Solved: "∑k=110(sin⁡2πk11−icos⁡2πk11)=?"

Zack Mora

Answered question

2022-03-31

\sum_{k = 1}^{10} (\sin \frac{2 π k}{11} - i \cos \frac{2 π k}{11}) =

pastuh7vka

Beginner2022-04-01Added 13 answers

\sum_{k = 1}^{10} (\sin \frac{2 π k}{11} - i \cos \frac{2 π k}{11}) = - i \sum_{k = 1}^{10} e^{\frac{i 2 π k}{11}} = - i \times e^{\frac{i 2 π}{11}} \times \frac{{e^{\frac{i 2 π}{11}}}^{10} - 1}{e^{\frac{i 2 π}{11}} - 1}

= - i e^{\frac{i 2 π}{11}} \frac{e^{\frac{i 20 π}{11}} - 1}{e^{\frac{i 2 π}{11}} - 1}

- i e^{\frac{i 2 π}{11}} \frac{e^{\frac{i 20 π}{11}} - 1}{e^{\frac{i 2 π}{11}} - 1} = - i \frac{e^{\frac{i 2 π}{11}} e^{\frac{i 20 π}{11}} - e^{\frac{i 2 π}{11}}}{e^{\frac{i 2 π}{11}} - 1} = - i \frac{1 - e^{\frac{i 2 π}{11}}}{e^{\frac{i 2 π}{11}} - 1} = i

Tristatex9tw

Beginner2022-04-02Added 18 answers

- i e^{\frac{i 2 π}{11}} \frac{e^{\frac{i 20 π}{11}} - 1}{e^{\frac{i 2 π}{11}} - 1} = - i \frac{\exp (\frac{i 22 π}{11}) - \exp (\frac{i 2 π}{11})}{\exp (\frac{i 2 π}{11}) - 1}

= - i \frac{\exp (i 2 π) - \exp (\frac{i 2 π}{11})}{\exp (\frac{i 2 π}{11}) - 1}

= - i \frac{1 - \exp (\frac{i 2 π}{11})}{\exp (\frac{i 2 π}{11}) - 1}

= - i (- 1)

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