Context: Preparing for JEE and it is one of the practice problems.
Question:
If $\alpha = cos(\frac{8\pi}{11})+ i sin(\frac{8\pi}{11})$ then $\Re({\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5})$ is equal to
A) $\frac{1}{2}$
B) $-\frac{1}{2}$
C) $0$
D) None
Key: Option B
What I've tried:
$$\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5 = e^\frac{8\pi}{11}+e^\frac{16\pi}{11}+e^\frac{24\pi}{11}+e^\frac{32\pi}{11}+e^\frac{40\pi}{11}$$
Then I was struck here knowing that the given argument is not in form of $\frac{2k\pi}{n}$ so now I cant use $\text{n}^{th}$ root of unity
So what should I do now? Is there any way to convert in form of $\alpha = e^{i(\frac{2k\pi}{n})}$
where $k = {0, 1, ... n-1}?$
Or is there any other way to solve it?
Roots of unity is indeed the correct approach here, rather than geometric series.
$a^{11} = 1$ and $a \bar{a} = 1$. Thus the conjugate of $a$, $\bar{a} = \frac{1}{a} = \frac{1}{a} \cdot a^{11} = a^{10}$, which has the same real part as $a$. Now since:
$$1 + (a + a^2 + \cdots + a^5) + (a^6 + a^7 + \cdots + a^{10})$$ $$ = 1 + (a + a^2 + \cdots + a^5) + (\overline{a^5} + \overline{a^4}+ \cdots + \bar{a}) = 0,$$ it hence follows that the real part is $-1/2$.