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:
\(\ds \cos \theta + i \sin \theta\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!}\) | Definition of Complex Cosine Function and Definition of Complex Sine Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!} }\) | Sum of Absolutely Convergent Series | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {\dfrac {\paren {i \theta}^{2 n} } {\paren {2 n}!} + \dfrac {\paren {i \theta}^{2 n + 1} } {\paren {2 n + 1}!} }\) | Definition of Imaginary Unit | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {i \theta}^n} {n!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{i \theta}\) | Definition of Complex Exponential Function |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: $(2.18)$
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Euler's Formula
- For a video presentation of the contents of this page, visit the Khan Academy.
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- 2001: Christian Berg: Kompleks funktionsteori: $\S 1.5$