Loser66
  • Loser66
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Loser66
  • Loser66
|dw:1436370152512:dw|
Loser66
  • 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 \(1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4\\=5+5z+5z^2+5z^3+10z^4\)
Loser66
  • Loser66
I am stuck here. :(

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Loser66
  • Loser66
@Michele_Laino @misty1212 @ganeshie8
Empty
  • 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: \[5(1+z+z^2+z^3+z^4)=0\]
Loser66
  • Loser66
No, the answer is \(-5e^{3\pi i/5}\) but I don't know how to get it
Empty
  • 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?
Empty
  • Empty
I think what you were probably thinking was: \(z^4 = z^{-1}\)
Loser66
  • Loser66
|dw:1436372562924:dw|
Empty
  • 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.
Empty
  • 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!
Empty
  • Empty
If I skipped too many steps or you'd like me to explain it more, feel free to ask! :D
Loser66
  • Loser66
If I don't know the answer, how to derive?
Loser66
  • 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 =\(\huge5e^{8\pi i/5}= 5e^{5\pi i/5+3\pi i/5}\\\huge=5*e^{\pi i}*e^{3\pi i/5}=-5e^{3\pi i/5}\)
Loser66
  • Loser66
Thanks for the help. Much appreciate. I will post a new one. :) please help
Empty
  • Empty
Ok!

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