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Noninteger roots of complex numbers 
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#1
Mar2409, 11:55 AM

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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 noninteger 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 noninteger? I've been told that if n is noninteger there will be infinite solution. How could it be? 


#2
Mar2409, 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?



#3
Mar2409, 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? 


#4
Mar2409, 04:10 PM

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Noninteger 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?



#5
Mar2409, 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? 


#6
Mar2409, 04:44 PM

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Exactly right. That would mean (k1k2)/c=n. The left side is rational, the right side isn't. The roots are all distinct.



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