Question
The value of $$\sum\limits_{k = 1}^{10} {\left( {\sin \frac{{2k\pi }}{{11}} + i\cos \frac{{2k\pi }}{{11}}} \right)} $$ is
A.
$$i$$
B.
$$1$$
C.
$$ - 1$$
D.
$$ - i$$
Answer :
$$ - i$$
Solution :
$$\eqalign{
& \sum\limits_{k = 1}^{10} {\left( {\sin \frac{{2k\pi }}{{11}} + i\cos \frac{{2k\pi }}{{11}}} \right)} \cr
& = i\sum\limits_{k = 1}^{10} {\left( {\cos \frac{{2k\pi }}{{11}} - i\sin \frac{{2k\pi }}{{11}}} \right)} \cr
& = i\sum\limits_{k = 1}^{10} {{e^{ - \frac{{2k\pi }}{{11}}i}}} \cr
& = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - \frac{{2k\pi }}{{11}}i}} - 1} } \right\} \cr
& = i\left[ {1 + {e^{ - \frac{{2\pi }}{{11}}i}} + {e^{ - \frac{{4\pi }}{{11}}i}} + ..... + 11\,{\text{terms}}} \right] - i \cr
& = i\left[ {\frac{{1 - {{\left( {{e^{ - \frac{{2\pi }}{{11}}}}} \right)}^{11}}}}{{1 - {e^{ - \frac{{2\pi }}{{11}}i}}}}} \right] - i \cr
& = i\left[ {\frac{{1 - {e^{ - 2\pi i}}}}{{1 - {e^{ - \frac{{2\pi }}{{11}}i}}}}} \right] - i \cr
& = i \times 0 - i\,\,\,\,\,\,\,\,\,\,\,\left[ {\because {e^{ - 2\pi i}} = 1} \right] = - i \cr} $$