How to convince someone with calculus or plots that the curve $x y - 1 = 0$ is connected over complexes?

Question

Some curves defined by polynomial equations are disconnected over reals but not over complexes, e.g., $x y - 1 = 0$. How can we convince someone with background only on equations over reals that the curve drawn by above equation is connected over complexes? Is a plot or something possible, for example? It will be a 4d plot if x and y are expanded to real and imaginary parts. Any other plot, or algebraic way to show connectedness?

Answer 1

A solution over $\mathbb{C}$ is a pair $(z, 1/z)$ with $z \neq 0$. Consider another solution $(w, 1/w)$. There is a path from $z$ to $w$ in $\mathbb{C}$ which does not cross $0$ (this is the difference with $\mathbb{R}$), and this yields a path from $1/z$ to $1/w$ by inverting every point on the path. From this we get a path from $(z, 1/z)$ to $(w, 1/w)$ which lives inside the solution set.

You can draw the corresponding plots explicitly. It is especially easy on the unit circle, where inversion is just complex conjugation. For example, to go from $-1$ to $1$ in $\mathbb{C}$ you can take the upper semicircle of the unit circle. The "inverted" path (inverting every complex number on the path) is exactly the lower semicircle. This shows how to go from $(-1,-1)$ to $(1,1)$ (which wasn't possible over the reals) via the complex domain.

Answer 2

You can connect any two complex points $u, v$ with a spiral

$$u^{1-t}v^t$$ where $t$ runs from $0$ to $1$.

More precisely, in parametric polar coordinates,

$$\begin{cases}\theta=\theta_u(1-t)+\theta_vt,\\\rho=\rho_u^{1-t}\rho_v^t.\end{cases}$$

(You can very well eliminate $t$.)

Then the curve $(-\theta,\rho^{-1})$ corresponds to the inverse points and is a similar continuous spiral.