It looks like you have computed the "tetrahedra shapes" for the three ideal hyperbolic tetrahedra making up the knot complement. Each of these gives a (euclidian!) shape to the four "cusp triangles" that "cut off" the four ideal vertices of the tetrahedron. To compute the shape of the cusp, you need to understand how it is tiled by the (in total 12) cusp triangles. To do this by hand takes some work! To get started, draw one of the tetrahedra $t$ in the upper-half-space model (with a vertex at infinity), and consider how the other three tetrahedra, that also meet infinity, and that also meet $t$, lie in the model. Each of these has just one cusp triangle that separates the body of the tetrahedron from infinity. You want to use the shapes $x_i$, $y_i$, and $z_i$ to tile these outward... (say on the horoplane at height one).
EDIT: Very nice pictures! Here is a bit of a snappy session which may be helpful to you.
In[1]: M = Manifold("5_2")
In[2]: M.cusp_info()
Out[2]: [Cusp 0 : complete torus cusp of shape -2.49024466751 +
2.97944706648*I]
In[3]: M.set_peripheral_curves("shortest")
In[4]: M.cusp_info()
Out[4]: [Cusp 0 : complete torus cusp of shape -0.49024466751 +
2.97944706648*I]
The thing to notice here is that the cusp shape (that is the "shape of the fundamental parallelogram") depends on the choice of a basis for the cusp group $P$. The cusp group $P$ is a copy of $\mathbb{Z}^2$. Each element of $P$ gives a parabolic Mobius transformation. Depending on the generating set (for $P$) that we use, we will get different cusp shapes!
Since $5_2$ is a knot, the cusp group has one natural basis coming from the topological meridian (bounds a disk in $S^3$) and the topological longitude (bounds a Seifert surface in $S^3$). Since the complement of $5_2$ is a hyperbolic manifold (called m015 in the snappy census), the cusp group has another natural basis coming from the two elements of $P$ which have smallest (parabolic) translation.
If you ask snappy for the cusp shape of the hyperbolic manifold, it agrees with you, up to a sign.
In[5]: M.identify()
Out[5]: [m015(0,0), 5_2(0,0), K3_2(0,0), K5a1(0,0)]
In[6]: N = Manifold("m015")
In[7]: N.cusp_info()
Out[7]: [Cusp 0 : complete torus cusp of shape -0.4902446675 +
2.9794470665*I]
Best Answer
The complete list is unknown. In general, it is difficult to determine which manifolds have a prescribed cusp field.
However, there are a well-studied set of examples that namely arithmetic manifolds with (invariant) trace field that is quadratic imaginary. Let me focus first on the manifolds with invariant trace field $Q(i)$. These manifolds are commensurable with the Whitehead link complement and the Borromean rings complement. All of the examples listed in Sam Nead's answer fit this criterion. There are of course more manifolds of this type (as it is an infinite class).
In terms of computed examples, arithmetic data associated to 3-manifolds was studied by
Goodman, Oliver; Heard, Damian; Hodgson, Craig, Commensurators of cusped hyperbolic manifolds, Exp. Math. 17, No. 3, 283-306 (2008). ZBL1338.57016. (and
Coulson, David; Goodman, Oliver A.; Hodgson, Craig D.; Neumann, Walter D., Computing arithmetic invariants of 3-manifolds, Exp. Math. 9, No. 1, 127-152 (2000). ZBL1002.57044.). The data computed in the first reference is more relevant here, it is accessible on Craig Hodgson's webpage: https://researchers.ms.unimelb.edu.au/~snap/commens.html
Also, the tables of data at the end of:
Maclachlan, Colin; Reid, Alan W., The arithmetic of hyperbolic 3-manifolds, Graduate Texts in Mathematics. 219. New York, NY: Springer. xiii, 463 p. (2003). ZBL1025.57001.
Finally, there is a knot complement $S^3 \setminus 12n706$ often referred to as the Boyd knot complement (for David Boyd) is a knot complement which is not commensurable with the Whitehead link complement but has cusp field $Q(i)$. The knot complement decomposes into regular ideal tetrahedra and octahedra and is beautiful in its own right. Attached is a picture of the cusp neighborhood computed by SnapPy.
In terms of quadratic cusp shape, there are three other knot complements known to have this property, but those cusp shapes are in $Q(\sqrt{-3})$. One is the figure eight knot complement and the other two are the dodecahedral knot complements of Aitchison and Rubinstein:
Aitchison, I. R.; Rubinstein, J. H., Combinatorial cubings, cusps, and the dodecahedral knots, Topology ’90, Contrib. Res. Semester Low Dimensional Topol., Columbus/OH (USA) 1990, Ohio State Univ. Math. Res. Inst. Publ. 1, 17-26 (1992). ZBL0773.57010. 1: https://i.stack.imgur.com/DE0Aa.jpg