Question

Evaluate:

${\sum }_{k=1}^4[\sin \frac{2k\pi }{5}-i\cos \frac{2k\pi }{5}]$

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

Answer: (C)

$-i$

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

$\sum _{k=1}^4[\sin \frac{2k\pi }{5}-i\cos \frac{2k\pi }{5}]=-i\sum _{k=1}^4\left[cos\right(\frac{2k\pi }{5})+isin(\frac{2k\pi }{5}\left)\right]=-i\sum _{k=1}^4{e}^{i\frac{2k\pi }{5}}$

Solving the index part only which is

$i\frac{2k\pi }{5}=i\frac{2\pi }{5}(1+2+3+\cdots +4)$......[putting the values of $k$]

$=i4\pi$

So,

$-i{\sum }_{k=1}^4{e}^{i\frac{2k\pi }{5}}=-i{e}^{i4\pi }=-i(cos4\pi +isin4\pi )=-i$