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Suppose that we don’t have the concept of complex numbers. Please forget everything about complex numbers. Here I want to introduce you a special way to get the concept of complex numbers naturally.

Construction

Now, let us try to find the factorization of $x^n-1=0$.

Let $U_n(x)=x^n-1$.
As follows \begin{align} U_2(x)&=(x-1)(x+1)\\ U_3(x)&=(x-1)(x^2+x+1)\\ U_4(x)&=(x-1)(x+1)(x^2+1)\\ &\ \ \vdots \end{align}

Associating each factor of $U_n(x)$ with the distance $d(X,Y)$ between the point $X(x,0)$ and another point $Y$ within the coordinate plane $x$-$y$,we can easily discover that $x-1$ and $x+1$ of $U_2(x)$ correspond to two points -- $(1,0),(-1,0)$.

Furthermore, $x^2+x+1$ of $U_3(x)$ with completing the square is able to transform to $(x+1/2)^2+(\sqrt3/2)^2$,which means there exist two points -- $(-1/2,\sqrt3/2),(-1/2,-\sqrt3/2)$.

Considering the point $(1,0)$ corresponding to $(x-1)$ at the same time,it perfectly matches the three vertices of an inscribed equilateral triangle within the unit circle.So as the case of $U_4(x)$ (the four vertices of a square inscribed in a unit circle).

The rest can be done in the same manner.

Question

Then there is a natural thought that the factors of $U_n(x)$ can be corresponded to $n$ vertices of a regular $n$-sided polygon inscribed in a unit circle. But how can we prove this strictly?

qiu in mountains
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  • While this is a good idea, cyclotomic polynomials would be the wrong term to use here, since they are not always of the form x^n-1=0 – Illusioner_ Mar 31 '25 at 10:47
  • @ Illusioner_ You are right. I will edit the narrative. – qiu in mountains Mar 31 '25 at 10:55
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    Are you looking for something like this? https://mathrefresher.blogspot.com/2008/01/geometric-interpretation-of-roots-of.html – Derek Allums Mar 31 '25 at 11:03
  • Related: https://math.stackexchange.com/questions/3244132/what-is-a-simple-physical-situation-where-complex-numbers-emerge-naturally/3244234#3244234 – Ethan Bolker Mar 31 '25 at 11:12
  • @Ethan Bolker Thank you for your supplementary information! – qiu in mountains Mar 31 '25 at 12:12
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    @Derek Allums Thank you for your supplementary information! – qiu in mountains Mar 31 '25 at 12:12
  • Good idea,but I think you can describe more in details about how to introduce complex numbers. – Rick Z Mar 31 '25 at 14:21
  • "...which means there exist two points..." That is a huge gap, and the only way I can guess what it might mean and how it might make sense is because I know a sensible way to "multiply" a point by itself, which is basically equivalent to the complex numbers. – JonathanZ Mar 31 '25 at 14:43
  • I will also leave a comment asking for clarification: You say you want to get a "concept of complex numbers". What abstract properties do you want to preserve (e.g. field strcture, $\mathbb{R}$-vector space structure, existence of radicals, completeness, ...)? Depending on that the answer might change. And one immediate remark: If you want to e.g. preserve the field structure, then your construction is not yet enough - you need to somehow obtain points like $(1+i)$. – ClemensB Mar 31 '25 at 14:58
  • @JonathanZ You are right. I missed some narrative. Consider the distance between the point X(x,0) and another point Y. Y is the point generated by completing the square. Thus,$U_n(x)=\prod_{i=1}^{n}XY_i$, where $Y_i$ is a vertex of a regular n -sided polygon inscribed in a unit circle(remained to prove). – qiu in mountains Mar 31 '25 at 15:56
  • @Rick Z Thank you for your advice! I will ponder it further. – qiu in mountains Mar 31 '25 at 16:05
  • @ClemensB Thank you for your question. My initial thoughts were simply about the factorization. Furthermore, if we get this, then each vertex $Y_i$ of the regular n -sided polygon inscribed in a unit circle can be mapped to a primary factor, e.g. (1,0)$\mapsto$ (x-1). Then, those quadratic factor which can't be factorized in $ \mathbb{R}$ can be factorized somehow into two primary factors. In this procedure, $i^2=-1$ is constructed. The aspects you mentioned will be the direction of my next thinking. – qiu in mountains Mar 31 '25 at 16:31
  • Your descriptions give no real clue as to how the points $Y$ are derived. Why should $x-1$ correspond to $(1,0)$ while $x+1$ corresponds to $(0,1)$? For $U_3$, one can see how the two $Y$s of the quadratic are derived, but what happens for $U_5$ and higher, where no such algebraic process exists? You really need to nail these $Y$s down, not just give a hand-wavy description. – Paul Sinclair Apr 01 '25 at 19:43
  • @Paul Sinclair It looks to me like the coordinates of the point(s) corresponding to any polynomial are just the coordinates of the representation(s) of its roots on an Argand Diagram. Thus the points corresponding to a polynomial $\ p(x)=ax^2+bx+c\ $ with $\ b^2-4ac<0\ $ are $\ \left(-\frac{b}{2a},\frac{\sqrt{4ac-b^2}}{2a}\right)\ $ and $\ \left(-\frac{b}{2a},-\frac{\sqrt{4ac-b^2}}{2a}\right)\ .$ What isn't clear is how any construction of $\ \mathbb{C}\ $ qui in mountains might have in mind would be any different from something like $\ \mathbb{R}[x]/\langle p(x)\rangle\ .$ – lonza leggiera Apr 04 '25 at 12:46
  • And since the irreducible quadratic factors of $\ x^n-1\ $ are all of the form $\ x^2-2\left(\cos\frac{2k\pi}{n}\right)x+1\ ,$ the corresponding points are $\ \left(\cos\frac{2k\pi}{n},\sin\frac{2k\pi}{n}\right)\ $ and $\ \left(\cos\frac{2k\pi}{n},{-}\sin\frac{2k\pi}{n}\right)\ $—that is a point on the $\ n$-gon inscribed in the unit circle and its reflection in the $\ x$-axis. – lonza leggiera Apr 04 '25 at 13:29
  • @lonzaleggiera - and how does that result in the $x-1$ factor mapping to $(1,0)$, but the $x+1$ factor mapping to $(0,1)$, as described? – Paul Sinclair Apr 04 '25 at 15:41
  • @Paul Sinclair The polynomial $\ x-1\ $ has the single root $\ 1\ ,$ whose representation on an Argand diagram is $\ (1,0)\ .$ I didn't notice until now that my supposition doesn't work for qui in mountains's choice of $\ (0,1)\ $ for the point corresponding to the factor $\ x+1\ ,$ which would correspond instead to $\ (-1,0)\ $ if I were right. My apologies for not reading your previous comment more carefully. However, if qui in mountains were to look more closely at the factors of $\ x^4-1\ ,$ I believe he or she might want to reconsider which point gets assigned to the factor $\ x+1\ .$ – lonza leggiera Apr 04 '25 at 22:21
  • Or even better, the factors, $\ x-1, x+1, x^2+x+1, x^2-x+1\ ,$ of $\ U_6(x)\ .$ The factors $\ x-1\ $ and $\ x^2+x+1\ $ have already been made to correspond with the points $\ (1,0),$$,\left({-}\frac{1}{2},\frac{\sqrt{3}}{2}\right)\ $ and $\ \left({-}\frac{1}{2},{-}\frac{\sqrt{3}}{2}\right)\ ,$ and the same procedure of completing the square used for the factor $\ x^2+x+1\ $ will produce $\ \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\ $ and $\ \left(\frac{1}{2},{-}\frac{\sqrt{3}}{2}\right)\ $ for the points corresponding to $x^2-x+1\ .$ – lonza leggiera Apr 05 '25 at 00:59
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    These five points lie at the vertices of a regular hexagon inscribed in the unit circle. The sixth vertex of this hexagon lies at $\ (-1,0),$ not $\ (0,1)\ ,$ however. So, if qui in mountains wants the factors of $\ U_n(x)\ $ to "corresponded to $\ n\ $ vertices of a regular $n$-sided polygon inscribed in a unit circle" as stated in his or her question, the point corresponding to the factor $\ x+1\ $ needs to be $\ (-1,0)\ ,$ not $\ (0,1)\ .$ – lonza leggiera Apr 05 '25 at 00:59
  • @lonza leggiera My god.This is my problem. I made a typo. You're right.The point corresponding to the factor x+1 is (−1,0).Thank you! – qiu in mountains Apr 27 '25 at 05:42

1 Answers1

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If $\ n=2r\ $ is even, then $$ U_n(x)=(x-1)(x+1)\prod_{j=1}^{r-1}\left(x^2-2\cos\left(\frac{j\pi}{r}\right)x+1\right)\tag{1}\label{1} $$ factorises over $\ \mathbb{R}\ $ into a product of two linear factors, $\ (x-1)\ $ and $\ (x+1)\ ,$ and $\ r-1\ $ irreducible quadratic factors $\ x^2-$$\,2\cos\left(\frac{j\pi}{r}\right)x+$$\,1\ ,$ for $\ j=1,2,\dots,r-1\ .$

If $\ n=2r+1\ $ is odd, then $$ U_n(x)=(x-1)\prod_{j=1}^r\left(x^2-2\cos\left(\frac{2\pi j}{2r+1}\right)x+1\right)\tag{2}\label{2} $$ factorises into a product of a single linear factor, $\ (x-1)\ ,$ and $\ r $ irreducible quadratic factors $\ x^2-2\cos\left(\frac{2\pi j}{2r+1}\right)x+1\ ,$ for $\ j=1,$$\,2,$$\,\dots,r\ .$

If you apply your scheme of completing the square to irreducible quadratics of the form $\ x^2-2\cos\theta x+1\ $ (with $\ 0<\theta<\pi\ ),$ you get $$ x^2-2\cos\theta x+1=(x-\cos\theta)^2+\sin^2\theta\ , $$ and if you now use your scheme for associating pairs of points in the plane with irreducible quadratics, you the get the two points $\ (\cos\theta,\sin\theta)\ $ and $\ (\cos\theta,-\sin\theta)\ $ for this particular quadratic.

Thus, when you apply your scheme to the linear and quadratic factors in \eqref{1}, you get the $\ 2r\ $ points $ (1,0), (-1,0)\ $ and $\ \left(\cos\left(\frac{j\pi}{r}\right),\sin\left(\frac{j\pi}{r}\right)\right),$$\ \left(\cos\left(\frac{j\pi}{r}\right),-\sin\left(\frac{j\pi}{r}\right)\right)\ $ for $\ j=1,2,\dots,r-1\ ,$ which are the vertices of a regular $\ 2r$-gon inscribed in the unit circle.

Likewise, when you apply your scheme to the linear factor and quadratic factors in \eqref{2}, you get the $\ 2r+1\ $ points $ (1,0)\ $ and $\ \left(\cos\left(\frac{2\pi j}{2r+1}\right),\sin\left(\frac{2\pi j}{2r+1}\right)\right),$$\ \left(\cos\left(\frac{2\pi j}{2r+1}\right),-\sin\left(\frac{2\pi j}{2r+1}\right)\right)\ $ for $\ j=1,2,\dots,r\ ,$ which are the vertices of a regular $\ 2r+1$-gon inscribed in the unit circle.

lonza leggiera
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