Are my exceptions for exponentiation rules complete?

Question

I wanted to learn about cases when rules of exponentiations don't hold. Namely, when the base of exponent is any real number or an imaginary number (i.e. a complex number with the real part equal to zero) and exponent is any integer. Alas, I wasn't able to find information (like in proofs) about cases when exponetial rules don't hold under said possible values of exponents and bases. I had to resort to creating a Python script that just plugged in different numbers in exponentiation equations and reported when something went wrong (I heard that normally these things are proved in Abstract algebra, but I don't know Abstract algebra). This is basically my non-strict proof for rules of exponentiations for such bases and exponents. I learned following things:

0.I had to take stance on how to treat $0^0$, should it be considered as an error or not? Like you can avoid division by zero in some cases if you decide that $0^0$=1. But I have decided to treat $0^0$ as undefined, to play it safe. Thus my script treated all occurences of $0^0$ as sign that given equality broke down.

1.$x^{a}x^{b}=x^{a+b}\space$ is wrong when x=0 & $(a\leq0\space \vee\space b\leq0\space)$

2.$\frac{x^{a}}{x^{b}}=x^{a-b}\space$ is wrong when x=0

3.$(x^{a})^b=x^{ab}\space$ is wrong when x=0 & $(a\leq0\space \vee\space b\leq0\space)$

4.$(xy)^a=x^ay^a \space$ is wrong when $a\leq0\space$&$(x=0\space \vee\space y=0)$

5.$(\frac{x}{y})^{a}=\frac{x^{a}}{y^{a}}\space$ is wrong when y=0 $\vee$ (x=0 & $a\leq0$)

All exceptions to the rules of exponentiation here lead either to division by zero or to $0^0$

I want to know one thing. Are my exceptions complete (given assumptions about possible values of exponents and bases that I mentioned before in bold)? By complete I mean, that they have 0 false positives and 0 false negatives. Or to put another way, are there any wrong [rules of exponentiation] equations that they let slip and any corretct equations that they mistake for incorrect ones?

Answer 1

These are the salient facts about exponentiation, particularly when complex numbers are involved:

• There is a function $\exp z := \sum_{n=0}^\infty \dfrac{z^n}{n!}$, where the "$z^n$" here is just multiplication of $n$ copies of $z$ together. $\exp z$ converges for all complex numbers $z$ to a complex value. (Yes, you tagged "precalculus", but exponentiation is defined in terms of this function when the exponent is not rational or the base is not real, and this function is only definable using limits. So you will have to allow this one exception to that requirement.)

• $\exp 0 = 1,\ e := \exp 1 \approx 2.718281828,\ \exp(\pi i) = -1$.

• The exponential function satisfies the property $$\exp(w + z) = \exp(w)\exp(z)$$for all complex numbers $w, z$.

• For all $z, \exp z \ne 0$.

• For any complex number $w \ne 0$, and any real number $\alpha$, there exists a unique complex number $z$ with $w = \exp z$ and $\alpha - 2\pi < \text{Im}(z) \le \alpha$.

By the last property, we can choose some $\alpha$, and then define $\log w := z$ to be the unique $z$ with $w = \exp z$ and $\alpha - 2\pi < \text{Im}(z) \le \alpha$. And then we can define $$w^z := \exp(z\log w)$$

Problems arise because the definition of $\log w$ depends on our choice of $\alpha$. While $\log w$ is well-behaved elsewhere, the value suddenly jumps by $2\pi i$ along the ray $z = t\exp {i\alpha}, t > 0$. Whereas the exponential rule $\exp(w + z) = \exp(w)\exp(z)$ is true everywhere, the corresponding logarithm rule is $$\log(wz) = \log w + \log z + 2k \pi i$$ where $k$ is whatever integer is needed to have $\alpha - 2\pi \le \text{Im}(\log w + \log z ) + 2k\pi \le \alpha$.

This jumping around of logarithm values then affects the value of $w^z$, causing it to jump as well. If $z = n$ is integer, then $n\log w$ will just change by $2n\pi i$, and since $\exp(2n\pi i) = 1$, the value of $w^n = \exp(n\log w)$ is not affected by the jump, nor by the choice of $\alpha$. It has a well-defined value, which agrees with the original definition of $n$ copies of $w$ multiplied together (or the inverse of $|n|$ copies multiplied together when $n < 0$).

But for non-integer $z, w^z$ will depend on the choice of $\alpha$, and will suddenly jump in value as $w$ crosses the ray $t\exp {i\alpha}, t > 0$. This ray is called a "branch cut", and the situation is actually more complicated than I've said. The branch cut need not be taken along a line. It could follow any non-self-intersecting curve from $0$ to $\infty$.

When a definitive value for $w^z$ is needed and no other requirements exist, we usually choose $\alpha = \pi$, placing the branch cut along the negative real axis. But in general, mathematicians choose not to have a fixed definition of $\log w$ and $w^z$. The reason for this is that there are plenty of calculations where $\alpha = \pi$ is a very poor choice. Indeed, no matter what branch cut we could choose, there will be plenty of cases where that choice will make life exceedingly difficult. So instead of having a fixed definition for $\log w$, we instead leave the meaning indeterminant, and allow mathematicians to choose whichever definition best suits the particular calculation at hand.

So the answer to your question depends on which branch cut you choose. In all cases the correct behavior may be obtained by applying the rules $$w^z = \exp(z\log w)\\\exp(w + z) = \exp(w)\exp(z)\\\log(wz) = \log w + \log z + 2k\pi i$$ where the correction factor is chosen to keep the logarithm in the appropriate branch.

It is actually not wise when dealing with complex exponentiation to directly apply any of your five rules, because it is hard to determine when a branch cut was crossed in them.