Complex Sine Function is Entire/Proof 2

Theorem

Let $\sin: \C \to \C$ be the complex sine function.

Then $\sin$ 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:

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

is entire.

Note that:

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

\(=\)

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

\(\ds \)

\(=\)

\(\ds \sin z\)

Euler's Sine Identity

Therefore, $\sin$ is an entire function.

$\blacksquare$