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 $\displaystyle f \left({z}\right) = \sum_{n \mathop =0}^\infty a_n \left({z - \xi}\right)^n$ 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 $z \in \C$ with $\left \vert{z - \xi}\right \vert < R$.

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

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

From Radius of Convergence of Derivative of Complex Power Series, it follows that $g$ has radius of convergence $R$.

Fix an $\epsilon > 0$ satisfying $\epsilon < R - \left\vert{ z-\xi }\right\vert$.

Define:
 * $\displaystyle M=\left({ R - \epsilon - \left\vert{ z-\xi }\right\vert }\right)^{-2} \sum_{n \mathop =2}^\infty \left\vert{ a_n}\right\vert \left({R - \epsilon}\right)^n$

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 $f' \left({z}\right) = g \left({z}\right)$, as desired.

Remark
The proof for real power series is identical.