Real part in complex summation

Question

If $\beta=\cos\frac{9\pi}{19}+i\sin\frac{9\pi}{19}$. Then Real part of $\sum^{9}_{k=1}(-1)^{k+1}\beta^k$ is

Try: $\beta^{19}=\cos(9\pi)+i\sin(9\pi)=-1$

$\beta^{19}+1=0\Rightarrow (\beta+1)(\beta^{18}-\beta^{17}+\beta^{16}-\beta^{15}+\cdots\cdots -\beta+1)=0$

Could some help me to solve it , thanks

Answer 1

I followed a long drawn out method.

For the odd powers: $\displaystyle \Re (\beta) = \displaystyle e^{\displaystyle i\frac{9\pi}{19}} = -\Re \displaystyle e^{\displaystyle i\frac{10\pi}{19}}$

Similarly $\Re (\beta^3) = -\Re e^{\displaystyle i\frac{8\pi}{19}}; \Re (\beta^5) = -\Re e^{\displaystyle i\frac{28\pi}{19}}; \Re (\beta^7) = -\Re e^{\displaystyle i\frac{6\pi}{19}}; \Re (\beta^7) = -\Re e^{\displaystyle i\frac{24\pi}{19}}$

For the even powers we have $\Re (\beta^2) = \Re e^{\displaystyle i\frac{18\pi}{19}};\Re (\beta^4) = \Re e^{\displaystyle i\frac{36\pi}{19}}; \Re (\beta^6) = \Re e^{\displaystyle i\frac{16\pi}{19}}; \Re (\beta^2) = \Re e^{\displaystyle i\frac{34\pi}{19}} $

The complements of the above $9$ powers of $\beta$ generate the remaining $9$ non-real roots of $z^{19}=1$

Since $\Re(z) = z + \overline{z}$, with $\omega = e^{\displaystyle i\frac{2\pi}{19}}$ given summation equals $-\frac{1}{2} \left(\omega+\omega^2+\ldots \omega^{18} \right) = \frac{1}{2}$

Answer 2

Hints:

1. $\beta = e^{9\pi/19}$

2. If $S=\sum\limits_{k=n}^{m} r^k$, then $rS=\sum\limits_{k=n+1}^{m+1} r^k$

3. Consider $r=-\beta$ and $S-rS$.