[z^n]^(1/m) = and != [z^(1/m)]^n

  • Thread starter cscott
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  • #1
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Homework Statement



Show that if m and n are positive integers, [itex]m \ne 0[/itex], and if n/m is an irreducible fraction, then the set of values of [itex]z^{n/m}[/itex] defined by [tex](z^{1/m})^n[/itex] is identical to the set of value of [itex](z^n)^{1/m}[/itex]

I need to prove the case of a reducible fraction as well, where the two expressions aren't equal.

The Attempt at a Solution



I've been staring at this for a day now and I don't see where to start this beyond messing with the expressions in polar form... hints? Thanks.

------

Side question:

[itex](8^{2/3})(8^{-2/3})[/itex]

Does finding all three roots of each factor and then multiplying them in all combinations give all possible results of the above expression? Thanks.
 

Answers and Replies

  • #2
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Don't forget that an complex equality like the one you have is, in fact, an equality between sets. Starting from the polar expression for z is a good idea, but remember that [tex]z^{1/m}[/tex] is the set of the m-th roots of z, and you must prove that each of its members, when raised to the power n, is a member of the set [tex]\left(z^n\right)^{1/m}[/tex], whose elements are the m-th roots of [tex]z^n[/tex].
 
  • #3
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I get (I'm going to use cis() notation):

[tex]z^n = r^n cis \left(n\theta \right)[/tex]
[tex]z^{1/m} = r^{1/m} cis \left (\frac{\theta}{m} + \frac{2k\pi}{m} \right)[/tex]

[tex](z^n)^{1/m} = r^{n/m} cis \left (\frac{n}{m}\theta + \frac{2k\pi}{m} \right)[/tex] k = 0,1,2,...,|m|-1

[tex](z^{1/m})^{n} = r^{n/m} cis \left (\frac{n}{m}\theta + \frac{2nk\pi}{m} \right)[/tex]

I don't think my expression for [itex](z^{1/m})^{n}[/itex] is right with the [itex]2nk[/itex] in it...
 

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