Complex Cosine Function is Entire

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then $\cos$ is entire.


Proof 1

By the definition of the complex cosine function, $\cos$ admits a power series expansion about $0$:

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

By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\cos$ is entire it suffices to show that this series is everywhere convergent.

From Radius of Convergence from Limit of Sequence: Complex Case, it is sufficient to show that:

$\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 2}!} \times \frac {\paren {2 n}!} {\paren {-1}^n} } = 0$

We have:

\(\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 2}!} \times \frac {\paren {2 n}!} {\paren {-1}^n} }\) \(=\) \(\ds \size {-1} \lim_{n \mathop \to \infty} \size {\frac {\paren {2 n}!} {\paren {2 n + 2} \paren {2 n + 1} \paren {2 n}!} }\) Definition of Factorial
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\frac 1 {\paren {2 n + 2} \paren {2 n + 1} } }\)
\(\ds \) \(=\) \(\ds 0\)

Hence the result.

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Complex_Cosine_Function_is_Entire&oldid=690027"