1

Let $a\in \mathbb C$ be non zero. I tried to prove that there exist exactly $n$ different complex numbers $z_k$ such that $z_k^n = a$.

Please could someone tell me if my proof is correct?

We write $a$ in polar coordinates: $a = r e^{i \varphi} = r e^{i \varphi + i 2 \pi} = r e^{i \varphi + i 2 \pi k}$.

Let $z_k = r^{1\over n} e^{{i \varphi + i 2 \pi k \over n}}$. Then it is clear that $z_k^n = a$. We therefore have to show two things: that there are no more than $n$ such numbers and that if $k \neq j$ then $z_k \neq z_j$ for $k,j \in \{0, \dots, n-1\}$.

(i) Assume there are more and let $z_1, \dots , z_{n+1}$ be $n+1$ such numbers. Then $p(z) = z^n - a$ is a polynomial of degree $n$ with $n+1$ roots. This is a contradiction.

This part is my main concern, is it enough to argue like this?:

(ii) If $z_k = z_j $ then we have $ e^{{i \varphi + i 2 \pi k \over n}} = e^{{i \varphi + i 2 \pi j \over n}}$. Equivalently, $ e^{{i 2 \pi k \over n}} = e^{{ i 2 \pi j \over n}}$. But if ${2 \pi k \over n} = {2 \pi j \over n}$ modulo $2 \pi$ then $k = j$ since ${k \over n}, {j \over n} <1$.

Anna
  • 1,757
  • but if you can use the fact that a polynomial of degree $n$ has exactly $n$ roots then why bother with the rest? – benji Mar 01 '15 at 05:39
  • Your proof looks fine, but if I may make a suggestion, I'd echo what @benji said. All you need to do after that is show that there are no repeated roots, which follows by examination of the derivative of $z^n-a$ at the root value. – G. H. Faust Mar 01 '15 at 05:41
  • @benji That's using the fundamental theorem of algebra, which seems rather overkill here. – Pedro Mar 03 '15 at 06:12
  • 1
    @benji in the argument it is only used that there are not more than $n$ roots. This is a lot simpler than the existence of $n$ roots. – quid Mar 03 '15 at 10:35
  • @PedroTamaroff I agree it can be solved without using the theorem but it seems that the proof is using it or rather "half" of it. In any case most likely the intention is to not use it. But then the OP is probably expected to not use "half" of it either.. – benji Mar 03 '15 at 23:59

1 Answers1

2

Your proof is correct.

What the comments are getting at is that by the Fundamental Theorem of Algera every non-zero polynomial (over the complex numbers) has as many roots (counted with multiplicity) as its degree.

If you know this you can argue: $z^n - a$ has $n$ roots counted with multiplicity. Since $z^n -a $ and $(z^n -a)' = n z^{n-1}$ are co-prime, the multiplicity of each root of $z^n-a$ is $1$ and thus it has $n$ distinct roots.

However, I doubt this is the intended solution. In addition, it is less constructive than your solution, which is thus also potentially preferable for that reason.

You could avoid any appeal to properties of polynomials by arguing (i) differently, but your argument seems also fine.

quid
  • 42,135