# Derivative of Complex Power Series/Proof 2

## 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

### Lemma

$\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$

$\Box$

Define:

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

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

Let:

$\displaystyle M = \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2}$

We use the Root Test to prove convergence of this series:

 $\ds \limsup_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2} }^{1/n}$ $=$ $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \lim_{n \mathop \to \infty} \paren {\paren {R - \epsilon}^{n - 2} }^{1/n} \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$ $\ds$ $=$ $\ds \dfrac {R - \epsilon} R$

The last equality follows from the lemma and:

$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$

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

It follows by the Triangle Inequality that:

$\cmod {z - \xi + h} \le \cmod {z - \xi} + \size h \le R - \epsilon$

By the Triangle Inequality, Difference of Two Powers, and Closed Form for Triangular Numbers, the following holds:

 $\ds \cmod {\dfrac {\map f {z + h} - \map f z} h - \map g z}$ $=$ $\ds \cmod {\sum_{n \mathop = 2}^\infty a_n \paren {\dfrac {\paren {z - \xi + h}^n - \paren {z - \xi}^n} h - n \paren {z - \xi}^{n - 1} } }$ $\ds$ $\le$ $\ds \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\dfrac {\paren {z - \xi + h}^n - \paren {z - \xi}^n} h - n \paren {z - \xi}^{n - 1} }$ Triangle Inequality $\ds$ $=$ $\ds \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\paren {\sum_{u \mathop = 0}^{n - 1} \paren {z - \xi + h}^u \paren {z - \xi}^{n - u - 1} } - n \paren {z - \xi}^{n - 1} }$ Difference of Two Powers $\ds$ $=$ $\ds \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\sum_{u \mathop = 0}^{n - 1} \paren {\paren {z - \xi + h}^u \paren {z - \xi}^{n - u - 1} - \paren {z - \xi}^{n - 1} } }$ $\ds$ $=$ $\ds \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {\sum_{u \mathop = 0}^{n - 1} \paren {z - \xi}^{n - u - 1} \paren {\paren {z - \xi + h}^u - \paren {z - \xi}^u } }$ $\ds$ $=$ $\ds \sum_{n \mathop = 2}^\infty \cmod {a_n} \cmod {h \sum_{u \mathop = 0}^{n - 1} \paren {z - \xi}^{n - u - 1} \sum_{v \mathop = 0}^{u - 1} \paren {z - \xi + h}^v \paren {z - \xi}^{u - v - 1} }$ Difference of Two Powers $\ds$ $\le$ $\ds \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{u \mathop = 0}^{n - 1} \cmod {z - \xi}^{n - u - 1} \sum_{v \mathop = 0}^{u - 1} \cmod {z - \xi + h}^v \cmod {z - \xi}^{u - v - 1}$ triangle inequality $\ds$ $\le$ $\ds \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \sum_{u \mathop = 0}^{n - 1} \paren {R - \epsilon}^{n - u - 1} \sum_{v \mathop = 0}^{u - 1} \paren {R - \epsilon}^v \paren {R - \epsilon}^{u - v - 1}$ by assumption and $\cmod {z - \xi + h} \le R - \epsilon$, which was proven earlier $\ds$ $=$ $\ds \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^{n - 2} \sum_{u \mathop = 0}^{n - 1} \sum_{v \mathop = 0}^{u - 1} 1$ $\ds$ $=$ $\ds \size h \sum_{n \mathop = 2}^\infty \cmod {a_n} \paren {R - \epsilon}^{n - 2} \sum_{u \mathop = 0}^{n - 1} u$ $\ds$ $=$ $\ds \size h \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \paren {R - \epsilon}^{n - 2}$ by Closed Form for Triangular Numbers $\ds$ $=$ $\ds M \size h$

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.