Question

The value of ${\sum }_{k=1}^4(sin\frac{2\pi k}{7}-icos\frac{2\pi k}{7})$ is

• A. -1
• B. $0$
• C. -i
• D. i

Answer 1

The correct option is D i

Let $z=cos\frac{2\pi }{7}+isin\frac{2\pi }{7}$, then by De Moiver's theorem $Z^k=cos\frac{2\pi k}{7}+isin\frac{2\pi k}{7}$
Now the given sum

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

$={\sum }_{k=1}^6\left[\right(-i\left)(cos\frac{2\pi k}{7}+isin\frac{2\pi k}{7})\right]$

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

which is a G.P. of which the first term is z, number of the terms is 6 and the common

ratio is $z=cos\frac{2\pi }{7}+isin\frac{2\pi }{7}\ne 1$ So summing up the G.P., we have

$S=(-i)\frac{z(1-z^6)}{1-z}=(-i)\frac{z-z^7}{1-z}=(-i)\frac{z-1}{1-z}=i$

$[\because z^7={(cos\frac{2\pi }{7}+isin\frac{2\pi }{7})}^7=cos2\pi +isin2\pi =1$