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

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 $$(z^{1/m})^n[/itex] is identical to the set of value of $(z^n)^{1/m}$ I need to prove the case of a reducible fraction as well, where the two expressions aren't equal. ## The Attempt at a Solution I've been staring at this for a day now and I don't see where to start this beyond messing with the expressions in polar form... hints? Thanks. ------ Side question: $(8^{2/3})(8^{-2/3})$ Does finding all three roots of each factor and then multiplying them in all combinations give all possible results of the above expression? Thanks. ## Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org Don't forget that an complex equality like the one you have is, in fact, an equality between sets. Starting from the polar expression for z is a good idea, but remember that [tex]z^{1/m}$$ 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 $$z^n$$.

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...