jmc

Let $x=ki,$ where $k$ is a real number. Then the given equation becomes $$(ki{)}^4-3(ki{)}^3+5(ki{)}^2-27(ki)-36=0,$$which simplifies to $$k^4+3i{k}^3-5k^2-27ik-36=0.$$The imaginary part must be 0, so $3i{k}^3-27ik=3ik(k^2-9)=0.$

Since $k=0$ does not work, we must have $k=\pm 3$. Therefore, the pure imaginary solutions are $\boxed{3i,-3i}.$