# Derivative of Complex Power Series/Proof 1

## Theorem

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

Let $\sequence {a_n}$ be a sequence in $\C$.

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

Let $R$ be the radius of convergence of the series defining $f \paren z$.

Let $\cmod {z - \xi} < R$.

Then $f$ is complex-differentiable and its derivative is:

- $\displaystyle f' \paren z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$

## 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:

\(\displaystyle \cmod {\dfrac {\map f {z + h} - \map f z} h - \map g z}\) | \(=\) | \(\displaystyle \cmod {\dfrac 1 h \paren {\sum_{n \mathop = 0}^\infty a_n \paren {z + h - \xi}^n - \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n} - \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \cmod {\sum_{n \mathop = 0}^\infty a_n \sum_{k \mathop = 0}^n \binom n k \paren {z - \xi}^k h^{n - k - 1} - \sum_{n \mathop = 0}^\infty \dfrac {a_n} h \paren {z - \xi}^n - \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1} }\) | Binomial Theorem | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \cmod {\sum_{n \mathop = 1}^\infty a_n \paren {\sum_{k \mathop = 0}^n \binom n k \paren {z - \xi}^k h^{n - k - 1} - \dfrac 1 h \paren {z - \xi}^n - n \paren {z - \xi}^{n - 1} } }\) | Difference of Absolutely Convergent Series | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \cmod {\sum_{n \mathop = 2}^\infty a_n \sum_{k \mathop = 0}^{n - 2} \binom n k \paren {z - \xi}^k h^{n - k - 1} }\) | by algebraic manipulations | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \cmod h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{k \mathop = 0}^{n - 2} \binom n k \cmod {z - \xi}^k \cmod h^{n-k-2}\) | Triangle Inequality | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \cmod h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{k \mathop = 0}^{n - 2} \binom n k \cmod {z - \xi}^k \paren {R - \epsilon - \cmod {z - \xi} }^{n - k - 2}\) | by assumption | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \cmod h \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{k \mathop = 0}^n \binom n k \cmod {z - \xi}^k \paren {R - \epsilon - \cmod {z - \xi} }^{n - k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \cmod h \paren {R - \epsilon - \cmod {z - \xi} }^{-2} \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^n\) | by the binomial theorem | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle M \cmod h\) |

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

$\blacksquare$

## Also see

- The proof for real power series is identical.