First slide

Algebra of complex numbers

Question

$\sum_{r = 1}^{8} [\sin (\frac{2 π r}{9}) + i \cos (\frac{2 π r}{9})] =$

Very Easy

A

-1
B

1
C

i
D

-i

Solution

$\sum_{r = 1}^{8} (\sin (\frac{2 π r}{9}) + i \cos (\frac{2 π r}{9})) = \sum_{r = 1}^{8} i (\cos (\frac{2 r π}{9}) - i \sin (\frac{2 r π}{9})) = - i$

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