Derivative of Complex Power Series/Proof 1

Proof
Define:
 * $\displaystyle \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

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

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

Define:
 * $\displaystyle M = \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^n$

Suppose that $\cmod h \le R - \epsilon - \cmod {z - \xi}$.

Then, by the Binomial Theorem and the Triangle Inequality:

Letting $h\to 0$ we see that $\map {f'} z$ exists and $\map {f'} z = \map g z$, as desired.

Also see

 * The proof for real power series is identical.