Complex Cosine Function is Entire/Proof 2

Theorem

Let $\cos: \C \to \C$ be the complex cosine function.

Then $\cos$ is entire.

Proof

Let:

\(\ds \map f z\)

\(=\)

\(\ds \exp z\)

\(\ds \map g z\)

\(=\)

\(\ds i z\)

\(\ds \map h z\)

\(=\)

\(\ds -i z\)

for all $z \in \C$.

By Complex Exponential Function is Entire, we have that $f$ is entire.

By Polynomial is Entire, we have that $g$ and $h$ are entire.

Therefore, by Composition of Entire Functions is Entire, we have that both $f \circ g$ and $f \circ h$ are entire.

By Linear Combination of Entire Functions is Entire, we then have that:

$\dfrac 1 2 \paren {f \circ g + f \circ h}$

is entire.

Then:

\(\ds \frac 1 2 \paren {\map {\paren {f \circ g} } z + \map {\paren {f \circ h} } z}\)

\(=\)

\(\ds \frac 1 2 \paren {\map \exp {i z} + \map \exp {-i z} }\)

\(\ds \)

\(=\)

\(\ds \cos z\)

Euler's Cosine Identity

Therefore $\cos$ is an entire function.

$\blacksquare$