$\ce{Cu^2+} $ has 9 electrons and the $\mathrm d$-orbital is almost completely filled (except for 1 electron vacancy), eventually forming $\mathrm{sp^3}$ hybrid orbitals, where each $\ce{NH3}$ donates a lone pair of electrons.
But it has been found to have a square planar geometry, hence completely contradicting the above reasoning. So, please explain what is the hybridization of the $\ce{Cu^2+}$ in the given complex?
I feel a bit like a broken record. Do not use hybridisation to describe coordination compounds; it is not helpful.
The cation you called tetraamminecopper(II) is more accurately described as tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$, because two aqua ligands are there, just further away. The four ammine ligands donate two electrons each, as do both the aqua ligands, totalling 21 electrons overall. Due to the strongly Jahn-Teller distorted geometry and other reasons, the bonding interactions are overall stronger than antibonding ones making the complex favourable.
The interaction between the metal and ligand involves $\mathrm{e_g, a_{1g}}$ and $\mathrm{t_{2u}}$ orbitals (when viewed as an undistorted octahedral complex). That would include two d-orbitals, three p and one s. If your life depended on it and you are the last person on Earth, you may attempt to extract a hybridisation from that. But don’t do it.
