Radius of Convergence of Geometric Progression in Complex Plane

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Theorem

Consider the complex power series:

$S = \displaystyle \sum_{k \mathop = 0}^\infty z^n$

The radius of convergence $S$ is $1$.


Proof

\(\displaystyle \lim_{n \mathop \to \infty} \dfrac {\cmod {z^n} } {\cmod {z^{n - 1} } }\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {z^n} {z^{n - 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \cmod z\)
\(\displaystyle \) \(=\) \(\displaystyle \cmod z\)

By the Ratio Test, it follows that:

$S$ is convergent for $\cmod z < 1$
$S$ is divergent for $\cmod z > 1$.

Hence the result by definition of radius of convergence.

$\blacksquare$


Sources