Question

Evaluate:

${\sum }_{k=1}^8[\sin \frac{2k\pi }{9}-i\cos \frac{2k\pi }{9}]$

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

The correct option is C $-i$

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

$\sum _{k=1}^8[\sin \frac{2k\pi }{9}-i\cos \frac{2k\pi }{9}]=-i\sum _{k=1}^8\left[cos\right(\frac{2k\pi }{9})+isin(\frac{2k\pi }{9}\left)\right]=-i\sum _{k=1}^8{e}^{i\frac{2k\pi }{9}}$

Solving the index part only which is

$i\frac{2k\pi }{9}=i\frac{2\pi }{9}(1+2+3+\cdots +8)$......[putting the values of $k$]

$=i8\pi$

So,

$-i{\sum }_{k=1}^8{e}^{i\frac{2k\pi }{9}}=-i{e}^{i8\pi }=-i(cos8\pi +isin8\pi )=-i$