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

[cscott]

Homework Statement

Show that if m and n are positive integers, $m \ne 0$, and if n/m is an irreducible fraction, then the set of values of $z^{n/m}$ defined by [tex](z^{1/m})^n[/itex] is identical to the set of value of $(z^n)^{1/m}$<br /> <br /> I need to prove the case of a reducible fraction as well, where the two expressions aren't equal.<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> 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.<br /> <br /> ------<br /> <br /> Side question:<br /> <br /> $(8^{2/3})(8^{-2/3})$<br /> <br /> Does finding all three roots of each factor and then multiplying them in all combinations give all possible results of the above expression? Thanks.[/tex]

 

[JSuarez]

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 $$z^{1/m}$$ 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 $$\left(z^n\right)^{1/m}$$, whose elements are the m-th roots of $$z^n$$.

 

[cscott]

I get (I'm going to use cis() notation):

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

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

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

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