Question

$\sum _{k=1}^6(\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7})$ equals

• A. 1
• B. i
• C. -i
• D. -1

Answer 1

The correct option is B i

Solution:

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

$=\sum _{k=1}^6-i(\cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7})$

$=-i\sum _{k=1}^6{e}^{\frac{i2k\pi }{7}}$

$=-i[e^{\frac{i2\pi }{7}}+e^{\frac{i4\pi }{7}}+e^{\frac{i6\pi }{7}}+e^{\frac{i8\pi }{7}}+e^{\frac{i10\pi }{7}}+e^{\frac{i12\pi }{7}}]$

$=-i\left[e^{\frac{i2\pi }{7}}\left(\frac{1-e^{\frac{i12\pi }{7}}}{1-e^{\frac{i2\pi }{7}}}\right)\right]$

$=-i\left[\left(\frac{e^{\frac{i2\pi }{7}}-e^{\frac{i14\pi }{7}}}{1-e^{\frac{i2\pi }{7}}}\right)\right]$

$=-i\left[\left(\frac{e^{\frac{i2\pi }{7}}-1}{1-e^{\frac{i2\pi }{7}}}\right)\right]$ $\because (e^{\frac{i14\pi }{7}}=1)$

$=-i\times (-1)$

$=i$

Hence, A is the correct option.