Cosine Function is Absolutely Convergent/Complex Case/Proof 1

Theorem

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

Let $\cos z$ be the cosine of $z$.

Then:

$\cos z$ is absolutely convergent for all $z \in \C$.

The definition of the complex cosine function is:

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

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

is convergent.

We have

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

$=$

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

$\ds $

$\leq$

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

$\ds $

$=$

$\ds \sum_{n \mathop = 0}^\infty \frac {\size z^n} {n!}$

changing indices

$\ds $

$=$

$\ds \exp \size z$

$\blacksquare$