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Non-integer roots of complex numbers

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\sqrt[n]{Z}[/tex] has exactly n distinct value for integer n.
    What can you say about non-integer n's ?

    2. Relevant equations

    3. The attempt at a solution
    I used Euler's formula to see clearly what the roots are if n is integer.
    As it is told i find [tex]\sqrt[n]{Z}[/tex] has n roots.
    But what if n is non-integer?
    I've been told that if n is non-integer there will be infinite solution.
    How could it be?
  2. jcsd
  3. Mar 24, 2009 #2


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    That's not quite true. If you can see what the roots are when n is an integer then you can probably figure out what the roots are when n is rational and there's only a finite number. The case where n is irrational is where you have an infinite number of solutions. Can you show that?
  4. Mar 24, 2009 #3
    If i take n, for instance 2.3 there will be several solution because at a certain k value i will notice that i've found the same root before.
    if i take n=2.34 there'll be more solution because cosine or sine of it will have different values from previous ones..and the number of solutions increase with the increase in number of fractions.

    If I think an irrotational number is infinitely long there will be infinitely different solutions.
    Is that the case or just a piece of nonsense i make up?
  5. Mar 24, 2009 #4


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    That's roughly true. Two roots corresponding to the integers k1 and k2 are going to be the same if 2*pi*k1/n and 2*pi*k2/n differ by a multiple of 2*pi. Can you show if n is irrational then that can never happen?
  6. Mar 24, 2009 #5
    c is an integer

    [tex](2{\pi}k1 - 2{\pi}k2)/n = c2\pi[/tex]

    [tex]k1 - k2 = cn[/tex]

    hence k1 and k2 is integer, n must be integer too to have same roots.
    that means if n is an irrational number this equation cannot be satisfied namely all roots we can find are different so we can say there are infinitely many distinct roots, is this correct?
  7. Mar 24, 2009 #6


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    Exactly right. That would mean (k1-k2)/c=n. The left side is rational, the right side isn't. The roots are all distinct.
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