Sum of Complex Exponentials of i times Arithmetic Sequence of Angles/Formulation 3
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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:
- $\ds \sum_{k \mathop = p}^q e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {\paren {p + q} \alpha} 2} + i \map \sin {\theta + \frac {\paren {p + q} \alpha} 2} } \frac {\map \sin {\paren {q - p + 1} \alpha / 2} } {\map \sin {\alpha / 2} }$
Proof
First note that if $\alpha = 2 \pi k$ for $k \in \Z$, then $e^{i \alpha} = 1$.
\(\ds \sum_{k \mathop = p}^q e^{i \paren {\theta + k \alpha} }\) | \(=\) | \(\ds e^{i \theta} e^{i p \alpha} \sum_{k \mathop = 0}^{q - p} e^{i k \alpha}\) | factorising $e^{i \theta} e^{i p \alpha}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{i \theta} e^{i p \alpha} \paren {\frac {e^{i \paren {q - p + 1} \alpha} - 1} {e^{i \alpha} - 1} }\) | Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{i \theta} e^{i p \alpha} \frac {e^{i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} } \paren {\frac {e^{i \paren {q - p + 1} \alpha / 2} - e^{-i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\) | extracting factors | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{i \paren {\theta + \paren {p + q} \alpha / 2} } \paren {\frac {e^{i \paren {q - p + 1} \alpha / 2} - e^{-i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\) | Exponential of Sum and some algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \cos {\theta + \frac {\paren {p + q} \alpha} 2} + i \map \sin {\theta + \frac {\paren {p + q} \alpha} 2} } \paren {\frac {e^{i \paren {q - p + 1} \alpha / 2} - e^{-i \paren {q - p + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\) | Euler's Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \cos {\theta + \frac {\paren {p + q} \alpha} 2} + i \map \sin {\theta + \frac {\paren {p + q} \alpha} 2} } \frac {\map \sin {\paren {q - p + 1} \alpha / 2} } {\map \sin {\alpha / 2} }\) | Euler's Sine Identity |
$\blacksquare$