1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about nth root of unity

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine the nth roots of unity by aid of the Argand diagram...


    2. Relevant equations

    Is the nth root of unity where we have a complex number to the 'nth' power equal to 1? For example,

    $$(2+i)^n=1$$

    3. The attempt at a solution
    None yet, still trying to translate the question.
     
    Last edited: Apr 22, 2014
  2. jcsd
  3. Apr 22, 2014 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It's more like:

    Solve zn = 1 for z , where z is complex.
     
  4. Apr 22, 2014 #3
    Yes. For example, [itex]-\frac{1}{2} + i\frac{\sqrt{3}}{2}[/itex] is one of the 3rd roots of unity because [itex]\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^3 = 1[/itex].
     
  5. Apr 22, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    In math in English, 'unity' is a quaint way of saying 'one'.
     
  6. Apr 22, 2014 #5

    Mark44

    Staff: Mentor

    The only integer n for which this is true is n = 0.

    2 + i is a complex number whose magnitude is √5. Multiplying it by itself is effectively rotating it by a certain amount, and increasing the magnitude. In polar form, this is easier to see.
    2 + i = √5e, where θ = tan-1(1/2).
    (2 + i)n = [√5e]n = (√5)neinθ
     
  7. Apr 22, 2014 #6
    Okay, this makes much more sense now, so we are looking for x and y values that will have a magnitude (modulus z, or r) of 1 and we can use a generating function for complex numbers like so:

    $$\left( \frac{1}{k} \right)+i\left( \frac{\sqrt{k^2-1}}{k} \right)$$

    or this,

    $$\left( \frac{\sqrt{k^2-1}}{k} \right)+i\left( \frac{1}{k} \right)$$

    I worked really hard with Latex on those big parentheses only to decide I prefer them without :)
     
  8. Apr 22, 2014 #7

    Mark44

    Staff: Mentor

    This looks really nice - I like the big parens - but I don't see how this is answers your original question. An nth root of unity is a complex number z, such that zn = 1. If you write z = 1e, then you have zn = 1neniθ.

    My comment followed your question about whether (2 + i)n could equal 1.
     
  9. Apr 22, 2014 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How is k related to n ?
     
  10. Apr 22, 2014 #9

    ChrisVer

    User Avatar
    Gold Member

    the parenthesis are wrong... [itex]z^{n}≠|z|^{n}[/itex]
    in other words, writing [itex]z^{2}[/itex] you mean [itex]zz[/itex] not [itex]zz^{*}[/itex]

    So if you take the 2nd root for example (n=2) of the formula you provided:
    [itex](\frac{1}{k}+i \frac{\sqrt{k^{2}-1}}{k})(\frac{1}{k}+i \frac{\sqrt{k^{2}-1}}{k})≠1[/itex]...

    If you want to use the form for [itex]z=x+iy[/itex]
    then
    [itex](x+iy)^{n}=1[/itex]
    can be terrible...but not impossible to show....
    Using the Binomial theorem:
    [itex](x+iy)= \sum_{k=0}^{n} (n k) x^{k} (iy)^{n-k}[/itex]
    http://www.dummies.com/how-to/content/how-to-expand-a-binomial-that-contains-complex-num.html
    and proceed accordingly....(probably you'll have to end up with sins and cos, as a taylor expansions?)

    Of course that's totally tedious... it's faster to follow the already proposed idea of writing [itex]z=r e^{i\theta} (r=1)[/itex]
    from the last you can also deduce that [itex]x^{2}+y^{2}=r^{2}=1[/itex]
    or better put that [itex](x,y)[/itex] belong on the unitary "circle" (in fact it's a circle for n going to infinity, otherwise it's a "circle" created by n-points). So here again you see the sin,cos solutions (which must be on the complex plane)....
     
    Last edited: Apr 22, 2014
  11. Apr 25, 2014 #10
    Thanks! Maybe they aren't so bad then :)

    So I got this from your suggestions,

    $$1=\cos \left( \frac{2Pik}{n} \right) + i\sin \left( \frac{2Pik}{n} \right)$$

    What a remarkable answer, and so versatile! I used algebra to write another solution but so far have only been able to get one answer for the cubic.

    It's not really, went off on a tangent :P

    Yup, I messed up and was headed the wrong direction but have it now. I love binomial theorem, I wish they went over it in our engineering program, at least we have breaks for playing with such delightful things.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Question about nth root of unity
  1. Nth roots of unity (Replies: 1)

  2. Roots of Unity (Replies: 1)

  3. Roots of unity (Replies: 1)

  4. Roots of unity (Replies: 9)

Loading...