Question

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

Answer 1

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
\(1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4\\=5+5z+5z^2+5z^3+10z^4\)

Answer 2

I am stuck here. :(

Answer 3

@Michele_Laino @misty1212 @ganeshie8

Answer 4

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:

\[5(1+z+z^2+z^3+z^4)=0\]

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

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.

Answer 9

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!

Answer 10

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

Answer 11

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

Answer 12

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
=\(\huge5e^{8\pi i/5}= 5e^{5\pi i/5+3\pi i/5}\\\huge=5*e^{\pi i}*e^{3\pi i/5}=-5e^{3\pi i/5}\)

Answer 13

Thanks for the help. Much appreciate. I will post a new one. :) please help

Answer 14

Ok!