Suppose $a, b$, and $c$ are three complex numbers with product 1. Assume that none of $a, b$, and $c$ are real or have absolute value 1. Define 
$$ p=(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \quad \text{and} \quad q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}. $$
Given that both $p$ and $q$ are real numbers, find all possible values of the ordered pair $(p, q)$.
We show $(p, q)=(-3,3)$ is the only possible ordered pair. ## 【T First solution.   $$ \begin{aligned} p+3 & =3+\sum_{\text {cyc }}\left(\frac{x}{y}+\frac{y}{x}\right)=3+\frac{x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)}{x y z} \\ & =\frac{(x+y+z)(x y+y z+z x)}{x y z} \\ q-3 & =-3+\sum_{\text {cyc }} \frac{y^{2}}{z x}=\frac{x^{3}+y^{3}+z^{3}-3 x y z}{x y z} \\ & =\frac{(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)}{x y z} . \end{aligned} $$  It follows that  $$ \begin{aligned} \mathbb{R} & \ni 3(p+3)+(q-3) \\ & =\frac{(x+y+z)\left(x^{2}+y^{2}+z^{2}+2(x y+y z+z x)\right)}{x y z} \\ & =\frac{(x+y+z)^{3}}{x y z} \end{aligned} $$  Now, note that if $x+y+z=0$, then $p=-3, q=3$ so we are done.  Scale $x, y, z$ in such a way that $x+y+z$ is nonzero and real; hence so is $x y z$. Thus, as $p+3 \in \mathbb{R}$, we conclude $x y+y z+z x \in \mathbb{R}$ 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.  Construction As we saw in the setup, $(p, q)=(-3,3)$ will occur as long as $x+y+z=0$, and no two of $x, y, z$ to share the same magnitude or are collinear with the origin. This is easy to do; for example, we could choose $(x, y, z)=(3,4 i,-(3+4 i))$. Hence $a=\frac{3}{4 i}$, $b=-\frac{4 i}{3+4 i}, c=-\frac{3+4 i}{3}$ satisfies the hypotheses of the problem statement.