# Power series at $z=0$ implies power series at any $z$ in the disc of convergence

This should be a very simple exercise requiring some algebraic manipulation, but I can not seem to get it right! Its statement is the following: If $f$ ($\mathbb{C}\to\mathbb{C}$) is a power series centred at the origin $z=0$, then $f$ has a power series expansion around any point of its disc of convergence.

So let $z_0$ be a number in the disc of convergence of $f$. Then $$f(z)=f(z-z_0+z_0)=\sum_{n=0}^\infty a_n(z-z_0+z_0)^n = \sum_{n=0}^\infty a_n \sum_{k=0}^n {n\choose k}(z-z_0)^{n-k}z_0^k.$$ Now I should rearrange this to get $(z-z_0)^n$ and some new coefficients $b_k$ not depending on $z$ and I don't know how to do it!

Any help will be appreciated!

• Just combine all the coefficients of $(z-z_0)^m$ coming from all $(n,k)$ such that $n-k=m$. But that doesn't solve the problem: the hard part of this problem is proving that the resulting power series actually converges to $f$ in a neighborhood of $z_0$. Sep 1, 2017 at 18:42
• Possible duplicate: math.stackexchange.com/questions/2188444/… Sep 1, 2017 at 18:48

• Throw \$around your math, like \$z_0\$which comes out as$z_0\$. Oct 26, 2018 at 8:39