## Loser66 one year ago if $$z=e^{2\pi i/5}$$ then $$1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=??$$ Please, help

1. Loser66

|dw:1436370152512:dw|

2. Loser66

then $$z^4=-z\\z^5=z^0=1\\z^6=z\\z^7=z^2\\z^8=z^3\\z^9=z^4=-z$$ so I have $$1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4\\=5+5z+5z^2+5z^3+10z^4$$

3. Loser66

I am stuck here. :(

4. Loser66

@Michele_Laino @misty1212 @ganeshie8

5. Empty

Split it up into "vectors in equilibrium" and by that I mean think in terms of: $1+z+z^2+z^3+z^4=0$ But first we can subtract 5 off of all the exponents and combine them: $$1+z+z^2+z^3+4+4z+4z^2+4z^3+5z^4$$ $$5+5z+5z^2+5z^3+5z^4$$ Wait a sec, we can just factor out a 5 to get: $5(1+z+z^2+z^3+z^4)=0$

6. Loser66

No, the answer is $$-5e^{3\pi i/5}$$ but I don't know how to get it

7. Empty

Also $$z^4 \ne -z$$. Hmm oh I see maybe I have messed up in reading it, can you check for me that you've written it correctly as well?

8. Empty

I think what you were probably thinking was: $$z^4 = z^{-1}$$

9. Loser66

|dw:1436372562924:dw|

10. Empty

Ahhh ok in your original question it doesn't look like there's a $$z^4$$ term thank you! Ok let me see now about this new problem.

11. Empty

The only difference I see is that we have a new term, $$5z^4$$ so that will result in the same answer as the last one plus this. So if they are saying that the answer is $$-5e^{i \frac{3 \pi}{5}}$$ we can check by substituting in $$-1 = e^{i \pi}$$, so let's do that: $-5e^{i \frac{3 \pi}{5}} = 5e^{i \pi}e^{i \frac{3 \pi}{5}} =5 e^{i \pi \frac{3+5}{5}}=5e^{i \frac{2 \pi}{5}4}=5z^4$ Hey it works now!

12. Empty

If I skipped too many steps or you'd like me to explain it more, feel free to ask! :D

13. Loser66

If I don't know the answer, how to derive?

14. Loser66

ok, I got it, hehehe but from the last sentence, I have 5(1+z+z^2+z^3+z^4) +5z^4 = 5z^4 =$$\huge5e^{8\pi i/5}= 5e^{5\pi i/5+3\pi i/5}\\\huge=5*e^{\pi i}*e^{3\pi i/5}=-5e^{3\pi i/5}$$

15. Loser66