The complex map $z \mapsto z^{1/2}$ is a multivalued function on the complex plane—you can get two different square roots from it $\pm z^{1/2}$ depending on what you want.
How you bring the square root into complex analysis
In many ways the positive square root is extremely “well-behaved”, like you can get all of its derivatives at the point $z=1$ assuming the positive square root, just by looking at its real derivatives. Those derivatives can be used as the basis for a Taylor series which converges on a unit disk around that point, and you can plausibly regard all of those points on that unit disk as having a "positive square root" given by that Taylor series and it's all very intuitive. For example if $a, b, ab$ are all in that unit disk then indeed you have $(ab)^{1/2} = a^{1/2} b^{1/2}$ just like you'd expect. Express any point as $r~e^{i\theta}$ in this disk and that Taylor series will indeed give you $+\sqrt{r}~e^{i\theta/2}$ just like you'd expect. The function is smooth on that disk and everything.
OK but that's just the unit disk $|z - 1| < 1.$ How do we extend the positive square root further? Well, complex analysis has a really nice aspect where something being once-differentiable in a neighborhood of a point, makes it infinitely-often-differentiable on that neighborhood. Differentiable implies smooth and smooth implies convergence to Taylor series. This allows a technique called “analytic continuation” which works like so:
- So like we said: you get a Taylor series which converges to the square root on the unit disk centered at the point $z_0=1$.
- Analytic continuation just says, "let's use this Taylor series to get some new derivatives for a point at $z_1=e^{i\pi/4}$, say, that is in that disk." And you get a new series that is convergent in a unit disk about $z_1$. This includes points not in our original set!
- Use this to get new derivatives at a point $z_2 = e^{i\pi/2}$, and use that to get new derivatives at $z_3 = e^{i3\pi/4}$, and so on. You can “walk” your smooth Taylor expansion around the complex plane, filling it all up with that function.
Analytic continuation is a really powerful method to push a valid definition across the whole complex plane from something little, in our case we took the definition on the real line and that Taylor expansion gives us a way to evaluate an arbitrary square root of some other complex number.
Here's the deal: when you do this analytic continuation procedure with $z \mapsto z^{1/2}$ and you make a full rotation about the complex plane, you finally come to evaluate this at $z_{8} = 1$ and now you find out that $z_{8}^{1/2} = -1.$ So analytic continuation works amazingly except that you don't get global consistency for these multivalued functions: when you walk around in a circle you discover that you have now gotten a different answer for the function at that point. And moreover the procedure was kind of hinting that to you when it said "your radius of convergence is never going beyond 0." Zero was an obstacle for the analytic continuation.
Riemann surfaces and branch cuts
The usual way to understand this phenomenon is to say that this function is nice and smooth at every point other than $\{0\}$ on some natural domain called a "Riemann surface", but that this natural domain isn't the complex plane $\mathbb C$. Instead it is something like $\{0, 1\} \times \mathbb C,$ two copies of the complex plane, which are each called "the branch sheets" of the Riemann surface. Crucially they are knit together into a "helix" structure, so there is an imaginary line called a "branch cut" drawn from 0 to infinity (exactly where is a matter for convention but most folks in practice take the negative real line) on all of the branch sheets, and when you cross over that line in this Riemann surface, you teleport onto the other sheet. So $e^{i(\pi+\delta\theta)}$ is on sheet 0 for $\delta\theta \le 0$ but is on sheet 1 for $\delta\theta > 0.$ This is the natural domain on which $z\mapsto z^{1/2}$ is nice and smooth and Taylor-expansions converge and all that. Analytic continuation allowed you to explore the full Riemann surface and did not care about your desire for consistency over just $\mathbb C,$ and when the analytic continuation just stretched out from $z_5$ across the negative real line, the "disk of convergence" poked out from sheet 0 into sheet 1 following the Riemann surface.
How this answers your question
Taken this way, $z=e^{\pi i}$ is right on the branch cut line, we can say that conventionally this is on the same branch sheet as $z=1$. But $z=e^{2\pi i}$ is an instruction to walk from the point $(1, 0)$ along the unit circle all the way around: this passes over that branch cut once and ends up at the other side. Thus in this structure $e^{2\pi i} \ne 1$, because they are on different branch sheets and they have different values for the square root ... in fact $1^{1/2}$ on the first branch sheet = 1, as we chose by fiat when we started doing the "positive square root," but $(e^{2\pi i})^{1/2}$ on the second branch sheet = -1 because we have crossed over onto this other branch sheet where all the values are negatives of what they were on the original.
[fake-proofs] "-1=1"yields over 400 results, at least a few of which are likely relevant (if not duplicate). – Blue Jul 11 '22 at 20:53