Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$
We rewrite $w$ and $z$ in polar form:
\begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*}
The equation $i \cdot w^r = z^s$ becomes
\begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*}
for some integer $k.$
Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that
\begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*}
Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.
We apply casework to the values for $s+3k:$


$s+3k\equiv0\pmod{3}$
There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv0\pmod{3},$ so there are $33$ values for $s.$ 
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
  $s+3k\equiv1\pmod{3}$
There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $s\equiv1\pmod{3},$ so there are $34$ values for $s.$ 
There are $9\cdot34=306$ ordered pairs $(r,s)$ in this case.
  $s+3k\equiv2\pmod{3}$
There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv2\pmod{3},$ so there are $33$ values for $s.$ 
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.

Together, the answer is $264+306+264=\boxed{834}.$
~MRENTHUSIASM