Complex Cosine Function is Unbounded
Theorem
The 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:
| \(\ds \cos z\) | \(=\) | \(\ds \dfrac {\map \exp {i \paren {i p} } + \map \exp {-i \paren {i p} } } 2\) | Euler's Cosine Identity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\exp p + \map \exp {-p} } 2\) | simplifying: $i^2 = -1$ | |||||||||||
| \(\ds \) | \(>\) | \(\ds \dfrac {\exp p} 2\) | as $\map \exp {-p} > 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\map \exp {\ln {2 K} } } 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds K\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \cmod {\map \cos {i p} }\) | \(>\) | \(\ds K\) |
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:
- $\cos \dfrac \pi 2 = 0$
Therefore, $\cos$ is 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 function is false.
Hence $\cos$ is unbounded.
$\blacksquare$