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I know that there is a Theorem which says that a Polynom of Degree n has at most n Solutions, however we have not proved it yet in our class. Is there Maybe another explaination for this Special case?

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    $\begingroup$ induction, starting from a line. next power of polynomial is the previous one multiplied by another monic, degree one polynomial, so QED. $\endgroup$ – K.K.McDonald Nov 30 '18 at 22:55
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Additions in $\color{blue}{\textrm{blue}}$

$\color{blue}{\textrm{Suppose that}~ z^n-1 = 0~ \textrm{has}~m > n~\textrm{solutions}~ a_1, a_2, \ldots, a_m}$.


Suppose that $a_1$ is a solution of $\require{enclose}\enclose{horizontalstrike}{z^n - 1 =0}$. Then: $\color{blue}{\textrm{Consider the solution}~a_1~\textrm{. Then:}}$

$$z^n-1 = (z-a_1)p_1(z) = zp_1(z)-a_1p_1(z),$$

where $p_1(z)$ is a polynomial. Since the degree of $zp_1(z)-a_1p_1(z)$ should be equal to $n$, then $p_1(z)$ has degree $n-1$.

Now, suppose that $\require{enclose}\enclose{horizontalstrike}{a_2}$ is another solution. Then: $\color{blue}{\textrm{Now, consider the solution}~a_2.~\textrm{Then:}}$

$$z^n-1 = (z-a_1)(z-a_2)p_2(z)=(z^2 \ldots)p_2(z).$$

This time, $p_2(z)$ should have degree $n-2$.

In general, given $k$ solutions $a_1, a_2, \ldots, a_k$, we ca write:

$$z^n-1 = p_k(z)\prod_{i=1}^{k}(z-a_k),$$

where the degree of $p_k(z)$ is $n-k$. Of course, this can be reiterated up to $p_n(z)$, which has degree $n-n = 0$, i.e. it is $p_n(z) = p,$ a constant. Formally:

$$z^n - 1 = p\prod_{i=1}^{n}(z-a_n).$$

Then, you have $n$ solutions $a_1, a_2, \ldots, a_n$. Of course, if some $a_i$ coincide, then you have at most $n$ solutions.

$\color{blue}{\textrm{In conclusion, the number of solutions cannot be}~m>n, \textrm{but}~ m\leq n}$.

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    $\begingroup$ @gimusi they don't assume that a polynomial of degree $n$ has at most $n$ roots. They proof it for this particular one. Granted the argument they use for the special case essentially shows the full result in that sense this answer is not a different explanation. However it seems more plausible to me than a misunderstanding that OP wanted an argument that does not presuppose that result, and just that rather then an substantially different argument. (What they may not have realized is how easy the proof of that result is.) $\endgroup$ – quid Dec 2 '18 at 0:43
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    $\begingroup$ @the_candyman No in my opinion you don't need to remove that, it's up to the asker evaluate things, revise its work and then ask for clarifications if needed. You could revise it a littele bit giving some clarification of course but also that it's up to you. Bye $\endgroup$ – gimusi Dec 2 '18 at 13:04
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    $\begingroup$ @the_candyman What is really sad for the reliability of the site is that some answers (maybe wrong or maybe correct) have been downvoted but other answers (I suppose not adressing the OP) have been upvoted or not downvoted. The only explanation I can find is a targeted downvoting and I can also accept that. But in any case we should work in such way that a good answer remains here for the asker and for futur users. $\endgroup$ – gimusi Dec 2 '18 at 13:46
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    $\begingroup$ @the_candyman Now it seems fine to me. $\endgroup$ – gimusi Dec 2 '18 at 20:17
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    $\begingroup$ @the_candyman I'm here mostly to learn and I'm always happy when we can discuss honestly about our thoughts and effort! Thanks for your efforto to improve your answer. Bye $\endgroup$ – gimusi Dec 2 '18 at 20:20
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By the factor theorem , $f(z)$ is divisible by $(z-r)$ for each root $r$.

If there were more than $n$ roots, $r_1,\dots,r_k$, then $f(z)=p_k(z)(z-r_1)\dots(z-r_k)\implies \operatorname{deg}f\ge k\gt n$.

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  • $\begingroup$ I think you've got a missing equals sign (I couldn't insert it for you since I can't save a 1-character edit!) $\endgroup$ – timtfj Dec 1 '18 at 0:27
  • $\begingroup$ You probably mean after $p_k(z)$. Actually I intended it that way: I'm trying not to assume anything. That $f$ is the product of the $(x-r_i)$ might not be clear. For instance, I don't want to use FTA. (Maybe a little silly. ) $\endgroup$ – Chris Custer Dec 1 '18 at 0:38
  • $\begingroup$ Ah, I see. Thanks $\endgroup$ – timtfj Dec 1 '18 at 1:28
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Because it is a non-constant, single-variable, polynomial with complex coefficients of degree $n$ and the fundamental theorem of algebra says it has $n$ roots in $\mathbb{C}$. Another way to say that is complex numbers is algebraically closed. By a successive factorization argument you can then show that the polynomial has exactly $n$ roots.

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    $\begingroup$ But the FTA is just what the OP has already mentioned, we are looking for a different approach to show that in this particular case $z^n=1$. $\endgroup$ – gimusi Dec 1 '18 at 19:35
  • $\begingroup$ The FTA is about existence of roots of polynomials with complex coefficients. The fact that the number of roots cannot exceed the degree is a completely independent algebraic fact. $\endgroup$ – egreg Dec 2 '18 at 21:23
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There's an explanation if you represent them in polar co-ordinates and consider that multiplying two complex numbers involves adding their arguments (ie angles) and multiplying their magnitudes (distances from the origin). It turns out they need to have $1$ as their magnitude and be multiples of$\frac{360°}{n}$ apart on the resulting circle, so only $n$ of them will fit.

(This is effectively the same as gimusi's answer.)

Edit: In fact, there are exactly $n$ of them, and they're equally spaced round the circle. The reason should be obvious if you pick one of the candidate angles and multiply it by $n$.

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    $\begingroup$ Thanks. I don't feel I understand a piece of maths until I can explain it in words, so I'm trying to do that! I think the symbols are just a means of manipulating the cincepts, and I'm trying to get better at understanding the conceots. $\endgroup$ – timtfj Dec 1 '18 at 1:18
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First fact: for complex (nonzero) polynomials $f(z)$ and $g(z)$, the degree formula holds: $$ \deg(f(z)g(z))=\deg f(z)+\deg g(z) $$ where $\deg$ denotes the standard polynomial degree.

Proof. If we write $$ f(z)=az^m+f_0(z),\qquad g(z)=bz^n+g_0(z) $$ where $f_0$ and $g_0$ group together the lower degree terms, $a\ne0$ and $b\ne0$, then $$ f(z)g(z)=abz^{m+n}+h(z) $$ where again $h(z)$ has degree less than $m+n$. Thus $f(z)g(z)$ has degree $m+n$. QED

Second fact (basic and well known: if $a$ is a root of the polynomial $f(z)$, then $f(z)$ is divisible by $z-a$.

Now we prove by induction on the degree of $f(z)$ the following statement.

Let $f(z)$ be a nonzero polynomial with coefficients in $\mathbb{C}$. Then the number of distinct roots of $f$ cannot exceed the degree of $f$.

The statement is obvious for polynomials of degree $1$. Assume we know it for polynomials of degree $n-1$. Let $f(z)$ have degree $n$ and let $a_1,a_2,\dots,a_m$ be its pairwise distinct roots. By the second fact, we have $f(z)=(z-a_m)g(z)$ and $g(z)$ has degree $n-1$. Now, for $k=1,\dots,m-1$, $$ f(a_k)=(a_k-a_m)g(a_k)=0 $$ and, since $a_k-a_m\ne0$, we conclude $g(a_k)=0$. Therefore $a_1,\dots,a_{m-1}$ are pairwise distinct roots of $g(z)$. By the induction hypothesis, we have $$ m-1\le n-1 $$ and therefore $m\le n$. QED

Note. This proof applies with no change to polynomials having coefficients in an arbitrary domain.

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We have that

$$z^n=1\iff z^n=e^{i2k\pi}$$

and then

$$z_k=e^{i\frac{2k\pi}n} \quad k=0,\ldots,n-1$$

and those are the $n$ roots of unity.

It is trivial to see that the solution is periodic since $e^{i2kπ/n}$ for $k=k_0$ and $k=k_0+n$ represents the same complex number.

Refer also to roots of unity for details.

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    $\begingroup$ This doesn't really explain anything at all. You've shown $n$ solutions, but there's zero reasoning about why there aren't $n + 1$ solutions. In other words, your $\iff$ basically is just a rephrasing of what the asker already knows. $\endgroup$ – T. Bongers Dec 1 '18 at 0:27
  • $\begingroup$ I certainly agree that it's easy to see that $e^{2\pi i k / n}$ satisfy $z^n - 1 = 0$, but I stand by my statement that you haven't shown that these are the only roots of unity. In other words: you have not included any explanation for why a polynomial of degree $n$ has at most $n$ roots. So you've only got half a solution, compared to the other four answers that do address this issue. $\endgroup$ – T. Bongers Dec 1 '18 at 6:02
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    $\begingroup$ So for the briefest possible explanation of my issue with this answer: The question asks why there are at most $n$ roots of unity. You've (not quite, but almost) shown that there are at least $n$ roots of unity. $\endgroup$ – T. Bongers Dec 1 '18 at 6:03
  • $\begingroup$ @T.Bongers I'm assuming that the asker already knows the topic and in particular the complex roots of unity. Once we have given a suggestion for that he/she is able to conclude by him/herself the doubt "Is there Maybe another explaination for this Special case?". Anyway now I've also add what I've already pointed out in the comments. $\endgroup$ – gimusi Dec 1 '18 at 8:17
  • $\begingroup$ My informal "only $n$ of them will fit" and gimusi's remark about the solution being periodic are saying the same thing. But neither of us has quite said explicitly that a repeated root only counts as one root (ie that we're not using the trick of turning the complex plane into a Riemann surface, if I've remembered its name right.) $\endgroup$ – timtfj Dec 1 '18 at 13:10

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