# 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.

• Google calculator gives the same answer.
– user66360
Dec 15, 2013 at 17:07
• Welcome to the wonderful world of ambiguity. For non-integer exponents, the expression $a^x$ has multiple values. Wolfram Alpha gave you one, you found another. Dec 15, 2013 at 17:08
• And why isn't there any standard for it?
– CODE
Dec 15, 2013 at 17:09
• There is no clear way which option to choose. In different situations, you want different answers to the same question. Dec 15, 2013 at 17:11
• The problem is that functions of the form $f(x)=a^x$ are multi-valued functions in the complex plane. More precisely, this means $1^{1/3}$ has not one answer, not two answers, but a total of three answers! This corresponds to the fact that $f(x)=x^3-1$ has three roots. Dec 15, 2013 at 17:11

## 6 Answers

Interesting question. This is a subtle point so I will say a lot.

1. 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.

2. 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.

3. 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.

4. 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 $a, b$.

5. 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.

6. 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.

7. 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.

8. 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."

• about 8) we can see it as a function with the codomain C^3, no? Dec 15, 2013 at 17:58
• No, because there is no natural ordering. You can see it as a function with values in $C^3 \text{mod} S_3$ if you like, though. For the more usual way to make 8 rigorous, google the Riemann surface of the square root (or cube root in this case) function. Dec 15, 2013 at 18:04
• thank you, it makes sense Dec 15, 2013 at 18:06
• For the record, if anyone is still reading, if I bump into a symbol $(-1)^(2/3)$ without any context, in my opinion the least bad interpretation is that the answer is $1$. Dec 17, 2013 at 12:32

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, $$\left(e^{5 \pi i}\right)^\frac{2}{3} = e^{\frac{10}{3} \pi i} = e^{-\frac{2}{3} \pi i}.$$

If we look for real answer in two ways answer is $1$ for complex cases see other answers $$(-1)^{\frac{2}{3}}=(-1)^{2\cdot\frac{1}{3}}=((-1)^2)^{1/3}=1^{1/3}=1$$ $$(-1)^{\frac{2}{3}}=(-1)^{\frac{1}{3}\cdot2}=((-1)^{1/3})^2=(-1)^2=1$$

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}

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 $\cos\theta+i\sin\theta=\cos(2\pi/3)+i\sin(2\pi/3)=-0.5+i0.866...$

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: $$(-1)^t = e^{i t \pi} = \cos(i t \pi) + i\sin(i t \pi)$$