Derivative of Complex Power Series/Proof 1

Theorem
Let $\xi \in \C$ be a complex number.

Let $\langle a_n\rangle$ be a sequence in $\C$.

Let $\displaystyle f \left({z}\right) = \sum_{n=0}^\infty a_n \left({z - \xi}\right)^n$ be a power series in a complex variable $z \in \C$ about $\xi$.

Let $R$ be the radius of convergence of the series defining $f \left({z}\right)$.

Let $\left \vert{z - \xi}\right \vert < R$.

Then:
 * $\displaystyle f’ \left({z}\right) = \sum_{n=1}^\infty n a_n \left({z - \xi}\right)^{n-1}$

Proof
Define:
 * $\displaystyle g\left(z\right)=\sum_{n=1}^\infty na_n\left(z-\xi\right)^{n-1}$

Fix an $\epsilon>0$ satisfying $\epsilon<R-\left\vert z-\xi\right\vert$. Define:
 * $\displaystyle M=\left(R-\epsilon-\left\vert z-\xi\right\vert\right)^{-2}\sum_{n=2}^\infty\left\vert a_n\right\vert\left(R-\epsilon\right)^n$

Suppose that $\left\vert h\right\vert\le R-\epsilon-\left\vert z-\xi\right\vert$. Then, by the binomial theorem and the triangle inequality:

Letting $h\to 0$ gives $f'(z)=g(z)$, as desired.

Remark
The proof for real power series is identical.