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

Printable View

- Oct 10th 2010, 10:19 PMprasumcomplex 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

- Oct 11th 2010, 12:05 AMOpalg
Let $\displaystyle \omega = e^{2\pi i/11}$, so that $\displaystyle a = \omega^4$. The roots of the equation $\displaystyle z^{11}=1$ are $\displaystyle z=\omega^k$ for $\displaystyle 0\leqslant k\leqslant10$. 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 $\displaystyle a,\ a^2,\ a^3,\ a^4$ and $\displaystyle a^5$, together with their complex conjugates. It follows that the real part of $\displaystyle a+a^2+a^3+a^4+a^5$ must be –1/2.

- Oct 11th 2010, 02:58 AMprasum
how do you know that re(a+a^2+a^3+a^4+a^5)=-1/2

- Oct 11th 2010, 04:42 AMemakarov
It was said above that $\displaystyle -1=w^1+\dots+w^{10}=a^1+\dots+a^5+\overline{a^1+\do ts+a^5}$ (here the bar denotes complex conjugation). Also note that the real part of $\displaystyle z+\bar{z}$ is twice the real part of $\displaystyle z$.