Complex Analysis Math GRE: 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 = ? $

Question

So I came across this problem while studying for the Math GRE, and I was wondering what's a quick trick to solve this (like in the amount of time you have for a problem on the GRE subject test) :

If $$z = e^{2\pi i/5}$$ then $$1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9 = ? $$

Thanks for your help.

Would it be something related to the sum of finite geometric series? — Dispersion, Oct 24 '17 at 02:09
that sounds like an interesting idea, but I wouldn't be too sure on how to do it! knowing the GRE guys there is probably a nice trick for this... — Ali H., Oct 24 '17 at 02:12

score 4 · Accepted Answer · answered Oct 24 '17 at 02:14

4

Hint: consider $$x^5=1$$ now what will be the roots of this equation?

$1, z, z^2,z^3,z^4$

Now the sum of roots of the equation is 0. Use this to solve your expression

answered Oct 24 '17 at 02:14

avz2611

3,658

So something like 1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4 =5+5z+5z^2+5z^3+10z^4 = 5(1+z+z^2+z^3+z^4)+5z^4 – Ali H. Oct 24 '17 at 02:21
oh I see that makes a lot of sense, so would it be -5e^{(3i(pi))/5} ? – Ali H. Oct 24 '17 at 02:23
Thanks again for your hint and explanation – Ali H. Oct 24 '17 at 02:28

score 3 · Answer 2 · answered Oct 24 '17 at 02:11

3

Hint

Just use the fact that $1+z+z^2+z^3+z^4 =0$ over and over.

answered Oct 24 '17 at 02:11

spaceisdarkgreen

58,508

Now I see what you meant thanks to @avz2611; your hint is very good as well thank you! – Ali H. Oct 24 '17 at 02:25
@AliHeydari This is one of the 'main facts' about roots of unity. Another way to see it is that we have $x^n-1 = (x-1)(1+x+x^2+\ldots +x^{n-1}).$ – spaceisdarkgreen Oct 24 '17 at 02:32

Complex Analysis Math GRE: 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 = ? $

2 Answers2