Question

Evaluate:

${\sum }_{k=1}^{10}\left[\begin{array}{c}\sin \frac{2k\pi }{11}-i\cos \frac{2k\pi }{11}\end{array}\right]$

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

The correct option is C $-i$

Since, $cos\theta +isin\theta =e^{i\theta }$

$\sum _{k=1}^{10}[\sin \frac{2k\pi }{11}-i\cos \frac{2k\pi }{11}]=-i\sum _{k=1}^{10}\left[cos\right(\frac{2k\pi }{11})+isin(\frac{2k\pi }{11}\left)\right]=-i\sum _{k=1}^{10}{e}^{i\frac{2k\pi }{11}}$

Solving the index part only which is

$i\frac{2k\pi }{11}=i\frac{2\pi }{11}(1+2+3+\cdots +10)$......[putting the values of $k$]

$=i10\pi$

So,

$-i{\sum }_{k=1}^{10}{e}^{i\frac{2k\pi }{11}}=-i{e}^{i10\pi }=-i(cos10\pi +isin10\pi )=-i$