Euler's Formula/Proof

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Theorem

Let $z \in \C$ be a complex number.

Then:

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


Proof

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

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

$\blacksquare$


Sources