Question about $y=x^{2/3}$ when $x<0$ Apologies upfront if this is a very basic question.
I don't understand why the function: $$y(x)=x^{2/3}$$ yields complex numbers with an imaginary part when $x<0$.
I can rewrite the above equation as: $$y(x)=x^{2/3}=\sqrt[3]{x^2}$$
So for example, if $x=-2<0$, the above rewritten formula gives: $$y(x)=\sqrt[3]{(-2)^2}=\sqrt[3]{4}$$
which is a real number.
Where am I going wrong? Thanks.
 A: Note that $\sqrt[3]{4}$ is not a real number if you are working on complex numbers.
Indeed $\sqrt[3]{4+0i}$ has three solutions in $\mathbb{C}$.

A: The thing is that any nonzero complex number has $n$ distinct $n$-th roots, so that $\sqrt[n]{x}$ is not uniquely defined for $n\geq 2$, in general.
However, every positive real a number has a positive, real $n$-th root.
Hence, in many contexts we write $\sqrt[n]{x}$ to denote the function on the positive reals that yields this positive real root.
When $n$ is odd, something similar holds for negative real $x$: every such $x$ has a negative real $n$-th root.
Similarly, for these contexts $\sqrt[n]{x}$ is used to denote the function on the reals that yields the real $n$-th root.
Now we're $\sqrt[n]{x}$ used to writing as $x^{1/n}$, and this happens in these contexts I've pointed out above.
But what if instead of $x^{1/n}$ we're dealing with $x^{c/n}$? Do we consider that as $\sqrt[n]{x^c}$ or ${\left(\sqrt[n]{x}\right)}^c$?
The fact of matter is that sometimes these yield different answers, and sometimes these are undefined.
Consider $(-2)^{2/4}$.
Is that $(-2)^{1/2}=\sqrt{-2}$?
Is that ${\left(\sqrt[4]{-2}\right)}^2$?
Is that $\sqrt[4]{(-2)^2}=\sqrt[4]{4}$?
See what we're getting at?
The thing is, except for very few particular cases of exponent $p/q$ $($those when $q=1$, or when $p=1$ and $q$ is odd$)$, there is no reasonable single-valued definition for $x^{p/q}$ on all of the reals.
Thus, unless you're only concerned about the positive reals, your best bet is realizing that this is indeed a more subtle topic: that $x^{p/q}$ is not uniquely defined, and that the theory for dealing with this involves complex numbers.

For complex numbers $z$ and $w$, one usually defines $z^w$ as $\exp(z\,\text{Ln}(w))$, where $\exp$ is the exponential and $\text{Ln}$ is the complex logarithm.
The complex logarithm is not like the real one.
The real logarithm can be defined as the inverse of the exponential function $\exp:\mathbb{R}\longrightarrow[0,+\infty)$.
However, the complex logarithm is not injective; it is periodic with period $2\pi i$, and hence does not admit an inverse, at least not in the classical, single-valued sense.
What we do is take branches of the logarithm, that is, we choose some specific values of its multi-valued inverse as our single-valued inverse.
However, this choice is arbitrary, and moreover it cannot be done continuously throughout all of the complex plane.
This latter observation is why we talk about branch cuts: these are parts of the complex plane we remove from the domain of our chosen single-valued inverse.
We (once again arbitrarily!) choose some points of our would-be domain and decide that our single-valued inverse will not be defined on those points.
We do so in order for the resulting inverse to be continuous (and more) in what remains of its domain.
Notice how much arbitrary choice is involved in this?
In general, these choices must be specificed beforehand!
A: Interesting that you choose to say that 
$y(x)=x^{2/3}=\sqrt[3]{x^2}$
You could have said that
$y(x)=x^{2/3}=\sqrt[3]{x}^2$ instead, no?
Just asking yourself if both formulas are equal should give you hints as to why we can speak about solutions with imaginary parts. 
The equality symbol can both time be put in question... of course. 
A: By the definition of the cubic root, $y$ is a solution of
$$y^3=x^2>0.$$
In the reals, $$y=\sqrt[3]{x^2}$$ is the only solution.
In the complex, there are three distinct solutions, 
$$\sqrt[3]{x^2},\\\omega\sqrt[3]{x^2},\\\omega^2\sqrt[3]{x^2}$$
where $\omega$ is a complex cubic root of the unit (such as $\dfrac{-1+i\sqrt3}2$).
