Complex Cosine Function is Unbounded

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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.


\(\ds \cos z\) \(=\) \(\ds \dfrac {\exp \paren {i \paren {i p} } + \exp \paren {-i \paren {i p} } } 2\) 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.


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.


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