Complex Numbers Ques 93
The value of $\sum _{k=1}^{6} (\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7})$ is
$(1998,2 M)$
(a) $-1$
(b) $ 0$
(c) $-i$
(d) $i$
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Answer:
Correct Answer: 93.(d)
Solution:
Formula:
Summation of series using complex number:
- $\sum _{k=1}^{6} (\sin \frac{2 k \pi}{7}-i \cos \frac{2 k \pi}{7})$
$ =\sum _{k=1}^{6}-i (\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}) $
$ =-i (\sum _{k=1}^{6} e^{\frac{i 2 k \pi}{7}})=-i e^{i 2 \pi / 7}+e^{i 4 \pi / 7}+e^{i 6 \pi / 7} $ $ +e^{i 8 \pi / 7}+e^{i 10 \pi / 7}+e^{i 12 \pi / 7} $
$ =-i (e^{i 2 \pi / 7} \frac{\left(1-e^{i 12 \pi / 7}\right)}{1-e^{i 2 \pi / 7}} )$
$ =-i (\frac{e^{i 2 \pi / 7}-e^{i 14 \pi / 7}}{1-e^{i 2 \pi / 7}}) \quad\left[\because e^{i 14 \pi / 7}=1\right] $
$ =-i (\frac{e^{i 2 \pi / 7}-1}{1-e^{i 2 \pi / 7}})=i$