What is the value of $(-1)^{\frac {2}{2}}$?

Question

When working with real numbers, should we interpret it as

$$\left( (-1)^{2} \right) ^{\frac 12}=1$$

or rather as

$$\left( (-1)^{{\frac 12}} \right) ^2= \text {not real/undefined}$$

or rather as $$(-1)^{\ 1} = -1$$

What about when we work with complex numbers?

EDIT: The reason why I called the second case "undefined" is because when we are only working with reals, $(-1)^{\frac 12}$ is not real. I don't think we can assume that further algebraic manipulations that work on real numbers also work on our $(-1)^{\frac 12}$.

Answer 1

It depends on what the person who wrote down $(-1)^{\frac22}$ meant by writing that.

The notation $A^B$ has a well-defined meaning that everybody agree on in the following cases

• $B$ is a natural number.
• $B$ is a (possibly negative) integer, and $A\ne 0$.
• $B$ is real and $A$ is positive.
• $A$ is $e$, and $B$ is any complex number.

The definitions of exponentiation that work in each of these cases are different, but fortunately they agree about the result whenever some combination of $A$ and $B$ fits into more than one of them.

Outside of these cases, ambiguity creeps in. In the general case of complex $A$ and $B$, one can treat $A^B$ as a multi-valued expression, but in some contexts it can be useful to single out a single one of the multiple values as the one we mean. (Indeed this is what happens in the two last of the cases above, where the canonical meaning of $A^B$ is only one of the multiple values the general case allows).

In particular, the rule that $A^{BC} = (A^B)^C$ holds in the three first cases cases, but not necessarily for a wider generalization that one might wish to use in a particular case.

For your particular case $(-1)^{\frac 22}$ does not immediately fall into any of the above categories. If we observe that $\frac22=1$, then we can rewrite it to $(-1)^1$ which does match the first two cases and gives the result $-1$.

However, the fact that somebody wrote the expression as $(-1)^{\frac22}$ suggests that they may have meant something different from that, and then there's no way out of either asking them, or try to figure out for yourself which interpretation of the notation makes sense in the context you see it in.

Answer 2

The rule $$a^{m/n} = (a^m)^{1/n} = (a^{1/n})^m$$ typically comes with the restriction that $a \ge 0$ if $n$ is even, so that rule can't really be applied to $(-1)^{2/2}$. I think the right way would be to simplify the exponent as much as possible first, then evaluate: $$(-1)^{2/2} = (-1)^1 = -1$$

The question about complex numbers is a little vague in its current form. By "work with complex numbers" do you mean what if we allow complex numbers so that $(-1)^{1/2} = i$ and isn't considered undefined? Do you mean what do we do if the base is complex and non-real? The exponent?