Proving that (z^m)^(1/n)=(z^(1/n))^m

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In summary, the conversation discusses proving the equation (z^m)^{1/n}=(z^{1/n})^m, where m and n do not have a common divisor. The attempt at a solution involves using complex numbers and considering the more general problem of (z^m)^n =? (z^n)^m or (z^m)^n =? z^(mn). The conversation also addresses the issue of z^(1/n) being a multivalued function and the need to define a branch. The question at hand is how to show that the set of values represented by (z^(1/n))^m is equivalent to the set of values represented by (z^m)^(1/n), which is equivalent
  • #1
estro
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Homework Statement



I want to prove something that seems trivial at first:
[itex] (z^m)^{1/n}=(z^{1/n})^m [/itex], where m and n don't have a common divisor.

The Attempt at a Solution



When I'm using [itex] z=re^{i\theta}[/itex], I arrive that above is true if and only if the following is true:
[itex] e^{ \frac{2\pi \theta mk}{n}}=e^{\frac{2\pi \theta k}{n}} [/itex]
and this confuses me.
 
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  • #2
estro said:
When I'm using [itex] z=re^{i\theta}[/itex], I arrive that above is true if and only if the following is true:
[itex] e^{ \frac{2\pi \theta mk}{n}}=e^{\frac{2\pi \theta k}{n}} [/itex]
and this confuses me.
Then you are doing something wrong.

The equality is true even for complex m,n, so you can consider the more general problem (z^m)^n =? (z^n)^m or (z^m)^n =? z^(mn) and save some fractions.
 
  • #3
This is what puzzlies me:

[itex] (z^{\frac {1} {n}})^{m} = r^{\frac {m}{n}}e^{\frac {m\theta + 2m \pi k} {n}i} =
r^{\frac {m}{n}}e^{\frac {m\theta} {n}i}e^{\frac {2m \pi k} {n}i}[/itex]

[itex] (z^m)^{\frac {1} {n}} = r^{\frac {m}{n}}e^{\frac {m\theta + 2 \pi k} {n}i} =
r^{\frac {m}{n}}e^{\frac {m\theta} {n}i}e^{\frac {2 \pi k} {n}i}[/itex]

How is this possible?
 
  • #4
You are using n and m with two different meanings, and you don't need that +2pi k there.
 
  • #5
Sorry, but I'm not sure I understand.
m and n are constants while k goes from 0 to n-1.

I do need +2kpi, as I need to prove that both represent the same set of values.
 
  • #6
estro said:
Sorry, but I'm not sure I understand.
m and n are constants while k goes from 0 to n-1

I think you are right to to be confused about this. z^(1/n) isn't even a function. It's multivalued. You need to define a branch. If you are naively using the polar form, the answer will depend on the argument you pick for z.
 
  • #7
So how can I prove that both expression represent the following:
https://dl.dropboxusercontent.com/u/27412797/q1.png [Broken]
 
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  • #8
estro said:
So how can I prove that both expression represent the following:
https://dl.dropboxusercontent.com/u/27412797/q1.png [Broken]

You don't do it by assuming that (z^(1/n))^m=(z^m)^(1/n). That's a complex numbers mistake. You can only safely take (z^a)^b=(z^b)^a to be true if a and b are integers. You know from deMoivre that all of those numbers are nth roots of z^m. What I think they want you to show is that that is all of them. For example putting k=n doesn't give you a new root, and that for no two different k in k=0,...n-1 are they the same root.
 
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  • #9
The question is formulated like this:
Show that set of values represented by (z^(1/n))^m are the same set of values represented by (z^m)^(1/n).

And both these equivalent to set: https://dl.dropboxusercontent.com/u/27412797/q1.png [Broken]
 
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  • #10
estro said:
The question is formulated like this:
Show that set of values represented by (z^(1/n))^m are the same set of values represented by (z^m)^(1/n).

And both these equivalent to set: https://dl.dropboxusercontent.com/u/27412797/q1.png [Broken]

Alright. The full question has it make a little more sense. I'll give you an example of what can go wrong. Take z=1, m=2 and n=2. Then the values of (1^2)^(1/2) are 1^(1/2) which is +1 or -1. The values of (1^(1/2))^2 are (+1)^2 or (-1)^2 which are both 1. So the value sets aren't equal. Because 2 and 2 aren't relatively prime. That's the problem you are dealing with.
 
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  • #11
So I should show that
[itex]e^{km\frac {2\pi i} {n}} = e^{k\frac {2\pi i} {n}} [/itex] because m mod(n) != 0?
 
  • #12
estro said:
So I should show that
[itex]e^{km\frac {2\pi i} {n}} = e^{k\frac {2\pi i} {n}} [/itex] because m mod(n) != 0?

Even if that were true, why? I think you should think about it some more. Figure out why it works if n=2 and m=3, those have no common divisor. Choosing z=1 is perfectly general. You've already factored out |z|. You just need to think about the angles.
 
  • #13
e^(2pi*(m/n)k) represent same angle when m and n have common divisor, but when they do the whole expressions becomes equal to 1.But I'm still not sure how to translate this into proving what I need to prove. I'm not even sure why the above is true...
 
  • #14
I think I'm closer to the solution as I understand now that:

[itex] e^{2pi \frac {mk}{n} i} = e^{2pi \frac {mk(mod(n))}{n} i} [/itex]
 
  • #15
estro said:
I think I'm closer to the solution as I understand now that:

[itex] e^{2pi \frac {mk}{n} i} = e^{2pi \frac {mk(mod(n))}{n} i} [/itex]

That's a good start. So you know k=0,1,2,...n-1 are the only possible solutions. k=n would just repeat k=0, k=n+1 would just repeat k=1 etc. Now I think you just want to show that they are all different solutions if m and n have no common divisor. This is really turning into a number theory problem.
 

1. What does the expression (z^m)^(1/n) mean?

The expression (z^m)^(1/n) is a mathematical notation for raising a complex number z to the power of m, and then taking the nth root of the result.

2. How is (z^m)^(1/n) different from z^(1/n)^m?

The order of operations in mathematics is important, and in this case, (z^m)^(1/n) means taking the nth root of z^m first, and then raising the result to the power of m. On the other hand, z^(1/n)^m means taking the nth root of z first, and then raising the result to the power of m.

3. Can you provide an example to prove that (z^m)^(1/n)=(z^(1/n))^m is true?

Yes, let's take z = 4, m = 3, and n = 2. Substituting these values into the expression, we get (4^3)^(1/2) = (64)^(1/2) = 8. Similarly, (4^(1/2))^3 = (2)^3 = 8. Therefore, we can see that both sides of the equation are equal, proving that the statement is true.

4. How does this expression relate to the laws of exponents?

This expression is a specific case of the power of a power rule, which states that (a^m)^n = a^(mn). By substituting z^m for a and 1/n for n, we get the expression (z^m)^(1/n) = z^(m/n). This shows that the two expressions are equivalent.

5. What implications does this expression have in scientific research or experiments?

This expression is commonly used in scientific research and experiments that involve complex numbers and their powers. It allows scientists to simplify calculations and make accurate predictions by manipulating the order of operations in complex equations involving powers. It also helps in understanding the properties of complex numbers and their operations.

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