Hint: Let the lead coefficient be $a_n\ne 0$ and let the constant term be $a_0\ne 0$. Then the usual Rational Roots Theorem tells you that the list of candidates is limited to numbers of the form $c/d$ where $c$ divides $a_0$ and $d\ge 1$ divides $a_n$. To use it, one then, at least in principle, tests every candidate.
In your situation, the result is exactly the same. The list of candidates is limited to numbers of the form $ci/d$, with the same conditions on $c$ and $d$.
Remark: Breaking up into real and complex parts is very often a more efficient way of ruling out candidates. The point of the above answer is that the procedure provided by the Rational Roots Theorem can be word for word extended to our new situation. The proof is essentially the same as the usual proof of the Rational Roots Theorem.