Question

Let $a=cos\frac{2\pi }{7}+isin\frac{2\pi }{7},A=a+a^2+a^4$ and $B=a^3+a^5+a^6$, then

Answer 1

Given, $a\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7},A=a+a^2+a^4,B=a^3+a^5+a^6$

Now,

$a^o+a^1+a^2+....+a^6=1\frac{(a^7-1)}{a-1}$ [$G.P$ Sum]

$a^7=\cos 2\pi +i\sin 2\pi =1$ (De morvie's theorem)

$\Rightarrow \sum _{i=0}^6=0$ ........... $\left(1\right)$

$A+B=a+a^2+a^3+.....+a^6=\sum _{i=0}^6{a}^1-a^0=0-1=-1$ .......... $\left(2\right)$

$\therefore$ Option $B$ is conect

Also, $AB=(a+a^2+a^4)(a^3+a^5+a^6)$

$=a^4+a^6+a^7+a^5+a^7+a^8+a^7+a^9+a^{10}$

$=a+a^2+a^3+a^4+a^5+a^6+3a^7[\because a^7=1a^8=a^{7}a=a]$

$=3-1=2$