Question

The value of $\sum _{k=1}^6[\sin \frac{2k\pi }{7}-i\cos \frac{2k\pi }{7}]$ is:

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

Answer: (D)

$i$

$\sum _{k=1}^6[\sin \frac{2k\pi }{7}-i\cos \frac{2k\pi }{7}]=$ $\sum _{k=1}^6[-i^2\sin \frac{2k\pi }{7}-i\cos \frac{2k\pi }{7}]$

$=-i\sum _{k=1}^6[\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7}]=-i\sum _{r=1}^6{e}^{i\frac{2k\pi }{7}}$

$=-i(e^{i\frac{2\pi }{7}}+e^{i\frac{4\pi }{7}}+..........+e^{i\frac{12\pi }{7}})=-i\left(\frac{e^{i\frac{2\pi }{7}}(e^{i\frac{12\pi }{7}}-1)}{e^{i\frac{2\pi }{7}}-1}\right)$, Summation of G.P.

$=-i\frac{e^{i2\pi }-e^{i\frac{2\pi }{7}}}{e^{i\frac{2\pi }{7}}-1}=-i\cdot \frac{1-e^{i\frac{2\pi }{7}}}{e^{i\frac{2\pi }{7}}-1}=-i(-1)=i$

Since $e^{i2\pi }=\cos 2\pi +i\sin 2\pi =1+0=1$