Non-integer roots of complex numbers

[uruz]

Homework Statement

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

Homework Equations

\sqrt[n]{Z}={|Z|}^{1/n}.(cos((\theta+2k\pi)/n)+isin((\theta+2k\pi)/n)

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?

 

[Dick]

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]

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]

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]

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?

 

[Dick]

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