# Complex Cosine Function is Unbounded

## Proof 1

Let $K \in \R_{>0}$ be an arbitrary real number.

Let $p = \ln {2 K}$.

Let $z = i p$, where $i$ denotes the imaginary unit.

Then:

 $\ds \cos z$ $=$ $\ds \dfrac {\exp \paren {i \paren {i p} } + \exp \paren {-i \paren {i p} } } 2$ Cosine Exponential Formulation $\ds$ $=$ $\ds \dfrac {\exp p + \exp \paren {-p} } 2$ simplifying $\ds$ $>$ $\ds \dfrac {\exp p} 2$ as $\exp \paren {-p} > 0$ $\ds$ $=$ $\ds \dfrac {\exp \paren {\ln {2 K} } } 2$ $\ds$ $=$ $\ds K$ $\ds \leadsto \ \$ $\ds \cmod {\cos \paren {i p} }$ $>$ $\ds K$

Thus for any $K$ we can find $z \in \C$ such that $\cmod {\cos z} > K$.

Hence the result by definition of unbounded.

$\blacksquare$

## Proof 2

By Complex Cosine Function is Entire, we have that $\cos$ is an entire function.

Aiming for a contradiction, suppose that $\cos$ is a bounded function.

By Liouville's Theorem, we have that $\cos$ is a constant function.

However, by Cosine of Zero is One:

$\cos 0 = 1$

and by Cosine of Right Angle:

$\sin \dfrac \pi 2 = 0$

Therefore, $\cos$ is clearly not a constant function.

This contradicts the statement that $\cos$ is a constant function.

We hence conclude, by Proof by Contradiction, that our assumption that $\cos$ is a bounded is false.

Hence $\cos$ is unbounded.

$\blacksquare$