Why is tetraamminecopper(II) a square planar and not a tetrahedral species?
It is very convenient to use crystal field theory to discuss this. 

It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy. 

The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc. 

Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate. 

However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound. 

This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.