0

I was trying to answer the following question from Adam's Calculus and have a problem with the hint it provides. I have read the similar threads and my question does not concern the procedure of solving the question, but the hint the book provides. The book takes $w_1=|z|^{1/n}(\cos\frac{\theta}{n}+{i}\sin\frac{\theta}{n})$ to be the principal root, which in this case will be, based on $\theta=0$, $w_1=1$.

Show that the sum of the n nth roots of unity is zero. Hint: Show that these roots are all powers of the principal root.

First, I wrote the polar form of the roots: $$w_1=\cos0+i\sin0$$,$$w_2=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$$, $$...$$, $$w_n=\cos\frac{2(n-1)\pi}{n}+i\sin\frac{2(n-1)\pi}{n}$$

I can easily show that the $w_3,w_4,...,w_n$ are all powers of $w_2$, but not the powers of the principal root, $w_1$.

I have read these two threads: (1) and (2) and know the rest of the calculation. But I cannot understand the hint. How can, for example, $w_2={e}^{{i}\frac{2\pi}{n}}$ be a power of $w_1={e}^{{i}0}$?

Kaveh Rad
  • 101
  • 7
  • 2
    $w_1 = 1$. It's not the principal root. $w_2$ is. – Ethan Bolker Jun 20 '22 at 16:15
  • 1
    If you have read these two threads, everything should be fine. – Wuestenfux Jun 20 '22 at 16:16
  • @DietrichBurde Thanks. My question concerns the hint provided. In the text, the book takes $w_1$ the principal root and according to the formula, I get $w_1=1$ as the principal root. – Kaveh Rad Jun 20 '22 at 16:25
  • https://math.stackexchange.com/questions/1122110/sum-of-nth-roots-of-unity?noredirect=1&lq=1 – Christophe Leuridan Jun 20 '22 at 16:28
  • @EthanBolker But why does the book take $w_1=|z|^{1/n}(\cos\frac{\theta}{n}+{i}\sin\frac{\theta}{n})$ the principal root? I get $\theta=0$. Is this right? – Kaveh Rad Jun 20 '22 at 16:29
  • I doubt that's what the book says. I think the $\theta$ in the formula should be $2\pi$ for the principal root. In any case your question is answered here and in the duplicates. – Ethan Bolker Jun 20 '22 at 16:36
  • @EthanBolker Robert I don't know if you have access to the book, here is the reference: A. Adams, Christopher Essex, Calculus, A Complete Course (2018, 9th Ed.), Appendix I, A9. The threads you are referring to explain the solution. My question here concerns the hint given in this specific question of the book. – Kaveh Rad Jun 20 '22 at 16:49

2 Answers2

1

The principal root is usually taken as $\cos(2\pi/n)+i\sin(2\pi/n)$,so it's your $w_2$. Note that that way $w_1=w_2^0$

Guillermo García Sáez
  • 1
  • Thanks for the response. I was confused by the book taking $w_1=|z|^{1/n}(\cos\frac{\theta}{n}+{i}\sin\frac{\theta}{n})$ the principal root. – Kaveh Rad Jun 20 '22 at 16:22
0

Use Vieta's formula: If $z^{n}+a_{n-1}z^{n-1}+.....+a_{1}z+a_{0}=0$ then setting $a_{0}=-1$ and every other coefficient zero we obtain $z^{n}=1$ and $\sum_{1}^{n}z_{i}=-a_{n-1}=0$ and we are done!