I'm asking if there is any difference between writing $-1$ in polar form as $e^{-i\pi}$ or $e^{i\pi}$. I've always tought that it didn't, until i tried this exercise:
Find all the roots of $(-1)^{-3/4}$ Answer: $e^{\frac{3 \pi i}{4}}$ and $e^{-\frac{3 \pi i}{4}}$.
What i did was writing $-1 = e^{i\pi} = e^{i \pi + 2k i \pi}$ (for any natural $k$) and then applying the power, yelding: $$z^{-3/4} = \operatorname{exp}\Big[i\big(\frac{-3 \pi}{4} - \frac{3i k\pi}{2}\big) \Big]$$ The roots i found are: $$z_1 \rightarrow k = 0 \rightarrow z_1 = e^{\frac{-3i\pi}{4}}$$ So far so good. The problem is with the second root. The exponential's argument is: $$i\Big(-\frac{3 \pi}{4} - \frac{3 \pi}{2}\Big) = \frac{-9\pi i}{4}$$ And this is equivalent to $\frac{-i\pi}{4}$ (Taking out a full rotation of $-2 \pi$). Hence my second root would be: $$e^{i \frac{-\pi}{4}}$$ Which is obviously not equivalent to $e^{\frac{3 \pi i}{4}}$. However, if i choose to define $(-1) = e^{-i \pi}$, which is also valid, i find the first root to be: $$z_1 = e^{i \frac{3 \pi}{4}}$$ And then, the second root just subtracting $\frac{3 \pi}{2}$ from the argument: $$z_2 = e^{-i \frac{3 \pi}{4}}$$ So apparently there is a difference in defining $-1$ as either $e^{i \pi}$ or $e^{-i \pi}$, or there is a mistake in my calculations that i really can't see. If not, why does this difference exists and what is a good method to find the roots?
Note: I'm defining $0 < \operatorname{Arg}(z) < 2 \pi$.
One way or another you're going to write $-1=e^{i \pi (x_0+2k)}$, where $x_0$ is an arbitrary (but fixed) odd integer. Then you're going to apply the power of $-3/4$ in the exponent and get $e^{-i \pi(3x_0/4+3 k/2)}$. The point is that this quantity, as a function of $k$, has period $4$, not $2$. So you should get four different answers. The set of results will be the same regardless of which $x_0$ you picked. The assignment of them to values of $k$ will depend on $x_0$, however. If for instance you pick $x_0=1,k=-2,,\dots,1$, then you get the representation
$$e^{9i\pi/4},e^{3i\pi/4},e^{-3i\pi/4},e^{-9i\pi/4}.$$
Note that a faster way to do this would be to raise the $-3$ power first, obtaining $(-1)^{-3}=-1$, and then take fourth roots after that. This would have given the easier-to-read representation
$$e^{-3i\pi/4},e^{-i\pi/4},e^{i\pi/4},e^{3i\pi/4}$$
which is the same solution as before, just written with arguments between $-\pi$ and $\pi$.
In general, if you have a nonzero complex number $z$ and a reduced fraction $m/n$, there are $n$ values of $z^{m/n}$. It is important that I say reduced fraction here. For example, $(-1)^{2/4}$ has $2$ values, not $4$. By contrast, $((-1)^2)^{1/4}$ does have four values. This is why you should reduce the fraction first: if you do, then you don't need to deal with these issues about order of operations.