# Sine Function is Absolutely Convergent/Complex Case

## Theorem

The complex sine function $\sin: \C \to \C$ is absolutely convergent.

## Proof 1

The definition of the complex sine function is:

$\ds \forall z \in \C: \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$

By definition of absolutely convergent complex series, we must show that the power series

$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }$

is convergent.

We have

 $\ds \ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\size z^{2 n + 1} } {\paren {2 n + 1}!}$ $\ds$ $\le$ $\ds \sum_{n \mathop = 0}^\infty \paren{ \frac {\size z^{2 n + 1} } {\paren {2 n + 1}!} + \frac {\size z^{2 n } } {\paren {2 n }!} }$ Squeeze Theorem for Complex Sequences $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\size z^n} {n!}$ changing indices $\ds$ $=$ $\ds \exp \size z$ Taylor Series Expansion for Exponential Function

The result follows from Squeeze Theorem for Complex Sequences.

$\blacksquare$

## Proof 2

Radius of Convergence of Power Series Expansion for Sine Function shows that the radius of convergence of the complex sine function is infinite.

$\blacksquare$