If the circles do not intersect on the cartesian plane, the intersection points become complex numbers
To be precise: the coordinates of the points of intersection become complex numbers. More precisely, they will form a conjugate pair of complex points, i.e. if $(a+ib,c+id)$ is one such solution, then you know the other solution to be $(a-ib,c-id)$ (if your circles were real).
Can these complex intersection points be visualized spatially?
Connecting these points you get the radical axis of the circle, just as you get for real points of intersection. You can imagine parametrizing that radical axis with a single complex parameter. To make this more precise, you could say that the line connecting the centers will intersect the radical axis at a point $t=0$, and from there a unit distance step along the radical axis will take you to a point $t=1$. You have to arbitrarily choose a direction for the axis here. You could come up with a formula to compute the point on the axis for every $t\in\mathbb C$.
Now you could draw a 2d diagram of that radical axis alone. You'd have the real and imaginary component of $t$ as two coordinates. A complex pair of points would be mirror images with respect to the real axis. Furthermore, for reasons of symmetry, the real component of your points of intersection would have to be zero in this coordinate system. In other words, the real coordinate of the solutions in the original coordinate system is the point where the radical axis meets the line joining the circle centers.
Visualizing two points on the imaginary axis all by themselves is not that interesting. Things do become more interesting if your setup is dynamic in some way. For example in this slide of a talk on a project I'm involved with, one can see the $x$ coordinates of the points you get by intersecting a circle with a horizontal line which you can move around. As you move the line from an intersecting position to a non-intersecting one, the points will come closer together along the real axis, and after meeting at the origin they will drift apart along the imaginary axis. (The slide in question also illustrates how to avoid the singular situation, so the points there won't move through the center but curve around it instead.) The picture of two circles intersecting would look pretty much the same in the parameter space of the radical axis I described above.
1)Follows from 3. In 3) $a$ represents a direction (actually normal to a direction), and $b$ represents a distance from the origin -- or $\left |\frac {b}{a}\right|$ does.
2)In cartesian:
$\left|\begin {array} {} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right| = 0$ suggests $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ are colinear
-- or the pyramid formed by the 3 points in the plane $z = 1$ together with the origin forms a pyramid of zero volume.
$\begin {bmatrix} {} z_1&\bar z_1&1\\z_2&\bar z_2&1\\z_3&\bar z_3&1\end{bmatrix} = \begin {bmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{bmatrix}\begin {bmatrix} 1&1&0\\i&-i&0\\0&0&1\end{bmatrix}$
The right most matrix is non-sigular.
6)$\overline{(z-z_1)}(z-z_2) = k$ is the locus of points $z$ such that $\angle z_1zz_2$ is constant.
7)is the very definition of an ellipse. The distance from one focus to a point on the ellipse to the other focus is constant.
10)$|z - z_1| = k|z-z_2|\\
(z - z_1)\overline (z-z_1) = k^2 (z - z_2)\overline (z-z_2)\\
x^2 + y^2 + 2xx_1 + 2yy_1 + x_1^2 + y_1^2 = k^2(x^2 + y^2 + 2xx_2 + 2yy_2 + x_2^2 + y_2^2)\\
(k^2 - 1)(x^2 - a)^2 + (k^2 - 1)(y^2 - b)^2 = r^2$
Best Answer
Let's see. On a unit circle, you just have complex phases, so $$A=e^{ia}, B=e^{ib}, \ldots$$ This part is clear.
Then you have to find intersections. This part is the most involved. Let's see for $AB$ and $DE$. We have to solve $$X=A+t(B-A) = D+t'(E-D)$$ for real numbers $t$ and $t'$ which isn't any better than working with vectors. If you check this link, you will find quite a horrible formula for an intersection with complex numbers that comes from solving the above equations. It does simplify a little bit for unit circle, but not much. Let's just write it out:
$$X=2i\frac{\sin(b-a)(e^{id}-e^{ie})-\sin(e-d)(e^{ib}-e^{ia})}{(e^{-a}-e^{-b})(e^{d}-e^{e})-(e^{a}-e^{b})(e^{-d}-e^{-e})}$$
At the final step, you would need to prove collinearity, which means $XY$ and $XZ$ are real multiples of each other.
There may be a nicer complex-based solution that I can't see - looking forward to seeing another answer to point it out. Until then, I have my doubts.
As a general advice, what simplifies and what doesn't, when you use complex numbers:
What stays difficult: