1. ## complex no

If a=e^(i*8pi/11) then find re(a+a^2+a^3+a^4+a^5) I used demoivres theorem but it is getting too lenghty

2. Originally Posted by prasum
If a=e^(i*8pi/11) then find re(a+a^2+a^3+a^4+a^5) I used demoivres theorem but it is getting too lenghty
Let $\omega = e^{2\pi i/11}$, so that $a = \omega^4$. The roots of the equation $z^{11}=1$ are $z=\omega^k$ for $0\leqslant k\leqslant10$. The sum of all these roots is 0. If we exclude the root z=1 then the sum of the remaining roots is –1 (and therefore so is the sum of the real parts of these roots). If you draw a picture of these 10 complex roots on the unit circle in the complex plane, you will see that they consist of $a,\ a^2,\ a^3,\ a^4$ and $a^5$, together with their complex conjugates. It follows that the real part of $a+a^2+a^3+a^4+a^5$ must be –1/2.

3. how do you know that re(a+a^2+a^3+a^4+a^5)=-1/2

4. It was said above that $-1=w^1+\dots+w^{10}=a^1+\dots+a^5+\overline{a^1+\do ts+a^5}$ (here the bar denotes complex conjugation). Also note that the real part of $z+\bar{z}$ is twice the real part of $z$.