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I just opened a thread that sparked a new question that I now want to discuss (making a new thread however to not clog the other one).

Obviously looking at $(-1)^{\frac{1}{2}}$ one might have the number $i$ in mind. But as I've now learned defining complex powers has a pretty random aspect to it. So if I take a branch other than the principal branch of the logarithm, say $$L(Re^{i\varphi}):=\ln(R)+i\varphi+2\pi i,$$ I might as well write $(-1)^{\frac{1}{2}}=\exp(\frac{1}{2} L(-1))=\exp\left(\frac{3\pi i}{2}\right)$ which is completely different. Are my thoughts correct here? Is $e^{3\pi i/2}$ just another value that might be assigned to $(-1)^\frac{1}{2}$? And does this have any upside at all? What's the point of using different branches?

RedLantern
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    With $(-1)^{\frac 12}$, you want to find all numbers which, when squared, are equal to $-1$. The two numbers satisfying this property are $i$ and $-i$. And $e^{3\pi i/2} = -i$. – Bermudes Apr 21 '19 at 13:28
  • So writing $(-1)^\frac{1}{2}=i$ doesn't make any sense? – RedLantern Apr 21 '19 at 13:46
  • No it doesn't and i is written in that way only in elementary courses. In more advanced courses, we do something like this: consider the set of all pairs, (x, y) ofreal numbers. Define the sum of such pairs by (x, y)+ (u, v)= (x+ u, y+ v) and the product by (x, y)(u, v)= (xu- yv, uy+ xv). It is easy to show that all the usual arithmetic laws apply- that this is a "field". It is also easy to show that the pairs of the form (x, 0) are isomorphic to the real numbers and that (0, 1)(0, 1)= (-1, 0). We identify (1, 0) with 1 and (0, 1) with i so we can write (x, y)= x(1, 0)+ y(0, 1)= x+ iy. – user247327 Apr 21 '19 at 13:58

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The notation $A^B$ does not have a generally recognized definition that applies when $A=-1$ and $B=1/2$.

There are some fairly popular candidate definitions, though.

One is to consider that $A^B$ can be a "multi-valued expression" whose value can be every complex number that can be written as $\exp(Bz)$ where $z$ is a complex number that satisfies $\exp(z)=A$. According to this definition, the possible values of $(-1)^{1/2}$ are exactly $i$ and $-i$. Your $\exp(\frac{3\pi i}{2})$ happens to be one of these: it is equal to $-i$.

The trouble with this is that "multi-valued expression" is a confusing and sometimes ill-defined concept, and in particular the meaning of an equality between multi-valued expressions becomes very context-dependent ("it means whatever you need it to mean for what you're doing to make sense" is not really a good basis for rigorous reasoning). Depending on how you interpret the $=$, you could get $(-1)^{1/2}=i$ to be true or false or some kind of either/or. Not particularly enlightening.

A more careful and sophisticated way to handle the idea of "multi-valued" is Riemann surfaces. They are useful for giving a sensible meaning to ambiguous expressions that depend smoothly on some parameter -- but on the other hand doing it carefully means that simply writing down an expression such as $(-1)^{1/2}$ does not in itself point to a value. So it is not an answer to the immediate question.

A different definition would be to speak of the principal value of $A^B$, defined as $\exp(Bz)$ where $z$ is the particular solution of $\exp(z)=A$ whose imaginary part falls in the half-open interval $(-\pi,\pi]$. Then $(-1)^{1/2}=i$ becomes plain truth.

The trouble now is that the "principal part" operation lacks many of the nice properties we otherwise expect arithmetic operations to have. It is not continuous, for example. And the choice of $(-\pi,\pi]$ is pretty arbitrary. The definition at least gives the warm fuzzy feeling of having a definite answer to what the value of the expression is, but there's not much mathematical utility in having that answer.

And simply refusing to define a value for this expression seems to be at least as common as any of the above alternatives.

hmakholm left over Monica
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  • This is the only honest answer to this paradox, which comes up dozens of places on this site. Every other thread has numerous responses, all explaining why the fallacy is obviously wrong, yet each one giving a different "obvious" reason. This is the only honest answer: There isn't a really good candidate, let alone an agreed upon standard, for defining these expressions (as basic as their syntax may seem). – SRobertJames Jul 10 '23 at 12:17