The value of _k=1^10[(2k)/(11)+i(2k)/(11)] is:

Question

The value of $\displaystyle\sum_{k=1}^{10}\left[\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right]$ is:

Accepted Answer

$$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)} $$

$$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $$

$$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}}  - 1} \right\}$$

$$ = i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i$$

$$ = i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$ = i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$

$$ = i \times 0 - i$$

[as $$\,\,\,\,\,\,$$ $${e^{ - 2\pi i}} = 1$$ ]

$=  - i$

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