Why there are exactly $n$ distinct $n$-th roots?

Question

I was solving the following exercise: Given a complex number $z \neq 0$, write $z = re^{i\theta}$ where $\theta = \arg(z)$. Let $z_1 = Re^{i\alpha}$, where $R = r^{1/n}$ and $\alpha = \theta/n$, and let $\epsilon = e^{2\pi i/n}$, where $n$ is a positive integer.

$(a)$ Show that $z_1^n = z$, that is $z_1$ is an $n$-th root of $z$.

$(b)$ Show that $z$ has exactly $n$ distinct $n$-th roots:

$$z_1, \epsilon z_1,\epsilon^2 z_1,\cdots, \epsilon^{n-1}z_1$$

Well, item $(a)$ was very easy indeed. Now letter $(b)$ is confusing me.Showing that each of these is a $n$-th root was very simple. Indeed, if we consider $\epsilon^k z_1$, then we have:

$$(\epsilon^k z_1)^n = \epsilon ^{nk}z_1^n = \epsilon^{nk}z=e^{i2\pi k}re^{i\theta}=re^{i(\theta+2\pi k)} = re^{i\theta}$$

Where the last equality is because $k$ is integer, so $\cos(\theta + 2k \pi) = \cos \theta$ and $\sin (\theta+ 2k\pi) = \theta$. I can't see why there are exactly $n$ distinct roots, it seems to me that the problem is that for $k > n-1$ the roots start to repeat, but I don't know how to show it. For instance, for $k = n$, I cannot see why $\epsilon ^n z_1$ is a repeated root. Is that the problem? Is really about the roots start repeating? Can someone give a hint on how to show this?

Thanks very much in advance.

Answer 1

A nonzero polynomial can have no more roots in a (commutative) field than its degree. Thus when you've found $n$ complex roots of the $n$-th degree polynomial $X^n-z$, there can be no more.

The reason for this is that whenever $a$ is a root of $P\in K[X]$, you can factor $P=(X-a)Q$ (because of Euclidean division by $(X-a)$, the fact that $a$ is a root ensures the remainder is$~0$) and because every other root of $P$ must be a root of$~Q$ (this is the most subtle point; commutativity of the base field is needed here, as well as the absence of zero divisors). Then an immediate induction argument shows that there can not be more that $\deg P$ roots of $P$ in any field (or integral domain) containing the coefficients of$~P$.

In your concrete example the above gives you a factorisation $$ X^n-z=(X-z_1)(X-\epsilon z_1)(X-\epsilon^2 z_1)\ldots(X-\epsilon^{n-1}z_1), $$ and any root of $X^n-z$ must be a root of one of the factors on the right; clearly this means it has to be one of the $n$ roots you found.

Answer 2

Suppose there were more than $n$ of them. Let $r_1,\ldots,r_{n+1}$ be $n+1$ of them. Then all of the first-degree polynomials $w-r_1,\ldots,w-r_n$ would be factors of $w^n-z$. Therefore $(w-r_1)\cdots(w-r_{n+1})$ would be a factor of $w^n-z$. That would make $w^n-z$ a polynomial in $w$ with degree more than $n$. But it's not.

To see why $w-r$ must be a divisor of $w^n-z$ if $r$ is a solution for $w$ of the equation $w^n-z=0$, just divide $w^n-z$ by $w-r$: $$ w^n-z=(w-r)(\cdots\cdots)+a $$ where $(\cdots\cdots)$ is the quotient and $a$ is the remainder. If $a\ne0$, then it's not hard to see how a contradiction follows. (I've omitted some details of the reasoning in algebra, but they're not hard to fill in.)

Answer 3

Suppose you claim that the set $$\{z_1,\ldots,z_n\}$$ gives all $n$-th roots of a number $w$, and those are all there are.

Then you ought to prove that:

$(1)$ If $z_i=z_j$ then $i=j$; or $i\neq j\implies z_i\neq z_j$.

$(2)$ If $v$ is another $n$-th root, then $v=z_i$ for some $i=1,\ldots,n$.

You're are indeed right: the roots do start repeating, so there you have a starting point. This is because $\exp(ix)=\exp(ix+2\pi ki)$ for any integer $k$. If $v^n=w$, note that $n{\,{\rm arg }\,\,v}=w$ and $|v|^n=w$. Use this and $$e^{\theta i}=\cos \theta+i\sin\theta$$ to work out $(2)$.