \geq 0} a_n \sum_{k=0}^n \binom{n}{k} z^k (-z_0)^{n-k} = \sum_{n \geq 0} \sum_{k=0}^n a_n \binom{n}{k} z^k (-z_0)^{n-k} = \\ \sum_{n \geq 0} \left[ \sum_{k \geq n} a_k \binom{k}{n} (-z_0)^{k-n} \right]z^n = \sum_{n \geq 0} b_n z^n = T(z) $$ Then we get: $$ \sigma(S)^{-1} = \overline{lim} \sqrt[n]{|a_n|} = \overline{lim} \sqrt[n]{|b_n|} = \overline{lim} \sqrt[n]{\left| \sum_{k \geq n} a_k \binom{k}{n} (-z_0)^{k-n} \right|} = \sigma(T)^{-1} $$

EDIT2: I decided to look the power series as a limit for the polynomial series then I realize the first idea makes no sense. We can't define $b_n$ by the way I thought in EDIT1. Here an example:

$$ |z| < 1:\frac{1}{1-z} = \sum_{n \geq 0}z^n = S(z) \\ \\ |z-1|<1:\frac{1}{z} = \frac{1}{1-(1-z)} = \lim_{m \to \infty} \sum_{n = 0}^m(1-z)^n = \lim_{m \to \infty} \sum_{n = 0}^m \sum_{k=0}^n \binom{n}{k}1^{n-k}(-z)^k = \\ = \lim_{m \to \infty} \sum_{n = 0}^m \sum_{k=0}^n \binom{n}{k}(-1)^kz^k = \lim_{m \to \infty} \sum_{n = 0}^m \left[ \sum_{k = n}^m \binom{k}{n}(-1)^n \right]z^n =^{wrong} \sum_{n \geq 0} b_nz^n = T(z) $$

Answer 1

You know that $S(z)$ is holomorphic in the region $D(z_0,r) = \{ z \in \mathbb{C} \mid \lvert z - z_0 \rvert < r \}$. Therefore $S(z)$ can be expanded as a convergent power series at any $w \in D(z_0,r)$. If $\lvert z_0 \rvert < r$, this applies for $w = 0$. Write $T(z) = \Sigma_{n=0}^\infty b_n z^n$. Then you have $b_n = \frac{1}{n!}S^{(n)}(0)$ which gives the same result as in your question. For the radius $r_T$ of convergence of $T(z)$ we get $r_T \ge r - \lvert z_0 \rvert$. Note that $r_T > r - \lvert z_0 \rvert$ is possible. Clearly $S(z) = T(z)$ for $z \in D(z_0,r) \cap D(0,r_T)$.

But this is not the complete story. If $f : U \to \mathbb{C}$ is holomorphic (where $U \subset \mathbb{C}$ is open), then $f$ can be expanded as a convergent power series at any $w \in U$. For the radius $r(f,w)$ of convergence we get

$$r(f,w) \ge sup \{R \mid D(w,R) \subset U \} .$$

See e.g. Radius of convergence of Taylor series of holomorphic function.

This means that if $0, z_0 \in U$ and $r(f,0) + r(f,z_0) > \lvert z_0 \rvert$, and if $S(z)$ is the power series for $f$ at $z_0$ and $T(z)$ the power series for $f$ at $0$, then $S(z) = T(z)$ for $z \in D(z_0,r(f,z_0)) \cap D(0,r(f,0)) \ne \emptyset$. However, the coefficients $b_n$ of $T(z)$ cannot be expressed via the coefficients $a_n$ of $S(z)$ if $r(f,z_0) \le \lvert z_0 \rvert$.