Why do copper (II) complexes contain so many valence electrons?
I'll address $\ce{[Cu(NH3)4(H2O)2]++}$, but the reason is similar for $\ce{[Cu(NH3)4]++}$

The 9 d-electrons in the complex are in non-bonding and antibonding orbitals. 

See pages 6-9 of Ligand Field and Molecular Orbital Theory.  


  the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs. Therefore, the electrons provided by the ligands are largely confined to the ligands in the complex.


So, to a large degree, the 12 electrons from the ligands are in bonding orbitals, but the 9 d-electrons are in nonbonding or antibonding orbitals. 


  The d electrons of the metal centre...occupy the lower energy, non-bonding metal $t_{2g}$ set and the higher energy, antibonding $e_g$ set. Therefore, the metal based d electrons tend to reside largely on the metal atom. In summary, the frontier orbitals of the complex are the non-bonding, entirely metal based $t_{2g}$ orbitals and the antibonding $e_g$ orbitals, which are mainly metal in character anyway.