$z=e^{2\pi i/5}$ solves $1+z+z^2+z^3+z^4=0$

Question

What is the best way to verify that $$1+z+z^2+z^3+z^4=0$$ given $z=e^{2\pi i/5}$?

I tried using Euler's formula before substituting this in, but the work got messy real fast.

Hint: since $z \neq 1$, you can use the geometric sequence formula (this is a standard trick; keep it in your toolbox) — user217285, Apr 01 '16 at 05:45

score 1 · Answer 1 · answered Apr 01 '16 at 05:46

1

Hint: for $z \neq 1$, we have:

$$ \frac{z^5-1}{z-1}=1+z+z^2+z^3+z^4$$

Now see what happens when you let $z=e^{2\pi i/5}$

answered Apr 01 '16 at 05:46

ASKASK

score 1 · Accepted Answer · answered Apr 01 '16 at 05:47

1

Clearly(?) $z^5=1$ and $z\ne 1$. Also, $$z^ 5-1=(z-1)(1+z+z^2+z^3+z^4)$$

answered Apr 01 '16 at 05:47

Hagen von Eitzen

2 Answers2