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Homework Help: Factoring the equation z^(n)-1=0 where z is a complex number

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Hey, I am attempting to fully factorize z[itex]^{n}[/itex]-1=0 for all integers of n where n does not equal zero, and where z is a complex number in the form a+bi. The question asks to first factorize the equation when n=3,4,5. I know how to factorize when n=3 and 4, but I get stuck at n=5. I don't know where to begin.

    (Below is just an explanation of what the question is referring to)

    This question is from a set of finding patterns in complex numbers. We are asked to plot all solutions of z[itex]^{n}[/itex]-1=0 using DE MOIVRE'S THEOREM on an Argand diagram or a unit circle. This step was easy; it was noticed that for any integer value of n, there are n roots of unity such that when all roots are plotted on an Argand diagram and I connect all adjacent roots, I form a regular n-gon (for example, if n=5, then there are 5 roots of unity. When adjacent roots are connected, a regular pentagon is formed). (The connecting of the adjacent roots was NOT asked in the question. I just told you guys this to help you hopefully visualize what I am talking about).

    The question then asks to choose ONE root and form line segments from that one root to ALL other roots (including roots that are adjacent to that one root). Then, we had to find the distances of the line segments that were just formed.

    To cut to the chase of what I found in the end, I basically noticed that when all distances of line segments (for any value of n) are multiplied together, the result is exactly n. However, I do not know how to prove this conjecture, so any help would be appreciated.

    2. Relevant equations
    de Moivre's theorem, binomial expansion (not sure about this)

    3. The attempt at a solution

    I do not know where to begin. Apparently, the factorization of z[itex]^{n}[/itex]-1=0 has to do with the distances of the line segments that I talked about up there^ (I am not sure of this though, the two may even have no relation at all).

    My first reaction as to how to factorize the equation is to substitute z for a+bi. Therefore I would have (a+bi) [itex]^{n}[/itex]-1=0, and I could use binomial expansion from there, but I don't see how that would help me in any way.

    TL;DR How do I factorize z[itex]^{n}[/itex]-1=0 where z is a complex number and n is any integer other than zero, without the use of de Moivre's theorem?
  2. jcsd
  3. Apr 21, 2012 #2


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    Actually to factorize [itex]z^n-1=0[/itex] I wouldn't think of using De Moivre's Theorem.

    So what we need to solve is


    and 1 in mod-arg form is equivalent to [itex]1=1\cdot cis (0)[/itex] but you can also take any multiple of [itex]2\pi[/itex] in the argument and it's still the same value, so we have [itex]1=1\cdot cis (2k\pi)[/itex] for any integer k.

    So now we have to solve

    [tex]z^n=1\cdot cis(2k\pi)[/tex]

    Solving for z:

    [tex]z=1^{1/n}\cdot cis\left(\frac{2k\pi}{n}\right)[/tex]


    Thus in order to factorize [itex]z^n-1[/itex] it's trivial to just take the n distinct roots and get your linear factors that way, but if you wanted to take the problem further to avoid having complex factors, each two factors with a root and its conjugate can be multiplied together to give a quadratic with real coefficients.
  4. Apr 21, 2012 #3


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    Other than the obvious "z- 1" factor, to completely factor this you will have to know the roots. If you don't want to use DeMoivre, would you accept using the fact that the roots of [itex]z^n= 1[/itex] are the vertices of an n-gon lieing on the unit circle? Further, if you want only real coefficients you will need to use the fact, as Mentallic says, that non-real roots will will come in conjugate pairs. If n is even, 1 and -1 will be roots and the other n- 2 roots will be in (n-2)/2 conjugate pairs. if n is odd, 1 is a root and the other n- 1 roots will be in (n- 1)/2 conjugate pairs.
  5. Apr 21, 2012 #4
    I don't quite understand the conjugate pair part. Could you give me an example for factoring z[itex]^{3}[/itex]-1=0? I found the roots via de Moivre's to be 1, -[itex]\frac{1}{2}[/itex]+[itex]\frac{\sqrt{3}}{2}[/itex]i, and -[itex]\frac{1}{2}[/itex]-[itex]\frac{\sqrt{3}}{2}[/itex]i.

    Then would the factored form be (z-1)(z-(-[itex]\frac{1}{2}[/itex]+[itex]\frac{\sqrt{3}}{2}[/itex]i))(z-(-[itex]\frac{1}{2}[/itex]-[itex]\frac{\sqrt{3}}{2}[/itex]i))?

    If so, how could I use this method of factorization to prove my conjecture? My conjecture was that if I connect ONE root to ALL other roots on an Argand diagram and measure the distances of the line segments formed (assume the distances are d1, d2, d3, ... , dn), then the product of the distances, that is d1d2d3...dn is exactly n (n is the number of roots, or the value of n in the equation z[itex]^{n}[/itex]-1=0)
  6. Apr 21, 2012 #5
    I'm just curious how you would move the n inside of cis without using de Moivre's theorem.
  7. Apr 21, 2012 #6
    There's a proof for de Moivre's on the Internet that shows you how raising cisx to the nth power is the same as cisnx
  8. Apr 21, 2012 #7
    This was asked on one of the Math sections a few weeks ago. And I believe it also came up a couple of months ago. I can't find either thread. I don't believe anyone's ever posted a solution yet.

    Some handy facts that might help:

    If z1 and z2 are roots of unity (satisfy z^n = 1) then the length of the chord between them is |z1 - z2|.

    Other handy things to know are that if |z| = 1, then z * z' = 1, where z' is the complex conjugate of z.

    And, the non-real roots of z^n = 1 occur in conjugate pairs.

    For the problem you might as well put the base point (the point you're drawing all the chords from) at 1. So you have to find the product of |1 - z| as z ranges over all the other roots. But the roots come in conjugate pairs, so you have for each pair (1 - z)(1-z') = 1 - z - z' + 1 = 2 - (z + z').

    Now z + z' is the sum of two complex numbers that are on the same vertical line in the plane; that is, they have the same x-coordinate or the same real part. And their imaginary parts are additive inverses so they cancel. In other words z + z' = 2 * cos(t) where t is the angle k*2pi/n for some k.

    (In other words the n-th roots of 1 have complex arguments 2pi/n, 4pi/n, 6pi/n, etc. That's where the k*2pi/n comes from.)

    All these things are useful, but I haven't been able to put them together yet to solve this problem and although I've seen a lot of similar hints, I haven't seen a complete solution yet.

    Also FWIW, z^n - 1 = (z -1)(z^(n-1) + z^(n-2) + ... + 1). For example

    z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1)

    so the 5-th roots of 1 are 1 and the roots of z^4 + z^3 + z^2 + z + 1. In general it's going to be difficult to factor that for arbitrary n, so I don't know if that's the best way to go.

    (edit) Aha I found one of the recent threads on this problem. As you can see, nobody came near a solution.

    Last edited: Apr 21, 2012
  9. Apr 21, 2012 #8


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    It's possible for n=5, since after dividing by the trivial factor (z-1), you're left with a quartic, and you can *always* find exactly all the complex solutions of any quartic, using something like this: http://en.wikipedia.org/wiki/Quartic_function#The_general_case.2C_along_Ferrari.27s_lines. It's just tedious, though.

    I don't see how to do it for n=6 and above since after the division, you're left with a quintic or higher. Those are not guaranteed to be exactly solvable in surds.
  10. Apr 21, 2012 #9
    Thanks! I guess the real problem that confuses me is how the factorization of z^n - 1 = 0 can be used to prove the theorem that the product of the distances is equal to n...

    EDIT: Is there any way to express the distance of, for example, |z0-z2| in terms of |z0-z1|? Or just expressing the distance between any two roots in terms of ONE distance?
    Last edited: Apr 21, 2012
  11. Apr 21, 2012 #10
    Well, yes, because z2 is the next point around the circle from z1, so it's the next multiple of 2pi/n.

    In other words take n = 7. Starting from 1 and then going counterclockwise, the roots are at 2pi/7, 4pi/7, 6pi/n, 8pi/7, 10pi/7, 12pi/7, and finally 14pi/7 = 2pi corresponding to the root z = 1.

    But the first (n-1)/2 of these are all you need, because the other ones are their conjugates. If you draw the circle this is easy to visualize.

    The conjugate of 2pi/7 is 12pi/7. The conjugate of 4pi/7 is 10pi/n. The conjugate of 6pi/7 is 8pi/7.

    The distance from 1 to any root is the same as the distance from 1 to the conjugate of the root, right? The roots in the upper half plane are the mirror image of their respective conjugates below the real axis.

    By the way I'm just talking about this problem in general (ie rambling). I am not convinced that factoring z^6 + z^5 ... + 1 is a useful approach. (I could be wrong)

    So if you look at the geometry of 1, z1, and z2, you might see some relationship.

    The other progress I've made on this problem is that for each conjugate pair, you have

    (1 - z)(1 - z') = 2 - 2*cos(Arg(z)). Did you follow that logic in my earlier post? We know that zz' = 1, and z + z' = 2 * Re(z) [where Re(z) is the real part of z]

    So you can express the product (for n odd) as the product from 1 to (n-1)/2 of 2 - 2*cos(k*2pi/n). Presumably you should be able to get from that to the desired result, but I haven't been able to do it yet.

    The other case, n even, is the same except that you also have an additional factor of 2 because z = -1 is a root whose distance from 1 is 2.

    Hope some of this is helpful ... just some ramblings on this problem.

    I have another completely different idea about this too. We know that the distances from 1 to z and 1 to z' are the same, but they're different for each pair of conjugate roots.

    But what if we took the opposite pairs of roots? In other words (for n = 7 again) instead of pairing z1 and z1', we go the other way ... pair z1 with z3', z2 with z2', and z3 with z1'. Perhaps THOSE respective products are the (n-1)-th roots of n, so multiplying them out gives the desired product of n? Does that make sense as an approach?

    I don't think this is a particularly easy problem, what class is this for? Then again I never could do trig identities :-)
  12. Apr 21, 2012 #11
    Thanks, that made a lot of sense (surprisingly!). I'm gonna go work on this for a while now and report back if I make some progress (unlikely lol)

    By the way, this is for a grade 11 IB math class. I think I'm overthinking the assignment a bit too much.

    I was thinking if I could just use the distance formula as my proof. Basically, I came up with a simple trig function, where it will find the distance between any two roots if I input a value of n and a value of delta(k) <-- this is the difference between the k values of two roots, so for example if I wanted to find the distance between the first and second root z0 and z1 respectively, then delta(k) would be 1. If i wanted to find the distance between z0 and z3 (third), then the delta(k) value would be 3. The formula was 2sin(xpi/n) if I recall correctly, where x is delta(k). I simply derived this formula from the distance formula.

    Would it be enough for this to be counted as proof? I was thinking of using a multiplication summation series (the one with the pi symbol, not sure what it's called lol) where the expression that I'm multiplying is the sine function from above.
  13. Apr 21, 2012 #12
    Nevermind, it seems that the solution was online. Looks like factorization actually has something to do with the product of the line segments.


    However, I don't understand the part where it says "This is a polynomial equation of degree N-1 with the free coefficient equal to N and the roots ω1, ω2, ..., ωN-1. As is well known, the product of all roots of a polynomial equation is given by its free coefficient (up to a sign)". Could anyone elaborate on what it means by free coefficient?

    Thanks for all the help so far
  14. Oct 13, 2012 #13
  15. Nov 18, 2012 #14
    Hats off !
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