non-integer roots of complex numbers


by uruz
Tags: complex, noninteger, numbers, roots
uruz
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#1
Mar24-09, 11:55 AM
P: 8
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
[tex]\sqrt[n]{Z}={|Z|}^{1/n}.(cos((\theta+2k\pi)/n)+isin((\theta+2k\pi)/n)[/tex]


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?
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Dick
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#2
Mar24-09, 12:47 PM
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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?
uruz
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#3
Mar24-09, 03:28 PM
P: 8
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?

Dick
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#4
Mar24-09, 04:10 PM
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non-integer roots of complex numbers


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?
uruz
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#5
Mar24-09, 04:39 PM
P: 8
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?
Dick
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#6
Mar24-09, 04:44 PM
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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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