Why doesn't $2^2 = -4$?

Question

I was just curious because a number raised to the $\frac 1x$ where $x$ is an integer greater than $1$ has $x$ solutions, why can't a number to the $x$ where $x$ is an integer greater than $1$ also have $x$ solutions?

Answer 1

The notion of multivalued exponentiation makes the most sense once you learn complex exponentiation.

In complex numbers, $x^y$ with $x\neq 0$ has $q$ values if $y=p/q$ is a rational number in reduced form, and infinitely many value otherwise.

So $2^{2}$ has one complex value, $2^{3/7}$ has $7$ complex values, $2^{\sqrt{2}}$ and $2^{1+i}$ both have infinitely many complex values.

This is because $x^y=e^{y\log x}$ by definition, and $\log x$ has infinitely many values for all $x$, differing by multiples of $2\pi i$. So $e^{y\log x}$ will yield $q$ distinct values in the rational case and infinitely many in the case where the number is not a rational.

For example, if $y$ is an integer, then $q=1$. If we take two different values for $\log x$, $\ell$ and $\ell+k(2\pi i)$, then $x^{y}$ can be either $e^{y\ell}$ and $$e^{y\ell + ky(2\pi i)}=e^{y\ell}e^{2ky\pi i}=e^{y\ell}$$

So there is only one value. The infinite set of values for $\log x$ yield the same value for $e^{y\log x}$.

Amongst other things, this means that some exponentiation theorems are a little murkier. For example: $$\left(2^{\sqrt 2}\right)^{\sqrt 2}$$ has infinitely many values, and $2^{2}=4$ is only one of those values. Any value of $x^{yz}$ is a value of $\left(x^y\right)^z$, but not visa versa.

Similarly, given values $y,z$, any value of $x^{y+z}$ is one value of $x^{y}x^z$, but not visa versa.

On the other hand, any value of $x^zy^z$ is a value of $(xy)^z$, and visa versa.