0

Set $w=\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}$.

I have to calculate:

$$1 + \sum_1^9 w^n$$

I have calculated that the answer is 0.

However, I am supposed to arrive at this conclusion without calculations and am not sure how to. Any ideas? Hints are appreciated over solutions.

Kind regards.

symplectomorphic
  • 18,604
Danny Rancher
  • 103
  • 4
  • Something geometric. – Daniel Fischer Apr 18 '14 at 20:30
  • Hint: $z^{10}-1=(z-1)(1+z+z^2+\cdots+z^{9})$. – symplectomorphic Apr 18 '14 at 20:37
  • In general, the monic polynomial with roots $\cos \frac{2 k \pi}{n} + i \sin \frac{2 k \pi}{n}$ from $k = 1$ to $n-1$ is exactly $x^{n-1} + x^{n-2} + \dots + x + 1$. So $x^9 + x^8 + \dots + x + 1$ is the polynomial with the tenth roots of unity (Except for $1$) as roots, and so when you plug in $x = w$ you get $0$ (since it is a root) – MT_ Apr 18 '14 at 20:50
  • @symplectomorphic could you please explain how you arrived at the hint? I'm having trouble understanding the dividing by $(z - 1)$ – Danny Rancher Apr 20 '14 at 13:42
  • 1
    @DannyRancher: It's just the sum formula for a geometric sequence with terms $1, z, z^2, \ldots, z^9$. This sum is $\frac{a_1(1-r^n)}{1-r}$, where $a_1$ is the first term (1 in this case), $r$ is the common ratio ($z$ in this case), and $n$ is the number of terms (10 in this case). So we have $1+z+z^2+\cdots+z^9=\frac{1(1-z^{10})}{1-z}=\frac{z^{10}-1}{z-1}$. – symplectomorphic Apr 20 '14 at 23:58
  • @symplectomorphic ah, I was confirming your answer using the quotient remainder method... thank you again. – Danny Rancher Apr 22 '14 at 19:06
  • @DannyRancher: well, it wouldn't be hard to see it with synthetic division. – symplectomorphic Apr 23 '14 at 01:48

1 Answers1

1

Note that $\omega$ given is a primitive $10$th root of unity. Then we can easily prove that $\omega^n, n \ne 10$ gives us all other primitive $10$th roots of unity. Then from vieta, we know that the sum over all $10$th roots of unity is $0$ and $1$ is the other $10$th root of unity. So the answer follows.

Sandeep Silwal
  • 8,135