Question

If $a=\cos \left(\frac{8\pi }{11}\right)+i\sin \left(\frac{8\pi }{11}\right)$, then $\text{Re}\left(a+a^2+a^3+a^4+a^5\right)=$

• A. 0
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (B)

$a=\cos \left(\frac{8\pi }{11}\right)+i\sin \left(\frac{8\pi }{11}\right)$
$\Rightarrow a=e^{\frac{i8\pi }{11}}$
$\Rightarrow a$ is $11\text{th}$ root of unity and all roots are $1,a,a^2,\dots ,a^{10}$
Now, $a^{10}=\frac{a^{10}\cdot a}{a}=\frac{a^{11}}{a}=\frac{1}{a}=\bar{a}$
Similarly, $a^9=\overline{a^2},a^8=\overline{a^3},a^7=\overline{a^4},a^6=\overline{a^5}$,
We know that,
Sum of $n$ roots of unity $=0$
$1+a^1+a^2+a^3+\dots +a^{10}=0$
$\Rightarrow a+a^2+a^3+a^4+a^5+a^6+a^7+a^8+a^9+a^{10}=-1$
$\Rightarrow (a+\bar{a})+\left(a^2+\overline{a^2}\right)+\left(a^3+\overline{a^3}\right)+\left(a^4+\overline{a^4}\right)+\left(a^5+\overline{a^5}\right)=-1$
$\Rightarrow 2\text{Re}(a)+2\text{Re}\left(a^2\right)+2\text{Re}\left(a^3\right)+2\text{Re}\left(a^4\right)+2\text{Re}\left(a^5\right)=-1[z+\bar{z}=2\text{Re}(z)]$
$\Rightarrow 2\text{Re}\left(a+a^2+a^3+a^4+a^5\right)=-1$
$\Rightarrow \text{Re}\left(a+a^2+a^3+a^4+a^5\right)=-\frac{1}{2}$