If z is one of the fifth roots of unity, not 1... If z is one of the fifth roots of unity, not 1, show that:  
$1+z+z^2+z^3+z^4=0$  
Which wasn't too bad, but the next part is killing me: show that:
$z-z^2+z^3-z^4=2i(sin(2\pi/5)-sin(\pi/5))$
Can anyone help? Thanks!
 A: For the first part you should note that by the formula for the geometric series,
$$1 + z + z^2 + z^3 + z^4 = \frac{z^5 - 1}{z- 1}.$$
So if you put $z = e^{2\pi i/5}$ into $z^5 - 1$ you get $(e^{2\pi i/5})^5 =1$ so that $1 + z + \ldots z^4 = 0$. You can see also that if you put $z = (e^{2\pi i/5})^n$ for $n=2,3,4$  you see that $z^5 - 1$ is always zero.
For the second part you should be able to see that if $z = e^{2\pi i/5}$, then $$z - z^4 = 2i\sin(2\pi/5).$$
The reason is if you draw a pentagon in the complex plane with vertices at the points $$1, e^{2\pi i/5}, e^{4\pi i/5}, e^{6\pi i/5}, e^{8\pi i/5}$$
then the vertices $e^{2\pi i /5}$ and $e^{8 \pi i/5}$ have equal real part so subtracting one from the other gives the result above. You should be familiar with de Moivre's formula for this.
Now similarly if you calculate $z - z^4$ you should get 
$$z^3 - z^2 = 2i\sin(6 \pi/5).$$ 
But then $2i\sin(6 \pi/5) = -2i\sin(\pi/5)$ so adding this together with that found for $z - z^4$ gives
$$z +z^3 - z^2 -z^4 = 2i(\sin(2 \pi/5) - \sin(\pi/5)).$$
A: This way is certainly less direct than the previous solution, but more straightforward.
\begin{eqnarray}
z-z^2+z^3-z^4&=&-\sum_{k=1}^4(-1)^ke^{2ik\pi/5}=-\sum_{k=1}^4\left(-e^{2i\pi/5}\right)^k=-\frac{-e^{2i\pi/5}-(-e^{2i\pi/5})^5}{1-(-e^{2i\pi/5})}\\
&=&\frac{e^{2i\pi/5}-1}{e^{2i\pi/5}+1}=\frac{(e^{2i\pi/5}-1)(e^{-2i\pi/5}+1)}{(1+\cos2\pi/5)^2+\sin^22\pi/5}=\frac{e^{2i\pi/5}-e^{-2i\pi/5}}{2+2\cos2\pi/5}\\
&=&\frac{i\sin2\pi/5}{1+\cos2\pi/5}=\frac{2i\sin\pi/5\,\cos\pi/5}{2\cos^22\pi/5}=
\frac{i\sin\pi/5}{\cos\pi/5}=i\tan\pi/5.
\end{eqnarray}
So it remains to check that $\tan\pi/5=2(\sin2\pi/5-\sin\pi/5)$. After a simplification, this amounts to check that
$$
1=2\cos\pi/5\;(2\cos\pi/5-1),
$$
and this can be seen directly from the fact that $\cos\pi/5=(1+\sqrt5)/4$.
