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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,


    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


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    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


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


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    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


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    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


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


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    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]
    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]
    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.
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