Question

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

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

The correct option is D $i$

Let,

$Z=\cos \left(\frac{2\Pi k}{7}\right)+\iota \sin \left(\frac{2\Pi k}{7}\right)$

$Z^k=Z=\cos \left(\frac{2\Pi k}{7}\right)+\iota \sin \left(\frac{2\Pi k}{7}\right)$

To find,

${\sum }_{k=1}^6(-\iota )\left[\cos \left(\frac{2\Pi k}{7}\right)+\iota \sin \left(\frac{2\Pi k}{7}\right)\right]$

$=(-\iota ){\sum }_{k=1}^6{q}^k$

$=(-\iota )\frac{Z(1-Z^6)}{1-Z}$

$=(-\iota )\left(\frac{Z-Z^7}{1-Z}\right)$

$=(+\iota )\left(\frac{1-Z}{1-Z}\right)$

$=\iota$

Hence, $\iota$ is the answer.