<s>[INST] 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!
 [/INST] 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
$$z +z^3 - z^2 -z^4 = 2i(\sin(2 \pi/5) - \sin(\pi/5)).$$
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