Easy way to find roots of the form $qi$ of a polynomial Let $p$ be a polynomial over $\mathbb{Z}$, we know that there is an easy way to check if $p$ have rational roots (using the rational root theorem).
Is there an easy way to check if $p$ have any roots of the form $qi$ where $q\in\mathbb{Q}$ (or at least $q\in\mathbb{Z}$) ? ($i\in\mathbb{C}$) ?
 A: Hint $\ f(q\,i) = a_0\! -\! a_2 q^2\! +\! a_4 q^4\! +\cdots + i\,q\,(a_1\! -\! a_3 q^2\! +\! a_5 q^4\! +\! \cdots) = g(q) + i\,q\,h(q)$
A: 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. 
A: You can actually do better than the normal Rational Root Theorem. Note that $p$ has integer (and thus real) coefficients. So if $ci/d$ is a root of $p$, then so is $-ci/d$. So $p$ must have $d^2x^2+c^2$ as a factor. Then you can conclude from the Gauss's lemma proof of the Rational Root Theorem that $d^2$ — not just $d$ — is a factor of the leading coefficient of $p$, and similarly $c^2$ is a factor of the trailing coefficient of $p$.
Combining this with Bill's/Karolis's answer will probably mean you have very few candidates to check even for moderately complicated polynomials.
A: Given a polynomial $P$, $P(ix) = R(x) + iI(x)$ where $R$ and $I$ are both polynomials over $\mathbb{Z}$. $P(ix) = 0$ iff $R(x) = 0$ and $I(x) = 0$. $R$ and $I$ are easy to find. The rest is the same rational root theorem.
