[example: A Figure-Eight Does Not Enclose Points Uniformly]
Consider the figure-eight curve $\gamma: [0, 2\pi] \to \mathbb{C}$ given by $\gamma(t) = \sin(t) + i\sin(2t)/2$. This curve crosses itself at the origin and traces two loops, one traversed counterclockwise and one clockwise. A point inside the upper loop, say $z_0 = 0.3i$, is encircled once counterclockwise. A point inside the lower loop, say $z_1 = -0.3i$, is encircled once clockwise — that is, with winding number $-1$. But the origin itself, through which the curve passes, is neither strictly inside nor outside either loop in the naive sense.
This example already shows that "inside" and "outside" are not binary concepts for general closed curves. Different points can be surrounded different numbers of times, and the sign carries meaning: counterclockwise is positive, clockwise is negative.
To verify: as $t$ increases from $0$ to $\pi$, the curve $\gamma$ traces the upper loop counterclockwise. For $z_0 = 0.3i$, the vector $\gamma(t) - z_0$ starts pointing roughly rightward, rotates counterclockwise through a full turn, and returns to its starting direction — one full positive rotation, so $n(\gamma, 0.3i) = +1$. As $t$ increases from $\pi$ to $2\pi$, the curve traces the lower loop clockwise. For $z_1 = -0.3i$, the vector $\gamma(t) - z_1$ rotates clockwise through a full turn, giving $n(\gamma, -0.3i) = -1$. Neither loop winds around the other loop's interior point, so the outside of each loop contributes zero net rotation.
[/example]