Sine Function is Absolutely Convergent/Complex Case

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Theorem

The complex sine function $\sin: \C \to \C$ is absolutely convergent.


Proof 1

The definition of the complex sine function is:

$\ds \forall z \in \C: \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$

By definition of absolutely convergent complex series, we must show that the power series

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

is convergent.

We have

\(\ds \ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\size z^{2 n + 1} } {\paren {2 n + 1}!}\)
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = 0}^\infty \paren{ \frac {\size z^{2 n + 1} } {\paren {2 n + 1}!} + \frac {\size z^{2 n } } {\paren {2 n }!} }\) Squeeze Theorem for Complex Sequences
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\size z^n} {n!}\) changing indices
\(\ds \) \(=\) \(\ds \exp \size z\) Taylor Series Expansion for Exponential Function

The result follows from Squeeze Theorem for Complex Sequences.

$\blacksquare$


Proof 2

Radius of Convergence of Power Series Expansion for Sine Function shows that the radius of convergence of the complex sine function is infinite.

Then Existence of Radius of Convergence of Complex Power Series shows that the complex sine function is absolutely convergent.

$\blacksquare$


Sources

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