Euler's Formula/Real Domain/Proof 5

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Theorem

Let $\theta \in \R$ be a real number.

Then:

$e^{i \theta} = \cos \theta + i \sin \theta$


Proof

As Sine Function is Absolutely Convergent and Cosine Function is Absolutely Convergent, we have:

\(\displaystyle \cos \theta + i \sin \theta\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!}\) Definition of Complex Cosine Function and Definition of Complex Sine Function
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!} }\) Sum of Absolutely Convergent Series
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {\dfrac {\paren {i \theta}^{2 n} } {\paren {2 n}!} + \dfrac {\paren {i \theta}^{2 n + 1} } {\paren {2 n + 1}!} }\) Definition of Imaginary Unit
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {\paren {i \theta}^n} {n!}\)
\(\displaystyle \) \(=\) \(\displaystyle e^{i \theta}\) Definition of Complex Exponential Function

$\blacksquare$


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