I will start with monodromies in the context of analytic continuation and linear differential equations, since that is where I encountered them first. Here are the current prerequistes for this article: complex numbers, Taylor series.

The basic problem goes back to the square root, when extended to the complex numbers. The school rules, such as \( (xy)^a = x^a y^a \), no longer work, as exemplified by $$-1 = \mathrm{i}\cdot\mathrm{i} = \sqrt{-1}\sqrt{-1} \overset{?}{=} \sqrt{1} = 1.$$ This happens, because of the root ambiguity, but could be blamed on poor notation and intuition inherited from (positive) real numbers. For \(r = \sqrt{w}\) means only that \(r\) satisfies \(r^2 = w\), and in that sense we do have \(\mathrm{i}^2=\sqrt{1}\), because \((\mathrm{i}^2)^2 = 1\). It is only when we call \(r\) the number that satisfies \(r^2=w\) that problems begin. Keeping track of all the roots consistently is thus the first priority.

In the general, complex case we go back to the equality \(\exp(\alpha)\exp(\beta)=\exp(\alpha+\beta)\), and recall that \(\exp\) is periodic. So that although \((e^{\alpha})^n = e^{n\alpha}\) gives an immediate root formula \(\sqrt[n\,]{e^{\beta}} = e^{\beta/n}\), we could just as well take \((e^{\alpha})^n = e^{n\alpha+2\pi k\mathrm{i}}\), \(k\in\mathbb{Z}\), giving a whole family of roots \(e^{\beta/n}e^{2\pi\mathrm{i}k/n}\).

Why do we care about the exponents? Firstly, because this is a straightforward way of defining arbitrary powers via \(z^a := \exp(a\ln(z))\), and we know everything about the exponential functions, including how many different solutions (roots) to expect^1.1. According to the previous formula, there are only \(n\) essentially distinct values of \(k\) in the above formulae, e.g. \(\{0, 1,\ldots, n-1\}\), that give distinct complex roots.

Rephrasing the example with \(\exp\), we have \( (e^{\pi\mathrm{i}})^{1/2}(e^{\mathrm{i}\pi})^{1/2} = (e^{2\pi\mathrm{i}})^{1/2} = e^{\pi\mathrm{i}} = -1\), and all seems well. Except that, by periodicity, \(e^{2\pi\mathrm{i}} = e^0\), and clearly everyone must agree that \((e^0)^{1/2} = e^{0/2} = e^0 = 1\), so we're back where we started. If only there were some way to distinguish \(1=e^0\) from \(1=e^{2\pi\mathrm{i}}\)…

The second reason is that using functions shifts the focus from numeric (arithmetic) to analytic properties. We are not so much interested in how to compute the value of square root accurately (continued fractions of course!), but rather how the value changes as the argument moves around the complex plane. The representation \(z=\rho\,e^{\mathrm{i}\varphi}\), makes it painfully clear, that the modulus \(\rho\) plays little role, since it is non-negative; it's the argument \(\varphi\), whose loops give us the headache1.15.

On the one hand, the interval of \([0, 2\pi)\) is enough for \(\varphi\) to parametrize all the numbers, on the other, it makes intuitive sense to describe several revolutions around the origin, and use numbers such as \(4\pi\) or \(40\pi\) to quantify how the path winds. The paths \([0,2\pi) \ni \varphi\mapsto e^{\mathrm{i}\varphi}\) and \([0,10\pi) \ni \varphi\mapsto e^{\mathrm{i}\varphi}\) are different as functions, even though they are projected onto the same unit circle. (This hints at the fundamental group, which will come into play later.) And perhaps the roots “feel” the position on the full path (as given by the argument \(\varphi\)), not just the final value of \(z=e^{\mathrm{i}\varphi}\). So, let's look at the root, when followed along two different (half) loops.

Several choices have to be explained here. Why are both the roots initially \(1\) (\(\sqrt{w}\) could be \(-1\))? Because we are now paying attention to continuity — we wish for everything to lie close together initially, so if the arguments start at the same point, so should the roots. Secondly, why do the roots follow their arguments? Because we also expect the root, as a function, to behave nicely (analytically) around \(1\). Its Taylor expansion gives \(\sqrt{1+z}\approx 1+\frac{1}{2}z\), so that \(\mathrm{Im}(\sqrt{u})=\mathrm{Im}(u) >0\), while \(\mathrm{Im}(\sqrt{w})< 0\), and the respective roots must fall on either side of the real axis. Then the inevitable happens, when both numbers reach the negative semiaxis, and the imaginary parts no longer distinguish between them. And yes, we have long left the region of convergence of the Taylor series (going in opposite directions to make it even worse), so no wonder things break down. Let's try following just one path for as long as possible.

Once we distinguish between the sheets, things work wonderfully: $$ {\color{darkorange}-1} = \sqrt{\color{darkorange}-1}\sqrt{\color{darkorange}-1} = \sqrt{({\color{darkorange}-1})({\color{darkorange}-1})} = \sqrt{\color{blue}1} = {\color{darkorange}-1}, $$ whereas \(\sqrt{\color{darkorange}1}={\color{darkorange} 1}\), and there is no confusion, because each \(1\) gets its own (arithmetic) root. We can also clearly distinguish between \({\color{darkorange} \mathrm{i}}=\sqrt{\color{darkorange}-1}\) and \({\color{blue} -\mathrm{i}}=\sqrt{\color{blue}-1}\), for which, thankfully, \({\color{darkorange} \mathrm{i}}({\color{blue} -\mathrm{i}}) = {\color{darkorange} 1}\).

If a civilization lost the ability to print or see colour, all of the above formulae would still make sense, with the tiny modification that \(\sqrt{}\) be read as “some root of”; \(-\mathrm{i}\) is some root of \(-1\) after all. They (and present-day formalists) would still insist on introducing a rigorous notation, not to run afoul of arithmetic^1.2 (or even just intransitivity of the relation “equal to some”). They could consider orange and blue to correspond to two components of a single vector like so: $$ {\color{darkorange} 1} = \begin{bmatrix} 1\\ 0 \end{bmatrix}, \quad {\color{blue}1} = \begin{bmatrix} 0\\ 1 \end{bmatrix}.$$

1.1	The appearance of the logarithm might seem circular — isn't it defined as the inverse of \(\exp\) and multivalued on top of that? — but it captures precisely the ambiguity of \(\beta\) when representing \(x\) as \(e^{\beta}\), so that we can deal with it explicitly. If \(a\) is an integer, the ambiguitiy is irrelevant, otherwise we know exactly what choices are left. The usual perverse example is \(\mathrm{i}^{\mathrm{i}}\), for which \(\ln(\mathrm{i}) = (2k+\frac{1}{2})\pi\mathrm{i}\), so in the end we get \( e^{-\pi(4k+1)/2}\) — infinitely many distinct real values.
1.15	Those familiar with a littel bit of differential geometry will protest, that \(\rho\) is just as guilty, because the change of coordinates from \(z=a+\mathrm{i}b\) to \(z=\rho e^{\mathrm{i}\varphi}\) is singular at \(\rho=0\). But that is begging the question, as we are trying to get to the bottom of what precisely goes wrong. It is no coincidence, that we have chosen the loops to go around zero specifically, but at the same time, there is no ned for \(\rho\) to apprach zero for the problems to appear. Still, as is usually the case, it is not one or the other — the different perspectives are complimentary and entangled, such is the beauty of maths.
1.2	For consider the equality \(\sqrt{\color{blue}1} = {\color{orange}-1} = {\color{orange}-}\sqrt{\color{orange}1}\), which becomes \(\sqrt{1} = -\sqrt{1}\) when projected to black and white, and just \(1=-1\) on division. Or \(\sqrt{\color{blue}1}+\sqrt{\color{orange}1}=0\), leading to \(2\sqrt{1}=0\). Except of course, those simplifications are unjustified, for we can at best write \(\sqrt{1}/\sqrt{1}=1\) (some root of one divided by some root of one is minus one) in the former, and \(\sqrt{1}+\sqrt{1}=0\) (some root of one plus some root of one is zero) in the latter. It's the qualifier some doing all the heavy lifting (and confusing).