$\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?