Existence of Radius of Convergence of Complex Power Series/Divergence

Theorem
Let $\xi \in \C$.

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

Let $R$ be the radius of convergence of $S \left({z}\right)$.

Let ${B_R}^- \paren \xi$ denote the closed $R$-ball of $\xi$.

Let $z \notin {B_R}^- \paren \xi$.

Then $S \paren z$ is divergent.

Proof
Let $z \notin {B_R}^- \paren \xi$.

Then by definition of the closed $R$-ball of $\xi$:
 * $\cmod {z - \xi} > R$

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

By definition of radius of convergence, there exists $w \in \C$ such that:
 * $\cmod {w - \xi} < \cmod {z - \xi}$

and $S \paren w$ is divergent.

Then:

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

Also see

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