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Question: The value of \(\sum_{k = 1}^{10}\left( \sin\frac{2\pi k}{11} - i\cos\frac{2\pi k}{11} \right)\)is –...

The value of ∑k=110(sin⁡2πk11−icos⁡2πk11)\sum_{k = 1}^{10}\left( \sin\frac{2\pi k}{11} - i\cos\frac{2\pi k}{11} \right)is –

A

1

B

–1

C

I

D

– i

Answer

I

Explanation

Solution

Sol. We have,

∑k=110(sin⁡2πk11−icos⁡2πk11)\sum_{k = 1}^{10}\left( \sin\frac{2\pi k}{11} - i\cos\frac{2\pi k}{11} \right)= ∑k=110(–i2sin⁡2πk11−icos⁡2πk11)\sum_{k = 1}^{10}\left( –i^{2}\sin\frac{2\pi k}{11} - i\cos\frac{2\pi k}{11} \right)

= –i ∑k=110(cos⁡2πk11+isin⁡2πk11)\sum_{k = 1}^{10}\left( \cos\frac{2\pi k}{11} + i\sin\frac{2\pi k}{11} \right) = –i∑k=110ei2πk11\sum_{k = 1}^{10}e^{i\frac{2\pi k}{11}}

= – i [∑k=010ei2πk11−1]\left\lbrack \sum_{k = 0}^{10}{e^{i\frac{2\pi k}{11}} - 1} \right\rbrack

= – i (sum of 11th roots of unity – 1)

= – i (0 – 1) = i.