SATHEE: Complex Numbers 5 Question 11

Complex Numbers 5 Question 11

12.

The value of $\sum_{k = 1}^{6} \sin \frac{2 π k}{7} - i \cos \frac{2 π k}{7}$ is

$(1998, 2 M)$

(a) -1

(b) 0

(d) $i$

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Answer:

Correct Answer: 12. (d)

Solution:

$\sum_{k = 1}^{6} \sin \frac{2 k π}{7} - i \cos \frac{2 k π}{7}$

$= \sum_{k = 1}^{6} - i \cos \frac{2 k π}{7} + i \sin \frac{2 k π}{7}$

$= - i \sum_{k = 1}^{6} e^{\frac{i 2 k π}{7}} = - i e^{i 2 π / 7} + e^{i 4 π / 7} + e^{i 6 π / 7}$ $+ e^{i 8 π / 7} + e^{i 10 π / 7} + e^{i 12 π / 7}$

$= - i e^{i 2 π / 7} \frac{(1 - e^{i 12 π / 7})}{1 - e^{i 2 π / 7}}$

$= - i \frac{e^{i 2 π / 7} - e^{i 14 π / 7}}{1 - e^{i 2 π / 7}} [∵ e^{i 14 π / 7} = 1]$

$= - i \frac{e^{i 2 π / 7} - 1}{1 - e^{i 2 π / 7}} = i$

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Complex Numbers 5 Question 11

12.

Answer:

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