Talk:Root of unity/Archive 1

message: Note that, contrary to first appearances, not all coefficients of all cyclotomic polynomials are 1, −1, or 0; the first polynomial where this occurs is Φ105, since 105=3×5×7 is the first product of three odd primes.
:It's not immediately clear to me why this property of 105 implies it's the first cyclotomic polynomial with coefficient not equal to 1, -1, or 0. Is this really easy to see, or something I've just forgotten? If it's pretty easy to show, then a quick justification would be good, if it's not very easy to show, I don't think the fact should be mentioned, it leaves the reader dangling.
It's because the pattern for cyclotomic polynomials is more complicated for numbers that are the product of three or more different odd primes and/or their powers (the smallest being, as you've stated, 105) than it is for numbers that are the product of two or fewer. Notice that a power of two doesn't make the polynomial more complicated; this simply means that the coefficients of odd powers of x will have the opposite sign (from -1 to +1 or +1 to -1). Glenn L 23:24, 23 June 2007 (UTC)
thread_title:  Cyclotomic polynomials 

message: The case  n=0  is included in the definition, and it is stated that there are  n  solutions to  z^n=1 . Actually the equation  z^0=1  has infinitely many solutions, rather than zero. I'll exclude  n=0  . Bo Jacoby 14:03, 20 September 2005 (UTC)
thread_title:  ''n''=0 

message: I'd be happier is the notation 1^{\frac{k}{m}} was introduced before it was used.  That is a highly non-standard notation; I've never seen anything like that before. I assume its historical in some way? Is there a reference for this notation? linas 16:10, 23 September 2005 (UTC)
replies: text: Hi Linas. You are right of cause, the definition must precede the use. I'll fix it. I have used \ 1^x instead of \ e^{2\pi i x} for many years as a simple ink-saving notation, but frankly I do not remember from where it came. Perhaps I have invented it myself. I tend to invent things now and then. If \ e^{2\pi i }=1, then why not write \ 1 instead of \ e^{2\pi i}  ? It is really no big deal. It could simplify the confusing notation for fourier transforms. The alternative interpretation, \ 1^x=1 , is uninteresting. Confusion is unlikely to occur because nobody will write \ 1^x meaning \ 1. See exponentiation. Bo Jacoby 10:14, 26 September 2005 (UTC)
replies: text: I've removed this notation.  I agree that it is nonstandard.  (I tend to think it is also misleading, because it gives the impression that there is a unique choice of primitive root 1^{\frac{1}{m}}, when in fact this is wholly arbitrary.)  Wikipedia should not be in the business of inventing major new notations from whole cloth.  —Steven G. Johnson 18:51, 26 September 2005 (UTC)
replies: text: 1^{\frac{1}{m}} is not an arbitrary primitive root, but is equal to e^{\frac{2\pi i}{m}} which is well defined. To the readers most of the material in Wikipedia is new, and of cause they don't mind. By deleting material just because is new to you, you prevent Wikipedia from exceeding your level. Think about it. Get wiser. Bo Jacoby 09:07, 27 September 2005 (UTC)
replies: text: Hey, please, be nice. Personal notations are not good here, and that is a solid matter of policy. Charles Matthews 10:52, 27 September 2005 (UTC)
replies: text: I'm sorry. But what is the policy of Wikipedia: that everybody can contribute, or that a censor removes anything that he hasn't seen before simply because he hasn't seen it before ? Bo Jacoby 06:39, 28 September 2005 (UTC)

text: The policy in cases of content disputes is that matters are resolved by discussion, in an atmosphere of mutual respect. The policy in the matter of innovation is that WP is not a place to innovate. Charles Matthews 07:35, 28 September 2005 (UTC)
thread_title: Notation

message: ===It can be proved in any number of ways===
#Is is really important to stress that there are more ways to prove it? I think not. 
#Nor do I think that it is true that it can be proved in essentially more than one way. Please provide an alternative proof.  Bo Jacoby 10:09, 27 September 2005 (UTC)
replies: text: Proof #2: the roots of unity are eigenvectors of the discretized Laplacian with periodic boundary conditions, which is Hermitian and therefore has orthogonal eigenvectors.  Proof #3: multiply the sum S of the roots of unity by one of the roots of unity z; by the rearrangement theorem you get the same sum in a different order; thus, Sz = S and therefore S = 0.   Proof #4: the group theoretic proof already mentioned, based on representation theory.  See also Charles' proofs below.  The existence of the various proofs of the orthogonality of the roots of unity is indeed important, because it means that this orthogonality appears in different ways in many branches of mathematics.  —Steven G. Johnson 23:16, 27 September 2005 (UTC)
replies: text: The proofs: geometric by barycentre of regular n-gon, algebraic by coefficient of xn − 1, complex number by ζS = S — I think these are different ways. By the way, the point is that we may include material from books, but don't have to. Charles Matthews 15:00, 27 September 2005 (UTC)

text: Well, you can cut it out. 'Logical independence' of proofs isn't a useful idea, I think. Most people recognise the idea that there can be proofs from different starting points. You are wrong about it being irrelevant. WP's articles aim for completeness, so certainly mentioning how one topic is seen in different fields is relevant. Charles Matthews 13:09, 29 September 2005 (UTC)
thread_title: Properties

message: #Is is really important to stress that there are more ways to prove it? I think not. 
#Nor do I think that it is true that it can be proved in essentially more than one way. Please provide an alternative proof.  Bo Jacoby 10:09, 27 September 2005 (UTC)
replies: text: Proof #2: the roots of unity are eigenvectors of the discretized Laplacian with periodic boundary conditions, which is Hermitian and therefore has orthogonal eigenvectors.  Proof #3: multiply the sum S of the roots of unity by one of the roots of unity z; by the rearrangement theorem you get the same sum in a different order; thus, Sz = S and therefore S = 0.   Proof #4: the group theoretic proof already mentioned, based on representation theory.  See also Charles' proofs below.  The existence of the various proofs of the orthogonality of the roots of unity is indeed important, because it means that this orthogonality appears in different ways in many branches of mathematics.  —Steven G. Johnson 23:16, 27 September 2005 (UTC)
replies: text: The proofs: geometric by barycentre of regular n-gon, algebraic by coefficient of xn − 1, complex number by ζS = S — I think these are different ways. By the way, the point is that we may include material from books, but don't have to. Charles Matthews 15:00, 27 September 2005 (UTC)

text: Well, you can cut it out. 'Logical independence' of proofs isn't a useful idea, I think. Most people recognise the idea that there can be proofs from different starting points. You are wrong about it being irrelevant. WP's articles aim for completeness, so certainly mentioning how one topic is seen in different fields is relevant. Charles Matthews 13:09, 29 September 2005 (UTC)
thread_title: It can be proved in any number of ways

message: Bo,
Please stop using the 1^{k\over n} notation until such time as you can provide a book reference for this notation. After the discussion above, I am surprised that you put it back in again. linas 20:33, 30 September 2005 (UTC)
replies: text: Well, Linas, I didn't find the discussion decisive. You requested a compromise, and I considered "the notation is convenient though nonstandard" a compromise. Wasn't PAR too in favour for the notatation? Bo Jacoby 06:32, 3 October 2005 (UTC)
replies: text: It's not a standard notation. It doesn't belong on WP, under policy. Charles Matthews 07:00, 3 October 2005 (UTC)
thread_title:  Notation, again 

message: Now please explain to me why the case n=0 is included when in fact it is not defined? My change of \mathbb{N}_0 to \mathbb{N} was revoked. Bo Jacoby 06:32, 3 October 2005 (UTC)
replies: text: Bo, no one has any problem with minor clarifications like this.  The problem is that you mix them in with large-scale changes of notation in the article, which are hard to undo except by reversion...then your littler changes get lost in the shuffle.  It also doesn't help that you react with moral outrage to every perceived slight.  —Steven G. Johnson 05:09, 6 October 2005 (UTC)
thread_title:  n=0 

message: Steven. The nth roots of unity can be arranged to form an n \times n matrix whose (j,k)th entry is 
:U_{j,k}=n^{-\frac{1}{2}} \omega_n^{jk}
is a lie. The elements of the matrix are not the roots of unity. I keep correcting it and you keep repeating it. Please argue in the discussion forum before you insist on removing my edits. Bo Jacoby 08:55, 5 October 2005 (UTC)
replies: text: Um, the article doesn't state that the entries of the matrix are roots of unity.  It clearly gives the sqrt(n) factor, and the matrix is indeed based on an arrangement of the roots of unity.  If you want to be pedantic about the word "arranged" (even though the equation makes things perfectly clear, whereas English is more flexible than you seem to think), fine, I replaced it with "used".  Throwing around accusations of "lies" only makes you look foolish.  —Steven G. Johnson 17:07, 5 October 2005 (UTC)
replies: text: By the way, your edit was reverted not because of the minor changes in the text, but because of an unhelpful wholesale change of notation.  This was clearly explained in my edit comments.  —Steven G. Johnson
The reader has bigger difficulties than the author in understanding the text. So the author should lend ear to the reader's comments. As a reader I changed several formulations which were simply wrong, such as the case n=0. But you reverted everything - not merely what you disagreed on, but everything. Do you realize know how many times you have done that? Please stop it. It is a backwards step instead of a forward step. I edited for a reason. A word that confuses rather than clarifies should be removed. The word arrange is explained in the Longman Dictionary of Contemporary English: to put into a correct, pleasing or desired order. The examples given are: To arrange flowers in a vase, and: The books are arranged on the shelves in alphabetical order. Please don't argue that it can ever mean divide by the square root of the number of items, and then copy them to fit into a square matrix. Any reader is pedantic about words until he gives up reading. Please don't argue ad hominem that I am specially pedantic and therefore should be taken less seriously. I would not be this foolishly indiplomatic if I had seen diplomacy work. But to your surprise I wish to say that I do enjoy working with you in improving WP. It is fun and it is worth while. Let's be friends.  Bo Jacoby 09:33, 6 October 2005 (UTC)

text: Bo, I apologize for any ad hominem comments.  On your part, please avoid charged language like "lies" etcetera.  As I mentioned above, the reason for the reverts were because you mix your small corrections in with wholesale changes of notation that I could not agree with, and the only practical way to undo such a large change is a revert.  I've tried to scan through the diff to see if you made other substantive changes that I should try to keep, but it is easy to miss little changes of wording.  —Steven G. Johnson 15:40, 6 October 2005 (UTC)
replies: text: (PS. If you arrange flowers in a vase, that doesn't mean that you haven't put anything else into the vase.  That doesn't mean I mind changing the wording, but I do think you overreacted. —Steven G. Johnson)
I agree that my charged language make me look like a fool and I am not happy about it. I apologize. Bo Jacoby 12:26, 7 October 2005 (UTC)
thread_title:  orthogonality 

message: Not quite there, but certainly this page is heading in the direction ... Charles Matthews 20:07, 5 October 2005 (UTC)
thread_title: [[Wikipedia:Lamest edit wars ever]]

message: It was mentioned that "It can be shown that they are located on the unit circle of the complex plane and that in that plane they form the vertices of a n-sided regular polygon with one vertex on 1.". Could somebody please show it? Thanks a lot.
:If x is an n th root of unity, then xn=1. Take the complex conjugate of that equation to see that the conjugate x* is a root of unity: (x*)n=1. Multiply these two equations to see that the positive real number xx*=|x|2 is a root of unity. So xx*=1, because 1 is the only positive real root of unity. So |x|=1. That means that x is on the unit circle. This was the first part of the assertion. Now consider the regular n-sided polygon with center in 0 and one of the vertices in 1. Consider the first vertex A in clockwise direction from 1. The triangle (0;1;A) is similar to triangle (0;A;A2) according to Complex_number#Geometric_interpretation_of_the_operations_on_complex_numbers. So A2 is the 2'nd vertex on the polygon. And (by induction) Ak is the kth vertex on the polygon. The numbers A, A2, ... , An=1. are n'th roots of unity if A is. This was the second part of the assertion. Bo Jacoby 15:29, 30 November 2005 (UTC)
replies: text: I suspect "show it" may have been a request for an illustration, although the proof might be good to add too. Doops |  talk 18:31, 30 November 2005 (UTC)
thread_title:  Geometrical Representation of Roots 

message: A common notation for e^{2 \pi i /n} I've noticed in many books is \zeta_n  or just \zeta. Should this be mentioned here or is it rarer notation than I think? I daren't edit this page for fear of starting a lame edit war :)
Moxmalin 22:44, 1 April 2007 (UTC)
replies: text: Edit wars are made by disrespectful editors who do not follow the rules: Help:Reverting#When_to_revert, not by those who in good faith improve the articles. Be bold and edit. Bo Jacoby 09:44, 2 April 2007 (UTC)
thread_title:  Notation 

message: Yes, it's pedantic -- I know; but as far as I know the convential definition requires unities to coincide (i.e. considers selection of unity to be an operation).  In any case, the interesting case is when the unity is stable.  --VKokielov 18:39, 4 June 2007 (UTC)
thread_title:  pedantry 

message: Because the roots of unity are algebraic, and because they are given by the trigonometric functions sine and cosine, it follows that the sine (or cosine) of any angle which is a rational fraction of the revolution 2π must be computable by a finite number of additions, multiplications, and square roots.
Even if I'm right, I don't want to stick it in out of context.  --VKokielov 17:12, 5 June 2007 (UTC)
replies: text: Square roots!..  I'm thinking of powers of two.  If the root isn't a power of two, then these needn't be square roots at all.  --VKokielov 17:39, 5 June 2007 (UTC)
replies: text:  I think that is true, but the fact that the numbers are algebraic is not enough - since not all algebraic integers are expressible by a finite number of additions, multiplications, and roots. (See Abel–Ruffini theorem). Fortunately, however, in the case of roots of unity, they can, (see referenced article above), but again, this is a property beyond that of just being algebraic.--Fangz 20:49, 20 June 2007 (UTC)
replies: text: This seems to be disproven by EJ's argument above (assuming its correctness, which I'm not qualified to check). Tesseran 06:16, 28 July 2007 (UTC)
thread_title:  My addition 

message: Quote from the subsection on Orthogonality: "The roots of unity appear as the eigenvectors of Hermitian matrices". I think this is incorrect and perhaps should be: "The roots of unity appear as the eigenvectors of Unitary matrices". The eigenvalues of Hermitian matrices are real numbers. Bo Jacoby 10:38, 1 August 2007 (UTC)
replies: text: No, it's correct as-is, although a little vague.  If one constructs any translation-invariant Hermitian matrix (invariant under cyclic shifts), e.g. a discrete Laplacian with periodic boundaries (as in the Strang reference), the eigenvectors are (or can be chosen to be) the roots of unity (i.e. it is diagonalized by the DFT).  This is a consequence of group representation theory, and is essentially an instance of Bloch's theorem.  (Actually, I don't think the matrix has to be Hermitian, although this was Strang's example.  It just has to be non-defective and translation-invariant.  However, from a Hermitian matrix like the discrete Laplacian where the eigenvalues are non-degenerate, it is easier to see orthogonality as a consequence.)  The eigenvalues of a Hermitian matrix are real. The eigenvectors are not in general. —Steven G. Johnson 17:25, 1 August 2007 (UTC)

text: As long as there is a reference for where to go for more information, brief mentions of more advanced topics are fine, and in fact desirable.  Encyclopedias, by their nature, should be encyclopedic, and have a very different goal from, say, a minimalist axiomatic development. —Steven G. Johnson 22:55, 1 August 2007 (UTC)
thread_title:  Hermitian matrices 

message: The specific application of roots of unity to Fourier analysis is described in detail by discrete Fourier transform, and we also have specific articles on aliasing etc.  This page should refer to those pages, but there is no reason to have a separate description of computing Fourier components, aliasing, etc. in this page.  Hence I am removing that information.  (Note that the longish discussion in the current "computing Fourier components" is merely a restatement of the fact that the matrix is unitary, and is especially redundant.)  —Steven G. Johnson 15:57, 7 August 2007 (UTC)
thread_title:  We have other pages on Fourier analysis 

message: I restored the examples because I believe they communicate the mechanics of basic computing with roots of unity and cyclotomic polynomials. Others may differ, of course, which is why I started this thread on the talk page. -Zahlentheorie (talk) 21:48, 26 December 2007 (UTC)
thread_title:  Examples 

message: Is it worth putting the following example (extracted from ) of a seventeenth root of unity?
 x + i\sqrt{1-x^2}
where
x =
\frac{
 -1 + \sqrt{17} + \sqrt{2(17 - \sqrt{17})} 
 + 
 2 \sqrt{17 + 3\sqrt{17} - \sqrt{2(17-\sqrt{17})} - 2\sqrt{2(17 + \sqrt{17})}}
}{16}
I haven't verified that this is correct, or that it is the simplest form possible.  It may be worth mentioning because it only involves taking square roots, which is due to a famous result of Gauss (on the constructibility of regular polygons by ruler and compass).  Perhaps also interesting is that the square roots being taken here are positive square roots of positive real numbers.  —Preceding unsigned comment added by DRLB (talk • contribs) 21:39, 4 January 2008 (UTC)
replies: text: See the formula in the article Heptadecagon. 03:38, 5 January 2008 (UTC)
thread_title:  Seventeenth root of unity 

message: * who will read this busy stuff. Remove all the heavy stuff - try to do with examples.
--Tangi-tamma (talk) 01:28, 6 April 2008 (UTC)
thread_title: Needs re-writing

message: I wrote a subsection on notation, but User:Stevenj reverted it immediately. (He is allergic to 11/n, no matter whether it means 1 or e2πi/n, and he shows no respect for the rules of WP:Reverting). Somebody else please write a subsection on notation. Bo Jacoby (talk) 11:16, 20 October 2008 (UTC)
replies: text: You have not been able to cite a single reputable source that uses the 11/n notation (i.e., \sqrt{1}=-1) for primitive n-th roots of unity.  As has been repeatedly pointed out to you, your repeated attempts to promote this notation of yours on Wikipedia is against policy.  Hence, it is perfectly justifiable to revert it on sight when you periodically try to sneak it back in.  —Steven G. Johnson (talk) 15:56, 20 October 2008 (UTC)
thread_title:  Notation 

message: I can't find an explicit calculation of the ith root of 1. It's not at Exponentiation#Complex roots of unity, either. FWIW, given that  = –i, then 1() is equivalent to 1–i, and thus to  as well. — Loadmaster (talk) 15:57, 4 March 2009 (UTC)
replies: text: e^{\frac{2\pi i}{i}}=e^{2\pi }\approx 535.4916555247\ldots  --84.228.133.241 (talk) 12:54, 29 October 2009 (UTC)
replies: text: Complex exponentiation is multivalued: 1^{1/i}=e^{2\pi ni/i}=e^{2\pi n}\, for any integer n. Note that this is exactly the same set of values as for 1i.—Emil J. 12:33, 29 March 2010 (UTC)
replies: text: ... because (1^i)^i=1^{(i^2)}=1^{-1}=1 Gandalf61 (talk) 12:39, 29 March 2010 (UTC)
thread_title:  ''i''th root of 1 

message: There is a mistake in the examples section it says the negative of a root is always also a root and gives the example: (-1/2+i Sqrt(3)/2) and (1/2+i Sqrt(3)/2) as the two primitive cube roots of 1, however, it should read that the root and its Complex Conjugate are always roots because (-1/2+i Sqrt(3)/2)^2 is the other primitive root and this is in fact (1/2-i Sqrt(3)/2). --Kyle  —Preceding unsigned comment added by 99.48.65.242 (talk) 22:13, 19 August 2010 (UTC)
replies: text: It doesn't say that. The article was correct. RobHar (talk) 23:07, 19 August 2010 (UTC)
thread_title: Untitled comment

message: :\sum_{k=0}^{n-1} z^k = \frac{z^n - 1}{z - 1} = 0 .  
How does that equal 0?
For example, if Z is 20 and n is 2.  Then 399/19 = 21, not 0.
Also  z^{n} - 1 = (z-1) * sum_{k=0}^{n-1} z^k is much better looking.
:It equals zero because z is assumed to be an nth root of unity, hence zn = 1. (Of course, 20 is not a root of unity). RobHar (talk) 05:02, 22 August 2010 (UTC)
thread_title: Summation Section

message: There are the two articles:
* Root of unity about complex roots of 1
* Primitive root modulo n about elements in Z/nZ that generate the multiplicative sub-group of Z/nZ
Now I like to write something about numbers a in Z/nZ with a^m = 1,
where m is chosen to be minimal.
I think this is called a primitive root of unity of order m in Z/nZ.
The Root of unity article mentions this briefly in the introduction.
But the article seems to be so much concerned with complex numbers that I hesitate to add a quite different topic to it.
Also adding to Primitive root modulo n may increase confusion about the term primitive root.
I tend to start a new article.
Any opinions? Suggestions for an article name?
HenningThielemann (talk) 22:04, 17 February 2011 (UTC)
replies: text: As you say, this article is about complex roots of unity. There doesn't appear to be an article discussing the concept in other fields, particularly those of characteristic different from 0. (Root of unity modulo n comes relatively close, but isn't the same.) It would be a nice addition if someone would start such a page. I came here looking for a standard definition of a primitive nth root of unity that worked in a field of characteristic p dividing n. I know one can just say it's a generator of the multiplicative group of nth roots of unity in the relevant field (or group of units, I suppose), but that seems to gloss over the inherent separability difference between this case and all others. This page seems cluttered enough that a brief blurb and a reference to a separate article would probably be best. 67.158.43.41 (talk) 16:41, 29 March 2011 (UTC)
thread_title:  Primitive roots of unity modulo n 

message: This article's section on cyclotomic polynomials is nearly the same length as the article Cyclotomic polynomial. It seems like this material should be very briefly summarized here and merged into that article. (IMO most of it isn't really on topic anyway--for instance, what does it matter to this topic that the first cyclotomic polynomial with a coefficient different from 0, 1, or -1 is the 105th?) 67.158.43.41 (talk) 16:52, 29 March 2011 (UTC)
replies: text: From looking at the history of the "Cyclotomic polynomial" article it looks like for many years it was simply a redirect to this article, so this was the original location for this content. In November 2008, someone created "Cyclotomic polynomial" as its own article and I guess no one removed the content here. I encourage you to prune (rather seriously) the content on cyclotomic polynomials present here! RobHar (talk) 19:05, 29 March 2011 (UTC)
thread_title:  Cyclotomic polynomials 

message: I was trying to verify my own results for the 7th roots of unity, calculating 2*cos(Pi/7) as the solution of the cubic equation,    x^3 + x^2 - 2x - 1 = 0  whose three roots are x=2\cos\frac{2n\pi}{7}.
I found a typo in the posted result online here.  The result I get, which appears correct, is
 \cos(\frac{2\pi,4\pi,6\pi}{7}) = \frac{-1 + \sqrt[3]{\frac{7+21\sqrt{-3}}{2}} + \sqrt[3]{\frac{7-21\sqrt{-3}}{2}}}{6}
The posted equation is incorrect.  The posted equation has a +1 instead of a -1, and a -7 instead of a +7.  The author solved  x^3 - x^2 - 2x + 1 = 0, which gives the solution for \cos(\frac{\pi,3\pi,5\pi}{7}) = \frac{1 + \sqrt[3]{\frac{-7+21\sqrt{-3}}{2}} + \sqrt[3]{\frac{-7-21\sqrt{-3}}{2}}}{6}, which is incorrect.
I didn't originally intend to calculate or verify the equations for the imaginary portion of the equation  i\sin(\frac{2n\pi}{7}), but then I couldn't resist finding a more brief equation.  \sin^2(\frac{2\pi}{7})=\frac{1-\cos(4\pi/7)}{2} which leads to \exp(\frac{2\pi i}{7})= \cos(\frac{2\pi}{7}) + \frac{i}{2}\sqrt{2 - 2\cos(\frac{4\pi}{7})}\;\;.  This is a general equation, applicable to all roots of unity radicals, which usually involve calculating 2\cos(2\pi/p).  By that time, 2\cos(4\pi/p) is also readily available, so it makes sense that this is probably the best approach to calculating the imaginary part of \exp(2\pi i/p).  For the case at hand, we are solving a cubic equation.  We use s1, and s2, which are complex cuberoots of 1.  The s1,s2 terms are to force the cube roots of the complex numbers on the right to be the approriate other cube root, corresponding to the equation on the left, so that the cubic equation result on the right is for 2\cos(4\pi/7).  Assuming my math is correct, this seems like a better formula for the real+imag portion of the desired roots of 7th root unity.  Does anyone know the author's of the sourced pdf's email address?  I would like to email the author the comments I just posted, if he is interested in updating/correcting his online paper.  We need to correct the error in the wikipedia page, even if the author's paper isn't corrected.  The link to the author's pdf is http://www.mindspring.com/~jimvb/TheSeventhRootofUnity.pdf  Here is my proposed update to the wikipedia page. If nobody has any suggestions or comments, I'll probably update the source page in a few days, once I think there are no further improvements desired.
Given the complex cuberoots of unity, s1=e^{2\pi i/3}, s2=e^{4\pi i/3}, than one of the seventh roots of unity is
e^{2\pi i/7}=
\frac{-1 + \sqrt[3]{\frac{7+21\sqrt{-3}}{2}} + \sqrt[3]{\frac{7-21\sqrt{-3}}{2}}}{6} +
\frac{i}{2}\sqrt{\frac{7 - s2\sqrt[3]{\frac{7+21\sqrt{-3}}{2}} - s1\sqrt[3]{\frac{7-21\sqrt{-3}}{2}}}{3}}
Sheldonison (talk) 19:48, 4 May 2011 (UTC)
replies: text: Thanks for noticing this. I've gone ahead and edited the article to reflect your corrections (with a slight notation change) (I also checked your answer myself). I've removed the source as a reference as it's clearly not very reliable. I added a "citation needed" tag for good measure. RobHar (talk) 05:31, 5 May 2011 (UTC)
thread_title:  7th roots of unity typo, suggestions 

message: My wording about exactly  th roots was a bit reckless, but I did not claim that  always has distinct roots. BTW I just learned about a (weaker) definition of cyclotomically closed field, which apparently is exactly what we need here. Incnis Mrsi (talk) 18:03, 5 April 2013 (UTC)
thread_title:  if ''n'' is not divisible by the characteristic of the field 

message: It would be nice to have a "Geometric interpretation" section that describes the fact that all n roots of unity lie at equidistant points on the unit circle (centered at the origin) in the cartesian/complex plane, i.e., they lie on the vertices of a regular n-gon inscribed within the unit circle. But I don't know where the most appropriate place is to add such a section in the article. — Loadmaster (talk) 19:00, 30 September 2013 (UTC)
replies: text: yes. what is the meaning of this? it seems special 24.98.133.72 (talk) 01:39, 19 August 2014 (UTC)
replies: text: Well, multiplying any combination of the nth roots of 1 always produces one of the (other) roots (or 1) on the complex unit circle. Does that qualify as "special"? There is no deeper meaning, since these are just some of the properties of the unit multiplier (1) in the complex field. — Loadmaster (talk) 19:15, 19 August 2014 (UTC)
thread_title:  Geometry 

message: Random observation while I was passing through: should the opening paragraph specify that n is positive? After all, raising any complex number (save 0) to the 0th power yields 1. 0 is obviously an integer, but that doesn't mean that the number is a root of unity. (The definition below does a better job, specifying that n must be positive.)
Aasmith (talk) 06:34, 16 September 2015 (UTC)
thread_title:  Positive power? 

message: In this case, what I don't understand is this assertion in the article:
:Generally, z ∈ R can be considered for any field R    
I'm sure this is a stupid question*, but does this mean that in general, a complex number is a real?  I guess it depends on the definition of "considered."
Even if not, why talk about complex numbers that are points on the real number line? The imaginary part is zero, so it's a degenerative, trivial complex number.
Right?
For someone who hates not understanding stuff, I sure do it a lot.
__________________
*I can't stand when people preface questions with that when asking me, but now I see why they do.  --Verdana♥Bold 07:57, 4 March 2017 (UTC)
replies: text: You are confusing , a variable, which, here, denotes an arbitrary ring, with the standard symbol for the field of real numbers, which is \mathbb R or . Nevertheless, I agree that the formulation was confusing, and not only because of the use of  (which was unnecessary). I have thus rewritten the paragraph. D.Lazard (talk) 08:57, 4 March 2017 (UTC)
thread_title:  I hate do not understanding stuff! 

message: Useful for generating functions http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf Wqwt (talk) 22:32, 1 September 2018 (UTC)
replies: text: This cannot be used in WP, as being original research, not regularly published. Moreover, this result fails to satisfy the WP:notability criteria, because of the lack of WP:secondary sources. D.Lazard (talk) 06:52, 2 September 2018 (UTC)
replies: text: If you consider brilliant.org a valid source then it shows up there too. I'm sure it's in some textbook somewhere, not only for contest math, but probably under a different name. It makes an appearance in Problems from the Book by Titu Andreescu. Wqwt (talk) 02:10, 3 September 2018 (UTC)
thread_title:  Roots of Unity Filter 

message: Since September, 28,  made 25 edits to the articles. A few of them introduce minor improvements, but most do not improve the article and consist mainly of introducing data for roots of unity from 9 to 12. Here are more details on these additions:
*Section "Cyclotomic polynomials" is intended for being a summary of the article of the same name (template . So adding details for these degrees makes no sense: Either they duplicate the content of the main article or, if they are useful, they should be added to the main article.
*In section "Algebraic expressions", no explicit formula is given in the edits. There are only general unsourced comments, that either duplicate the beginning of the section, or are not understandable for most readers (references without explanation to casus irreducibilis, or even are completely wrong (I had to revert twice the wrong assertion that eleventh roots of unity cannot expressed in radicals.
Thus I'll revert all these edits, and, when useful, I'll add a few words for clarification. Please, do not add data for higher degree, without getting a consensus here (see WP:CONSENSUS and WP:BRD). D.Lazard (talk) 16:20, 29 September 2018 (UTC)
thread_title:  Recent edits 

message: ... leave off the x- and y-axes!!!  Just show the roots of unity —  as the caption states.2600:1700:E1C0:F340:38C4:1D3E:151:63CC (talk) 03:25, 2 October 2018 (UTC)
replies: text: It makes no sense to show the roots of unity without showing where they are in the complex plane. However it may useful to clarify the caption by saying that the roots are the blue dots. This has been done. D.Lazard (talk) 08:35, 2 October 2018 (UTC)
thread_title:  In the first illustration ...