prove that a function is expressible as a power series 
I started by rearrange f(z), and expanded the terms in summation. 

Then, I did not get very far. It would be great if anyone can let me know what is needed to figure out bn. Thanks in advance. 
 A: You were kinda in the right track, I'll leave it as steps so you can prove it yourselv (fill the gaps), its not an easy theorem.
$1)$ Note that (as you already did): $$\sum_{n = 0}^{\infty} a_n z^n = \sum_{n = 0}^{\infty} a_n (z - z_0 + z_0)^n$$
$$= \sum_{n = 0}^{\infty} a_n \sum_{k = 0}^{n} \binom{n}{k} {z_{0}}^{n-k}(z-z_0)^n$$
$$= \sum_{n = 0}^{\infty} a_n \sum_{k = 0}^{\infty} \binom{n}{k} {z_{0}}^{n-k}(z-z_0)^n $$
Here the pass from the second to third equal is trivial because remember what happens to the binomial if $k > n$.
$2)$ If $c_n = a_n \sum_{k = 0}^{\infty} \binom{n}{k} {z_{0}}^{n-k}(z-z_0)^n$ prove that $\sum c_n$ is absolutely convergent. Hint (apply absolute value to $c_n$ and go backwards with the binomial theorem, then apply your hypothesis on the radius of convergence).
$3)$ Write each $c_n$ for each $n = 0,1,\dots$ as an infinite matrix where each $c_n$ is a row. 
$4)$ Because we proved the convergence is absolute we can rearrange the series, now every colum is the sum of some value (remember the $a_n$ and $z_0$ are fixed) multiplied by a power of $(z-z_0)$
Concluded that if you sum the colums and then the row that is changing the sums on $$\sum_{n = 0}^{\infty} a_n \sum_{k = 0}^{\infty} \binom{n}{k} {z_{0}}^{n-k}(z-z_0)^n $$
You'll get 
$$ \sum_{k = 0}^{\infty} \bigg ( \sum_{n = k}^{\infty} a_n\binom{n}{k} {z_{0}}^{n-k}\bigg)(z-z_0)^n $$
So $b_k =  \sum_{n = k}^{\infty} a_n\binom{n}{k} {z_{0}}^{n-k}$
