First slide

De-moivre's theorem

Question

The value of $\sum_{k = 1}^{10} (\sin \frac{2 k π}{11} + i \cos \frac{2 k π}{11})$ is

Difficult

A

$i$
B

$1$
C

$- 1$
D

$- i$

Solution

$\sum_{k = 1}^{10} (\sin \frac{2 k π}{11} + i \cos \frac{2 k π}{11}) = \sum_{k = 1}^{10} e^{i 2 k π / 11}$
$= 0 + i (- 1) = - i$

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