The value of -k-110sin2 - k-11-icos2 - k-11 is

∑ ^10_k=1 ( sin2π k/11 i cos2π k/11 )=∑ ^10_k=1 (i^2 sin2π k/11 i cos2π k/11 ) = i ∑ ^10_k=1 ( cos2π k/11+i sin2π k/11 )= i∑ ^10_k=1 e^i2π k/11 = i [ ∑ ^10_ ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The value of ...

Question

The value of $(\sum_{k = 1}^{10} (s i n \frac{2 π k}{11} - i c o s \frac{2 π k}{11}))$ is

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-i

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Solution

The correct option is C i
$\sum_{k = 1}^{10} (s i n \frac{2 π k}{11} - i c o s \frac{2 π k}{11}) = \sum_{k = 1}^{10} (i^{2} s i n \frac{2 π k}{11} - i c o s \frac{2 π k}{11}) = - i \sum_{k = 1}^{10} (c o s \frac{2 π k}{11} + i s i n \frac{2 π k}{11}) = - i \sum_{k = 1}^{10} e^{i \frac{2 π k}{11}} = - i [\sum_{k = 1}^{10} e^{i \frac{2 π k}{11} - 1}]$
= -i( sum of $11^{t h}$ roots of unity – 1)
= – i(0 – 1) = i

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The value of (∑10k=1(sin2πk11−icos2πk11)) is

The correct option is C i∑10k=1(sin2πk11−icos2πk11)=∑10k=1(i2sin2πk11−icos2πk11)=−i∑10k=1(cos2πk11+isin2πk11)=−i∑10k=1ei2πk11=−i[∑10k=1ei2πk11−1] = -i( sum of 11th roots of unity – 1) = – i(0 – 1) = i

The value of $(\sum_{k = 1}^{10} (s i n \frac{2 π k}{11} - i c o s \frac{2 π k}{11}))$ is