What is $(-1)^{\frac{2}{3}}$? Following from this question, I came up with another interesting question: What is  $(-1)^{\frac{2}{3}}$? 
Wolfram alpha says it equals to some weird complex number (-0.5 +0.866... i), but when I try I do this: $(-1)^{\frac{2}{3}}={((-1)^2)}^{\frac{1}{3}}=1^{\frac{1}{3}}=1$.If it has multiple "answers", should we even call it a "number"? Because if we don't, it would be a bit different from what we were taught in elementary school. I actually thought if it doesn't have a variable in it, it should be a number.  I'm a bit confused. Which one is correct and why? I would appreciate any help.
 A: As a number of commenters have pointed out, there are three possible values that satisfy $$x = (-1)^{\frac{2}{3}}.$$
These can be found by replacing -1 with three different ways of expressing -1 as a complex exponential, $e^{\pi i}, e^{-\pi i},$ and $e^{3 \pi i}$. Substituting in, we find that:
$$\begin{align}
\left(e^{\pi i}\right)^\frac{2}{3} & = e^{\frac{2}{3} \pi i} \approx -0.5 + 0.866025 i\\
\left(e^{-\pi i}\right)^\frac{2}{3} & = e^{-\frac{2}{3} \pi i} \approx -0.5 - 0.866025 i\\
\left(e^{3 \pi i}\right)^\frac{2}{3} & = e^{2 \pi i} = 1
\end{align}$$
There are even more ways to express -1, specifically $e^{(1 + 2n) \pi i}$, where $n \in \mathbb{Z}$ is any integer. This means that there are infinite ways to express -1, and all will give us a valid answer. However, if you try some, you'll notice that they are repeats of the three we have already seen. For example, $$\left(e^{5 \pi i}\right)^\frac{2}{3} = e^{\frac{10}{3} \pi i} = e^{-\frac{2}{3} \pi i}.$$
A: If we look for real answer in two ways answer is $1$ for complex cases see other answers
$$(-1)^{\frac{2}{3}}=(-1)^{2\cdot\frac{1}{3}}=((-1)^2)^{1/3}=1^{1/3}=1$$
$$(-1)^{\frac{2}{3}}=(-1)^{\frac{1}{3}\cdot2}=((-1)^{1/3})^2=(-1)^2=1$$
A: Interesting question. This is a subtle point so I will say a lot.


*

*There is a function defined and continuous on the set of all real numbers, which is
$$
x \mapsto x^{1/3}
$$
where the symbol $x^{1/3}$ denotes the unique real number whose cube is $x$. Similar for other fractions with $1$ for a numerator and an odd denominator.

*There is a function defined and continuous on the set of all positive real numbers, which is
$$
x \mapsto x^{1/2}
$$
where the symbol $x^{1/2}$ denotes the unique positive real number whose square is $x$. Without that caveat "positive" the symbol would be ambiguous, in contrast to the case with cube roots. Similar with other fractions with $1$ for a numerator and an even denominator.

*There is a function defined and continuous on the set of all real numbers, which is
$$
x \mapsto x^2
$$
which needs no more justification.

*There is a rule for exponents which works when $x$ is real and positive: when you see
$$
x\mapsto x^{ab}
$$
you may write this as $(x^a)^b$ or as $(x^b)^a$. In fact this rule works for any fractions $a, b$.

*Rule 4 does not continue to work when $x$ is a negative real number and $a$ or $b$ are allowed to be fractions. For example,
$$
(-1)^1 = (-1)^{(1/2)*2} \neq ((-1)^2)^{1/2} = 1.
$$
Note that the failure here has nothing to do with imaginary numbers; indeed, in the above equality, I never took a square root of a negative number. It's just that Rule 4 does not work when $x$ is allowed to be negative.

*For this reason, it's somewhat dangerous to try to define a symbol like $(-1)^{2/3}$. For instance, is it the same as $(-1)^{4/6}$? Note that either answer you give will be problematic; on the one hand $2/3$ is the same number as $4/6$, and so whatever definition we pick we had better get the same value; on the other hand, we shouldn't be speaking of taking even roots of negative numbers if we insist on working with only real numbers. 

*We can introduce complex numbers to get rid of the problem in (6). However, when we introduce complex numbers, it's no longer true that there is a unique number whose cube is a given real number. For instance, as you've discovered, by playing with Wolfram, there are cube roots of $1$ in the complex plane other than $1$ itself. Therefore, the function defined in (1) breaks down if we no longer insist that $x^{1/3}$ be real-valued.

*On the complex plane, it is best to think of $x^{1/3}$ as being a "multi-valued function," i.e. not a function at all, but something that takes in one value and returns multiple values. In fact, every complex number but $0$ has three distinct "cube roots."
A: If you read a bit below in Wolfram Alpha
you'll find all 3rd roots of $(-1)^2$:
$$
\begin{align}
\operatorname{e}^{\frac{2i\pi}{3}}&\approx 0.5+0.86603i\\
1&\quad\text{real root}\\
\operatorname{e}^{-\frac{2i\pi}{3}}&\approx 0.5-0.86603i\\
\end{align}
$$
A: The key is De Moivre's Theorem.  You want the cube roots of 1.
You might be able to prove that the square of $\cos\theta+i\sin\theta$ is $\cos2\theta+i\sin2\theta$, where $i$ is the square-root of -1.
Also, the cube of $\cos\theta+i\sin\theta$ is $\cos3\theta+i\sin3\theta$.
If we want the cube to equal 1, then $3\theta$ can be any multiple of $2\pi$ because $\cos2k\pi=1$ and $\sin2k\pi=0$ for any integer $k$.
Then $\theta = 2k\pi/3$.  For example, if $k=1$, we find $\theta=2\pi/3$, and
the cube-root is $\cos\theta+i\sin\theta=\cos(2\pi/3)+i\sin(2\pi/3)=-0.5+i0.866...$
A: How to choose which of the values to use?  One consideration: Perhaps I want $(-1)^t$ to be a continuous function of $t$, defined for all real $t$.  If I want that, then the conventional choice is:
$$
(-1)^t = e^{i t \pi} = \cos(i t \pi) + i\sin(i t \pi)
$$
