Projection of the $XY$ plane. Is there a way to project the infinite Complex plane to either the Poincare disk or the unit disk - for all values of $x + iy ?$
 A: Here is a bijective continuous function from the complex plane to the interior of the unit disk, i.e. $|w|<1$:
$$f(u)=\frac{u}{1+|u|}$$
where $|u|$ is the modulus (distance from the origin) of $u$. In other words, for real $\theta$ and nonnegative $r$,
$$f\left(re^{i\theta}\right)=\frac{re^{i\theta}}{1+r}$$
Any branch cut (such as $\arg(z)=0$ or $\arg(z)=\pm\pi$) can be chosen for the argument function: the choice will not change the function and will leave the function continuous. Even though $\theta$ is not well-defined at zero the function still is well-defined, with $f(0)=0$ whatever the choice of $\theta$.
This function projects any complex number to another one with the same argument but inside the unit circle. For example, the unit disk is projected to the disk with radius $\frac 12$. The inverse function from the interior of the unit disk to the complex plane is
$$f^{-1}(u)=\frac{u}{1-|u|}$$
or
$$f^{-1}\left(re^{i\theta}\right)=\frac{re^{i\theta}}{1-r}$$

I could even argue that this function is a "projection."
d
In this diagram, the eye is in the complex $xy$ plane, with the vertical axis the $z$ plane. We have place the eye so our complex point is on the positive horizontal axis, let's call it $r$. Then we take the intersection of the line from your complex point to the point $(0,0,1)$ with the cone with vertex at $(0,0,0)$ with upward slope $1$. We then project that point back down to the complex plane.
That final point is the value of our function. Thus our function $f$ is a projection of the convex plane onto the tip of a cone followed by a projection onto the interior of the unit disk.

Here is a graphic from the OP showing how the grid of the Cartesian coordinate system maps into the unit circle.

