[Combinatorics]
Problem: A complex number $z$ is selected uniformly at random such that $|z|=1$. Compute the probability that $z$ and $z^{2019}$ both lie in Quadrant II in the complex plane.

Solution: For convenience, let $\alpha=\pi / 4038$. Denote by

$$
0 \leq \theta<2 \pi=8076 \alpha
$$

the complex argument of $z$, selected uniformly at random from the interval $[0,2 \pi)$. Then $z$ itself lies in Quadrant II if and only if

$$
2019 \alpha=\frac{\pi}{2}<\theta<\pi=4038 \alpha
$$

On the other hand, $z^{2019}$ has argument 2019日, and hence it lies in Quadrant II if and only if there is some integer $k$ with

$$
\begin{gathered}
\frac{\pi}{2}+2 k \pi<2019 \theta<\pi+2 k \pi \\
\Longleftrightarrow(4 k+1) \cdot \frac{\pi}{2}<2019 \theta<(4 k+2) \cdot \frac{\pi}{2} \\
\Longleftrightarrow(4 k+1) \alpha<\theta<(4 k+2) \alpha .
\end{gathered}
$$

Because it is also true that $2019 \alpha<\theta<4038 \alpha$, the set of $\theta$ that satisfies the conditions of the problem is the union of intervals:

$$
(2021 \alpha, 2022 \alpha) \cup(2025 \alpha, 2026 \alpha) \cup \cdots \cup(4037 \alpha, 4038 \alpha)
$$

There are 505 such intervals, the $j^{\text {th }}$ interval consisting of $(4 j+2017) \alpha<\theta<(4 j+2018) \alpha$. Each interval has length $\alpha$, so the sum of the intervals has length $505 \alpha$. Thus the final answer is

$$
\frac{505 \alpha}{2 \pi}=\frac{505}{2 \cdot 4038}=\frac{\mathbf{5 0 5}}{\mathbf{8 0 7 6}} .
$$

Final answer: $\frac{505}{8076}$