$z^{\frac{4}{3}} = -2$ ; How to know which complex roots to keep from this equation So I recently came upon the following complex algebra problem:
$$
z^{\frac{4}{3}} = -2
$$
So, to solve it I have to find the z values that solve the following:
$$
z = (-2)^{\frac{3}{4}} 
$$
To do this I express -2 in exponential form:
$$
z = (2e^{i(\pi + 2\pi n)})^{\frac{3}{4}} 
$$
Then, I solve for that trying for $n=0,1,2,3$ and I come up with 4 roots:
$$
z_1 = 2^{\frac{3}{4}}e^{i\frac{\pi}{4}}
$$
$$
z_2 = 2^{\frac{3}{4}}e^{i\frac{3\pi}{4}}
$$
$$
z_3 = 2^{\frac{3}{4}}e^{i\frac{5\pi}{4}}
$$
$$
z_4 = 2^{\frac{3}{4}}e^{i\frac{7\pi}{4}}
$$
However, if I try to check these solutions for the original problem, only $z_2$ and $z_3$ succeed, while $z_1$ and $z_4$ do not solve the initial equation. Even plugging the original equation into Wolfram, gives me just those two roots.
I have been thinking about this over and over and don't understand where I'm going wrong or what is it that I'm failing to consider. Does anybody have any idea of where I'm going wrong?
Thank you in advance
 A: I presume you're treating $z^{4/3}$ as a multivalued function, and you're allowing any $z$ such that any branch of $z^{4/3}$ is $2$.
By definition, $z^{4/3} = \exp((4/3) \log(z))$ where $\log(z)$ is any branch of the logarithm of $z$. If $\text{Log}(z)$ is the principal branch (with imaginary part in $(-\pi, \pi]$), the other branches of $\log(z)$ are
$\text{Log}(z) + 2 \pi i n$ for arbitrary integers $n$,
and the corresponding branches of $z^{4/3}$ are
$\exp((4/3) \text{Log}(z) + (8 \pi i n/3))$.  There are three
possibilities, corresponding to the values of $n \mod 3$.
Now this is supposed to be $-2 = 2 \exp(\pi i)$.  Thus  for
$n \equiv 0 \mod 3$, $$2 = \exp((4/3) \text{Log}(z) - \pi i)$$
where $\text{Im}((4/3) \text{Log}(z) - \pi i) = 0$ and $\text{Re}((4/3) \text{Log}(z) = \text{Log}(2)$.  We get either $\text{Log}(z) = (3/4) \text{Log}(2) + 3 \pi i/4$, i.e. $z = 2^{3/4} e^{3 \pi i/4}$, or $\text{Log}(z) = (3/4) \text{Log}(2) - 3 \pi i/4$, i.e. $z = 2^{3/4} e^{-3\pi i/4}$.
(this $2^{3/4}$ being the real $3/4$ power).
For $n \equiv 1 \mod 3$, $$2 = \exp((4/3) \text{Log}(z) + 5 \pi i/3)= \exp((4/3) \text{Log}(z) - \pi i/3$$ where
$\text{Im}((4/3) \text{Log}(z) - \pi i/3 = 0$.  We get
$\text{Log}(z) = (3/4) \text{Log}(2) + \pi i/4$, or
$z = 2^{3/4} e^{\pi i/4}$.
For $n \equiv 2 \mod 3$, $$2 = \exp((4/3) \text{Log}(z) + 13 \pi i/3) = \exp((4/3) \text{Log}(z) + \pi i/3$$ where
$\text{Im}((4/3) \text{Log}(z) + \pi i/3 = 0$.  We get
$\text{Log}(z) = (3/4) \text{Log}(2) - \pi i/4$, or
$z = 2^{3/4} e^{-\pi i/4}$.
So there are indeed four solutions.  However,
if you try to verify these with Mathematica or most other computer algebra systems, they won't all work, as they like to use the principal branch rather than multivalued functions.
A: Please note that the complex function $f(z)=z^{\frac{1}{n}}$, $n \in \mathbb{N}, \, n \ge 2$ is a multivalued function. Writing the function in polar form,$$z=re^{i \theta } \qquad \rightarrow \qquad f(z)=(re^{i \theta })^{\frac{1}{n}}=r^{\frac{1}{n}}e^{i \frac{\theta }{n}},$$we can easily conclude that a point with arguments $\theta$, $\theta + 2\pi$, ..., $\theta + 2(n-1)\pi$ in the domain plane corresponds to $n$ distinct points with the arguments $\frac{\theta }{n}$, $\frac{\theta }{n}+\frac{2\pi }{n}$, ..., $\frac{\theta }{n}+\frac{2(n-1)\pi }{n}$ in the image plane. In other words, this function is a one-to-$\bf{n}$ correspondence.
With a similar argument, one can show that the function $f(z)=z^{\frac{4}{3}}$ is a one-to-three correspondence.
You have solved $z^{\frac{4}{3}}=-2$ correctly. However, please note that to check the solutions for the original problem, you should use the same representation of points you reached by solving the problem, that is,$$z_1 = 2^{\frac{3}{4}}e^{i\frac{3\pi}{4}}
$$$$z_2 = 2^{\frac{3}{4}}e^{i\frac{9\pi}{4}}$$$$z_3 =2^{\frac{3}{4}}e^{i\frac{15\pi}{4}}$$$$z_4 =2^{\frac{3}{4}}e^{i\frac{21\pi}{4}},$$which are clearly satisfy the original problem. Otherwise, you may get other values of $z^{\frac{4}{3}}$ not satisfying the original problem.
