Sine Function is Absolutely Convergent/Complex Case/Proof 1

Theorem
Let $z \in \C$ be a complex number.

Let $\sin z$ be the sine of $z$.

Then:


 * $\sin z$ is absolutely convergent for all $z \in \C$.

Proof
The definition of the complex sine function is:


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

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

The result follows from Squeeze Theorem for Sequences of Complex Numbers.