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.
Solution 1:
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."
Solution 2:
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, $$\left(e^{5 \pi i}\right)^\frac{2}{3} = e^{\frac{10}{3} \pi i} = e^{-\frac{2}{3} \pi i}.$$
Solution 3:
If we look for real answer in two ways answer is $1$ for complex cases see other answers $$(-1)^{\frac{2}{3}}=(-1)^{2\cdot\frac{1}{3}}=((-1)^2)^{1/3}=1^{1/3}=1$$ $$(-1)^{\frac{2}{3}}=(-1)^{\frac{1}{3}\cdot2}=((-1)^{1/3})^2=(-1)^2=1$$