Sum of Complex Exponentials of i times Arithmetic Sequence of Angles/Formulation 1

Theorem

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

Then:

$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$

Proof

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

\(\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} }\)

\(=\)

\(\ds e^{i \theta} \sum_{k \mathop = 0}^n e^{i k \alpha}\)

factorising $e^{i \theta}$

\(\ds \)

\(=\)

\(\ds e^{i \theta} \paren {\frac {e^{i \paren {n + 1} \alpha} - 1} {e^{i \alpha} - 1} }\)

Sum of Geometric Sequence: only when $e^{i \alpha} \ne 1$

\(\ds \)

\(=\)

\(\ds \frac {e^{i \theta} e^{i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\)

extracting factors

\(\ds \)

\(=\)

\(\ds e^{i \paren {\theta + n \alpha / 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\)

Exponential of Sum and some algebra

\(\ds \)

\(=\)

\(\ds \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \paren {\frac {e^{i \paren {n + 1} \alpha / 2} - e^{-i \paren {n + 1} \alpha / 2} } {e^{i \alpha / 2} - e^{-i \alpha / 2} } }\)

Euler's Formula

\(\ds \)

\(=\)

\(\ds \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }\)

Euler's Sine Identity

$\blacksquare$