jmc

Since $2i$ is a root, $$(2i{)}^4+a(2i{)}^3+5(2i{)}^2-i(2i)-6=0.$$Solving, we find $a=i,$ so the polynomial is $$z^4+i{z}^3+5z^2-iz-6=0.$$We can take out a factor of $z-2i,$ to get $$(z-2i)(z^3+3i{z}^2-z-3i)=0.$$We can check that $z=1$ and $z=-1$ are solutions of the cubic, so we can take out factors of $z-1$ and $z+1,$ to get $$(z-2i)(z-1)(z+1)(z+3i)=0.$$Therefore, the other roots are $\boxed{1,-1,-3i}.$