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) $
$\Rightarrow a=e^{\frac{i 8 \pi}{11}}$
$\Rightarrow a$ is $11 \text{th}$ root of unity and all roots are $1, a, a^{2}, \ldots, a^{10}$
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 $=0$
$1+a^{1}+a^{2}+a^{3}+\ldots+a^{10}=0$
$\Rightarrow a+a^{2}+a^{3}+a^{4}+a^{5}+a^{6}+a^{7}+a^{8}+a^{9}+a^{10}=-1$
$\Rightarrow (a+\bar{a})+\left(a^{2}+\overline{a^{2}}\right)+\left(a^{3}+\overline{a^{3}}\right)+\left(a^{4}+\overline{a^{4}}\right)+\left(a^{5}+\overline{a^{5}}\right)=-1$
$\Rightarrow 2 \text{Re}(a)+2 \text{Re}\left(a^{2}\right)+2 \text{Re}\left(a^{3}\right)+2 \text{Re}\left(a^{4}\right)+2 \text{Re}\left(a^{5}\right)=-1[z+\bar{z}=2 \text{Re}(z)]$
$\Rightarrow 2 \text{Re}\left(a+a^{2}+a^{3}+a^{4}+a^{5}\right)=-1$
$\Rightarrow \text{Re}\left(a+a^{2}+a^{3}+a^{4}+a^{5}\right)=-\frac{1}{2}$