The type of hybrid orbitals employed in the formation of [Cu(NH₃)₄]²⁺ ion is

The Cu atom is in form of Cu⁺² in the compound. 

The electronic configuration of Cu⁺² is , 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹,4s⁰

So, there would be a rearrangement of electrons in Cu²⁺ because of the NH₃​ ligand (which is a strong one). And the last electron in the d-orbital would be out waiting for the 'N' electrons to fill up first.

[because NH₃​ is a strong ligand and the electrons donated from it should have got a place first--this is only punctuation]

So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the e⁻ from 3d would take place. So, you can say the hybridisation here would be dsp². Square planar with one unpaired electron.