This problem asks me to:
"Making the formal substitution $z-a=(z-z_0)+(z_0-a)$' in the power series $\sum_{i=0}^{\infty}A_n(z-a)^n$ and gathering like terms, obtain a series $\sum_{n=0}^{\infty}C_n(z-z_0)^n$ and expressions for its coefficients in terms of $A_k$ and and $(z_0-a)^k,k=0,1\cdots $"
Here is my attempt at approaching it:
$$\sum_{n=0}^{\infty}A_n(z-a)^n=\sum_{n=0}^{\infty}A_n[(z-z_0)+(z_0-a)]^n=\sum_{n=0}^{\infty}A_n\sum_{i=0}^n \binom{n}{i}(z_0-a)^{n-i}(z-z_0)^i$$
I have no clue of how to proceed after that, but it seems like that I need to gather the terms in each power.
Edit:
Following the hint of @angryavian:
$$I=\sum_{i=0}^\infty A_n(z-a)^n$$$$=\sum_{n=0}^\infty A_n\sum_{i=0}^n\binom{n}{i}(z_0-a)^{n-i}(z-z_0)^i$$$$=\sum_{i=0}^\infty A_n\sum_{i=0}^\infty\binom{n}{i}(z_0-a)^{n-i}(z-z_0)^i$$$$=\sum_{n=0}^{\infty}\sum_{i=0}^{\infty}A_n\binom{n}{i}(z_0-a)^{n-i}(z-z_0)^i$$$$=\sum_{i=0}^{\infty}\sum_{n=0}^{\infty}A_n\binom{n}{i}(z_0-a)^{n-i}(z-z_0)^i$$
$$\implies I=\sum_{i=0}^{\infty}C_i(z-z_0)^i,C_i=\sum_{n=0}^\infty A_n\binom{n}{i}(z_0-a)^{n-i}$$