Sum of Complex Exponentials of i times Arithmetic Sequence of Angles/Formulation 2

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Theorem

Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.

Then:

$\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$


Proof

First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$.

\(\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} }\) \(=\) \(\ds e^{i \theta} e^{i \alpha} \sum_{k \mathop = 0}^{n - 1} e^{i k \alpha}\) factorising $e^{i \theta} e^{i \alpha}$
\(\ds \) \(=\) \(\ds e^{i \theta} e^{i \alpha} \paren {\frac {e^{i n \alpha} - 1} {e^{i \alpha} - 1} }\) Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$
\(\ds \) \(=\) \(\ds e^{i \theta} e^{i \alpha} \paren {\frac {e^{i n \alpha / 2} \paren {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } } {e^{i \alpha / 2} \paren {e^{i \alpha / 2} - e^{-i \alpha / 2} } } }\) extracting factors
\(\ds \) \(=\) \(\ds e^{i \paren {\theta + \paren {n + 1} \alpha / 2} } \paren {\frac {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\) Exponential of Sum and some algebra
\(\ds \) \(=\) \(\ds \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \paren {\frac {e^{i n \alpha / 2} - e^{-i n \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\) Euler's Formula
\(\ds \) \(=\) \(\ds \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }\) Euler's Sine Identity

$\blacksquare$


Sources

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