For the equation $x^2 + y^2 = 1,$ what does the graph look like extended into an imaginary space? In other words, let us first allow an imaginary domain for a two input function $f(x,y)$ where $x \in \mathbb{C}$ (complex numbers) and also $y \in \mathbb{C}.$
If we allowed a third axis to be imaginary, what would the graph look like?
 A: The difficulty in presenting graphs of such an equation is that it needs to be  depicted  in $ \ \mathbb{C}^2 \ \ , \ $ which requires four real dimensions.  If we adopt pathfinder's notation -- $ \ x \ = \ u + iv \ \ , \ \ y \ = \ r + is \ \ $ -- then we will only be able to present graphs of "slices" through the intersections of the hypersurfaces represented by $ \ u^2 + r^2 - v^2 - s^2 \ = \ 1 \ \ , \ \ uv + rs \ = \ 0 \ $ three dimensions at a time.
One approach we can take in representing the hypersurface for $ \  x^2 + y^2 \ = \ 1 \ $ is to "plot" the "real plane" (here, using the $ \ u-$ and $ \ r-$ axes) with the imaginary part of $ \ x \ $ (the $ \ v-$ axis) directed orthogonally.  We can then present this three-dimensional "component" of the hypersurface for various values of $ \ s \ \ , \ $ the imaginary part of $ \ y \ \ . \ $  (There is a corresponding set of "graphs" we could make using the $ \ u- \ , \ r- \ , \ s-  $ axes for different values of $ \ v \ \ , \ $ but we shall see that these are quite similar to those we will display.)
If we start with $ \ s \ = \ 0 \ \ , \ $ the complex-variable equations become $ \ u^2 + r^2 - v^2 \ = \ 1 \ \ , \ \ uv \ = \ 0 \ \ , \ $ which will appear as a hyperboloid of one sheet with its symmetry axis on the $ \ v-$axis and the union of the real ( $ \ v = 0 \ $ ) and the $ \ rv- \ ( \ u = 0 \ ) \ $ planes, respectively.  [This is shown in the image below; the "gap" along the $ \ \mathfrak{Re}(y) \ ( \ r = 0 \ ) \ $ axis is a software artifact.]

It is the intersection of these two hypersurfaces that is the "slice" of the hypersurface for $ \ x^2 + y^2 \ = \ 1 \ $ at $ \ s  \ = \ 0 \ \ , \ $ in which we are interested.  This "slice" is the union of the intersection of the hyperboloid with the real plane and the intersection of the hyperboloid with the $ \ rv-$plane.  This gives us the expected "real" unit circle $ u^2 + r^2 \ = \ 1 \ $ and the hyperbola $ \ r^2 - v^2 \ = \ 1 \ \ , \ $ which meet at $ \ (0 \ , \ \pm 1 \ , \ 0 \ , \ 0) \ $ [see image below].

As mentioned earlier, using "plots" with $ \ \mathfrak{Im}(x) \ = \ 0 \ \ $ instead merely leaves the hypersurfaces $ \ u^2 + r^2 - s^2 \ = \ 1 \ \ , \ \ rs \ = \ 0 \ \ , \ $ producing a "graph" that would just appear rotated $ \ 90º \ $ to the ones we will discuss here.
If we then let $ \ s \ $ "move away" from zero, the hyperboloids becomes $ \ u^2 + r^2 - v^2 \ = \ 1 + s^2 \ \ , \ $ which are co-axial to the one at $ \ s \ = \ 0 \ \ , \ $ with larger-diameter "necks" as $ \ |s| \ $ increases.  The second hypersurfaces become $ \ uv \ = \ -s·r \ \ , \ $ which are single surfaces with curvature, avoiding the $ \ r-$axis but connected at the origin, and no longer merely the union of orthogonal planes.  [The situation for $ \ s \ = \ +\frac12 \ $ is presented below; the "graph" for $ \ s \ = \ -\frac12 \ $ is similar, but appears as a "mirror-image" of the $ \ s \ = \ +\frac12 \ $ hypersurface. ]

The effect on the intersections we are following is to "fuse" remnants of the hyperbola with those of the "widening circle" at the hyperboloid's "neck".  What is produced is two anti-symmetric curves on either side of the $ \ rv-$plane (with the pair mirror-reversed by changing the sign of $ \ s \ ) \ . \ $ As we increase $ \ |s| \ \ , \ $ the curved portions of $ \ uv \ = \ -s·r \ \   $ become steeper, and thus these curves are "stretched steeper" as well.  [I have not been able to find an elementary parameterization for these intersection curves, so I had to "fabricate" their image to some extent.  It should serve to give an idea of what this "slice" looks like.]


A "visualization" of the complex $ \ x^2 + y^2 \ = \ 1 \ $ hypersurface would then be to contemplate how it is "swept out" by the various "slices" as $ \ s \ $ is varied.  The "views" with changing values of $ \ \mathfrak{Im}(x) \ $ or $ \ v \ $ would appear "rotated" relative to these, as we are "observing" the same structure from an orthogonal fourth-dimensional "direction".
[It would be "fun" to compare this with the hypersurface for the "real hyperbola" $ \ x^2 - y^2 \ = \ 1 \ \ , \ $ on the intersections of $ \ u^2 - r^2 - v^2 + s^2 \ = \ 1 \ $ and $ \ uv - rs \ = \ 0 \ \ , \ $ which is in some sense a "rotation" of the hypersurface we've examined here.]
