Let us denote $a = \frac{y}{x}$, $b = \frac{z}{y}$, $c = \frac{x}{z}$, where $x, y, z$ are nonzero complex numbers. Then
$$p+3 = 3 + \sum_{\text{cyc}} \left(\frac{x}{y} + \frac{y}{x}\right) = 3 + \frac{x^2(y+z) + y^2(z+x) + z^2(x+y)}{xyz} \\ = \frac{(x+y+z)(xy+yz+zx)}{xyz}.$$
$$q-3 = -3 + \sum_{\text{cyc}} \frac{y^2}{zx} = \frac{x^3 + y^3 + z^3 - 3xyz}{xyz} \\ = \frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{xyz}.$$

$$\mathbb{R} \ni 3(p+3) + (q-3) = \frac{(x+y+z)(x^2+y^2+z^2+2(xy+yz+zx))}{xyz} = \frac{(x+y+z)^3}{xyz}.$$

Scale $x, y, z$ in such a way that $x + y + z$ is nonzero and real; hence so is $xyz$. Thus, as $p + 3 \in \mathbb{R}$, we conclude $xy + yz + zx \in \mathbb{R}$ as well. Hence, $x, y, z$ are the roots of a cubic with real coefficients. Thus,
* either all three of $\{x, y, z\}$ are real (which implies $a, b, c \in \mathbb{R}$),
* or two of $\{x, y, z\}$ are a complex conjugate pair (which implies one of $a, b, c$ has absolute value $1$).
Both of these were forbidden by hypothesis.

Therefore, the only possibility is $x + y + z = 0$, which gives $p = -3$, $q = 3$.

Second solution, found by contestants

The main idea is to make the substitution
$$x = a + \frac{1}{c}, \quad y = b + \frac{1}{a}, \quad z = c + \frac{1}{b}.$$
Then we can check that
$$\begin{aligned} x+y+z &= p \\ xy+yz+zx &= p+q+3 \\ xyz &= p+2. \end{aligned}$$
Therefore $x, y, z$ are the roots of a cubic with real coefficients. As in the previous solution, we note that this cubic must either have all real roots, or a complex conjugate pair of roots. We also have the relation $a(y+1) = ab + a + 1 = x + 1$, and likewise $b(z+1) = y + 1$, $c(x+1) = z + 1$. This means that if any of $x, y, z$ are equal to $-1$, then all are equal to $-1$.
Assume for the sake of contradiction that none are equal to $-1$. In the case where the cubic has three real roots, $a = \frac{x+1}{y+1}$ would be real. On the other hand, if there is a complex conjugate pair (without loss of generality, $x$ and $y$) then $a$ has magnitude $1$. Therefore this cannot occur.

We conclude that $x = y = z = -1$, so $p = -3$ and $q = 3$. The solutions $(a, b, c)$ can then be parameterized as $(a, -1 - \frac{1}{a}, -\frac{1}{1+a})$. To construct a solution, we need to choose a specific value of $a$ such that none of the wrong conditions hold; when $a = 2i$, say, we obtain the solution $(2i, -1 + \frac{i}{2}, \frac{-1+2i}{5})$.

Third solution by Luke Robitaille and Daniel Zhu

The answer is $p = -3$ and $q = 3$. Let's first prove that no other $(p, q)$ work.

Let $e_1 = a + b + c$ and $e_2 = a^{-1} + b^{-1} + c^{-1} = ab + ac + bc$. Also, let $f = e_1 e_2$. Note that $p = e_1 + e_2$.

Our main insight is to consider the quantity $q' = \frac{b}{a} + \frac{c}{b} + \frac{a}{c}$. Note that $f = q + q' + 3$. Also,
$$\begin{aligned} qq' &= 3 + \frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} + \frac{bc}{a^2} + \frac{ac}{b^2} + \frac{ab}{c^2} \\ &= 3 + a^3 + b^3 + c^3 + a^{-3} + b^{-3} + c^{-3} \\ &= 9 + a^3 + b^3 + c^3 - 3abc + a^{-3} + b^{-3} + c^{-3} - 3a^{-1}b^{-1}c^{-1} \\ &= 9 + e_1(e_1^2 - 3e_2) + e_2(e_2^2 - 3e_1) \\ &= 9 + e_1^3 + e_2^3 - 6e_1e_2 \\ &= 9 + p(p^2 - 3f) - 6f \\ &= p^3 - (3p + 6)f + 9. \end{aligned}$$

As a result, the quadratic with roots $q$ and $q'$ is $x^2 - (f-3)x + (p^3 - (3p+6)f + 9)$, which implies that
$$q^2 - qf + 3q + p^3 - (3p+6)f + 9 = 0 \iff (3p+q+6)f = p^3 + q^2 + 3q + 9.$$

Claim — $f$ is not real.

Proof. Suppose $f$ is real. Since $(x - e_1)(x - e_2) = x^2 - px + f$, there are two cases:
* $e_1$ and $e_2$ are real. Then, $a, b$, and $c$ are the roots of $x^3 - e_1x^2 + e_2x - 1$, and since every cubic with real coefficients has at least one real root, at least one of $a, b$, and $c$ is real, contradiction.
* $e_1$ and $e_2$ are conjugates. Then, the polynomial $x^3 - \bar{e}_2x^2 + \bar{e}_1x - 1$, which has roots $\bar{a}^{-1}, \bar{b}^{-1}$, and $\bar{c}^{-1}$, is the same as the polynomial with $a, b, c$ as roots. We conclude that the multiset $\{a, b, c\}$ is invariant under inversion about the unit circle, so one of $a, b$, and $c$ must lie on the unit circle. This is yet another contradiction. $\square$

As a result, we know that $3p + q + 6 = p^3 + q^2 + 3q + 9 = 0$. The second miracle is that substituting $q = -3p - 6$ into $q^2 + 3q + p^3 + 9 = 0$, we get
$$0 = p^3 + 9p^2 + 27p + 27 = (p + 3)^3,$$
so $p = -3$. Thus $q = 3$.

It remains to construct valid $a, b$, and $c$. To do this, let's pick some $e_1$, let $e_2 = -3 - e_1$, and let $a, b$, and $c$ be the roots of $x^3 - e_1x^2 + e_2x - 1$. It is clear that this guarantees $p = -3$. By our above calculations, $q$ and $q'$ are the roots of the quadratic $x^2 - (f-3)x + (3f-18)$, so one of $q$ and $q'$ must be $3$; by changing the order of $a, b$, and $c$ if needed, we can guarantee this to be $q$. It suffices to show that for some choice of $e_1$, none of $a, b$, or $c$ are real or lie on the unit circle.
To do this, note that we can rewrite $x^3 - e_1x^2 + (-3 - e_1)x - 1 = 0$ as
$$e_1 = \frac{x^3 - 3x - 1}{x^2 + x},$$
so all we need is a value of $e_1$ that is not $\frac{x^3-3x-1}{x^2+x}$ for any real $x$ or $x$ on the unit circle. One way to do this is to choose any nonreal $e_1$ with $|e_1| < 1/2$. This clearly rules out any real $x$. Also, if $|x| = 1$, by the triangle inequality $|x^3 - 3x - 1| \ge |3x| - |x^3| - |1| = 1$ and $|x^2 + x| \le 2$, so $\left|\frac{x^3-3x-1}{x^2+x}\right| \ge 1/2$.

Therefore, the only possible values of $(p, q)$ are $(-3, 3)$.