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 \[z^4 + iz^3 + 5z^2 - iz - 6 = 0.\]We can take out a factor of $ z - 2i, $ to get \[(z - 2i)(z^3 + 3iz^2 - z - 3i) = 0.\]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} $.