Converting comments (about the original version of the question) to answer, as requested.
The cutting plane should be tangent to the Dandelin spheres. For the hyperbola case, looking from the side (as in your image), the "cutting line" should be externally tangent to the two "Dandelin circles".
It may be easier, in all cases, to start with the cone and cutting plane. Again, looking from the side, the "cone lines" and "cutting line" will determine, in the cone's interior, certain regions bounded by three sides (segments and/or rays). For an ellipse, there will be a finite triangle and an unbounded three-sided region; for a hyperbola, there will be two unbounded three-sided regions; for a parabola, there will be one unbounded three-sided region. Each "Dandelin circle" is tangent to the three sides of such a region.
Here's "unified" picture of a Dandelin configuration for both ellipses and hyperbolas (and parabolas as a limiting case). It consists simply of two circles, $\bigcirc K$ and $\bigcirc K'$, and their common tangents: the internal tangents meet at $A$, making an angle $\alpha$ with the line of centers; and the external tangents meet at $B$, making an angle $\beta$ with the line of centers. Necessarily, $\alpha > \beta$. (Exploration of the case where $\alpha=\beta$, which defines a parabola, is left to the reader.)
As indicated, $V=W$, $V'$, and $W'$ are points where an internal tangent meets an external. Internal tangent $\overleftrightarrow{VV'}$ touches the circles at $F$ and $F'$; external tangent $\overleftrightarrow{WW'}$ touches the circles at $G$ and $G'$. Note: because $|VF'|=|WG'|$ and $|VF|=|WG|=|V'F'|=|W'G'|$ (verification left to the reader), we have $|VV'|=|GG'|$ and $|WW'|=|FF'|$.
This allows us to restate the calculations made by OP:
- For an ellipse, we interpret $\overleftrightarrow{VV'}$ as the "cutting line"; that is, (the side-view) of the cutting plane. And $\overleftrightarrow{WW'}$ is a "cone line", a generator of the cone. Here, $V$ and $V'$ are the vertices, while $F$ and $F'$ are the foci, and we have
$$e = \frac{|FF'|}{|VV'|}=\frac{\color{red}{|FF'|}}{\color{blue}{|GG'|}}=\frac{|KK'|\cos\alpha}{|KK'|\cos\beta}= \frac{\cos\alpha}{\cos\beta}< 1 \tag{1}$$
- For an hyperbola, we interpret $\overleftrightarrow{WW'}$ as the "cutting line" and $\overleftrightarrow{VV'}$ as a "cone line". Here, $W$ and $W'$ are vertices, while $G$ and $G'$ are foci, and we have
$$e = \frac{|GG'|}{|WW'|}=\frac{\color{blue}{|GG'|}}{\color{red}{|FF'|}}=\frac{\cos\beta}{\cos\alpha}> 1 \tag{2}$$
It's worth reiterating that $(1)$ and $(2)$ say exactly that eccentricity of a conic is simply a ratio of internal and external tangent segments for these "Dandelin circles". I don't believe this is a new or particularly-profound observation, but I personally have never thought of things in quite this way. Nifty!
Also, as mentioned by OP, we have these relations involving the Dandelin radii, $r := |KF|=|KG|$ and $r':=|K'F'|=|K'G'|$:
$$r'+r = |KK'|\sin\alpha \qquad r'-r=|KK'|\sin\beta$$
These imply
$$\begin{align}
r &= \frac12|KK'|(\sin\alpha-\sin\beta) = |KK'| \cos\frac12(\alpha+\beta)\sin\frac12(\alpha-\beta) \\[4pt]
r' &= \frac12|KK'|(\sin\alpha+\sin\beta) = |KK'| \sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta) \\
\end{align}$$
so that
$$\frac{r'}{r} = \frac{\tan\frac12(\alpha+\beta)}{\tan\frac12(\alpha-\beta)} \tag{3}$$
You are absolutely right; the authors of these textbooks are mistaken, and your diagrams show why.
I happen to have watched a fair number of instruction videos on perspective drawing, and to my frustration, I haven't yet found a single one that addresses this mistake, even from the most meticulous and formal instructors.
This is somewhat understandable. Broadly speaking, artists establish rules based on aesthetics, intuition, and approximation; whereas mathematicians remain skeptical without proof. These are two valid mindsets for two different fields, and your question belongs to the overlap of the two. The danger of an artist learning the technical craft of perspective is that they might use rules of thumb indiscriminately without fully understanding when they apply. The danger of a mathematician learning to draw is that they might break out a straightedge and compass without considering whether that will actually contribute to a successful composition. It would be nice to find a happy middle ground between these two camps, but ellipses turn out to be surprisingly complex. A perfect geometric construction exists, but just isn't worth it in the the context of a larger drawing, so it's natural to look for a simpler way.
Note that the rule actually does work for isometric drawings. In other words, if all parallel lines remain parallel (as opposed to converging to a vanishing point), then it holds. There are other cases where it holds, as you allude to in the comments. Most importantly, the rule will be reasonably close when the drawing is within the cone of vision. In contrast, in your first example, the leftmost circle is outside the cone of vision and therefore quite distorted.
In summary, the "rule" should be considered a good approximation most of the time, but the more distortion there is, the more likely you should abandon it and use your instincts (or a straightedge and compass).
Best Answer
I'd say that depends.
If by complex plane you mean $\mathbb C^1$, the plane of complex numbers, then no, that doesn't really fit. So I'll ignore this interpretation for the rest of my answer.
If by complex plane you mean $\mathbb C^2$, a complex two-dimensional space, then I'd say that in that world, you have to note that the resulting object has four real dimensions. That makes it really hard to imagine, thus defying intuition.
On a purely algebraic level you could say that in this complex plane the distinction between ellipse and hyperbola becomes moot. Typically you'd distinguish these shapes by looking at the sign of some discriminant or other, but when the discriminant is complex-valued then the simple concept of a sign becomes the more complicated concepts of a complex argument, an angle. So both ellipse and hyperbola would be elements of a more general class of things, and there would be a continuous rotation of these general objects which have ellipse and hyperbola as special cases, special positions. About as special as an axis-aligned ellipse would be in the real plane, if you want a comparison.
Or you could take slices. Intersect your $\mathbb C^2$ with a real plane and look at the section that plane forms with an otherwise complex curve. If you define your plane as spanned by the real $x$ axis and the real $y$ axis, you get the conventional plane, but if you intersect with the real $x$ axis and the imaginary $y$ axis, or vice versa, then you get a world where the roles of ellipses and hyperbolas can be swapped.
You can also do a 3d cut. Take the real $x$ axis, the real $y$ axis and the imaginary $y$ axis as three spatial dimensions. Now you can visualize things to some degree. For example, if you take the unit circle (as a very special ellipse) you will find that you get two possible $y$ values for most $x$ values. For $-1<x<1$ those would be purely real, for $\lvert x\rvert>1$ purely imaginary. You can see aspects of the circle in such a drawing that are hard to grasp in 2d. But you need to still remember that you are looking at a somewhat arbitrary slice of a larger four-dimensional space, and are missing anything with a non-zero imaginary part for $x$.
Projecting 4d space into a 2d should be possible in theory as well, but I'd be worried you'd loose too much information for this to be of any use. Haven't tried, though. Since a quadric would be a 2d surface in 4d space, I expect a projection would be a boring region of the plane unless you add some form of lighting to give a more spatial impression. Tricky, but projecting from 4d to 3d first might allow you to delegate that to some existing tool.