Existence of Radius of Convergence of Complex Power Series/Absolute Convergence

Theorem
Let $\xi \in \C$.

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

Let $R$ be the radius of convergence of $S \paren z$.

Let $B_R \paren \xi$ denote the open $R$-ball of $\xi$.

Let $z \in B_R \paren \xi$.

Then $S \paren z$ converges absolutely.

If $R = +\infty$, we define $B_R \paren \xi = \C$.

Proof
Let $z \in B_R \paren \xi$.

By definition of the open $R$-ball of $\xi$:
 * $\cmod {z - \xi} < R$

where $\cmod z$ denotes the complex modulus of $z$.

By definition of radius of convergence, it follows that $S \paren z$ converges.

Suppose $R$ is finite.

Let $\epsilon = R - \cmod {z - \xi} > 0$.

Now, let $w \in B_R \paren \xi$ be a complex number such that $R - \cmod {w - \xi} = \dfrac \epsilon 2$.

Such a $w$ exists because, if at a loss, we can always let $w = \xi + R - \dfrac \epsilon 2$.

If $R = +\infty$, then let $w \in C$ be any complex number such that $\cmod {z - \xi} < \cmod {w - \xi}$.

Then:

From the $n$th Root Test, it follows that $S \paren z$ converges absolutely.

Also see

 * Existence of Interval of Convergence of Power Series for a proof of the same result in real numbers.