AIME I, 2022

Note that $w = \cos \frac{\pi}{6} + i \cdot \sin \frac{\pi}{6}$ is a primitive twelfth root of unity, so $w^3 = i$ and $w^r = w^{r+12m}$ for all integers $m$ and $r$. Furthermore, $z = \cos \frac{2\pi}{3} + i \cdot \sin \frac{2\pi}{3}$ is a primitive cube root of unity, so $w^4 = z$ and
$$i \cdot w^r = w^{r+12m+3} = z^s = w^{4s}.$$
Hence the given equation requires that there exists an integer $m$ such that $r + 12m + 3 = 4s$, so $r + 3 \equiv 4s \pmod{12}$. In particular, $r + 3 \equiv 0, 4, \text{ or } 8 \pmod{12}$.

* If $r + 3 \equiv 0 \pmod{12}$, then $r = 9 + 12m$ and $s = 3n$, where $0 \le m \le 7$ and $1 \le n \le 33$, accounting for $8 \cdot 33 = 264$ ordered pairs.
* If $r + 3 \equiv 4 \pmod{12}$, then $r = 1 + 12m$ and $s = 1 + 3n$, where $0 \le m \le 8$ and $0 \le n \le 33$, accounting for $9 \cdot 34 = 306$ ordered pairs.
* If $r + 3 \equiv 8 \pmod{12}$, then $r = 5 + 12m$ and $s = 2 + 3n$, where $0 \le m \le 7$ and $0 \le n \le 32$, accounting for $8 \cdot 33 = 264$ ordered pairs.
The requested number of ordered pairs $(r, s)$ is therefore $264 + 306 + 264 = 834$.