non-integer roots of complex numbers

by uruz
Tags: complex, noninteger, numbers, roots
 P: 8 1. The problem statement, all variables and given/known data $$\sqrt[n]{Z}$$ has exactly n distinct value for integer n. What can you say about non-integer n's ? 2. Relevant equations $$\sqrt[n]{Z}={|Z|}^{1/n}.(cos((\theta+2k\pi)/n)+isin((\theta+2k\pi)/n)$$ 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 $$\sqrt[n]{Z}$$ 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?
 Sci Advisor HW Helper Thanks P: 25,170 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?
 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?
 P: 8 c is an integer $$(2{\pi}k1 - 2{\pi}k2)/n = c2\pi$$ $$k1 - k2 = cn$$ 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?