Are all complex iterations in the form of $z_{n+1} = z_n + c$ fractals? I've learned that many fractals can be described like the below:
$$z_n \in F \text{ if } \limsup_{n\to\infty} |z_{n+1}| \leq i$$
Where $i$ is some number, usually I see 2, and $z$ is a member of the set. This is paired with some iteration on $z_n$ to make $z_{n+1}$. Classic example is the Mandelbrot set: $z_{n+1} = z_n^2 + c$, where $c$ is a complex number.
But some of the things I've read about fractals is they are self-similar, self-symmetric, have a finite area yet infinite perimeter.
I have a few examples (color scheme = more red as approaches the set, magenta very close to the set, blue touching the set, black inside the set):
A "tri-wing" vein set := $z_{n+1} = (-z_n^{-1})^2 + z_n$

A Julia set (k = -0.835 - 0.232i) := $z_{n+1} = z_n^2 + k$

These all are symmetric in some way, have finite area and infinite perimeter. But then I consider fractals like the Burning Ship ($z_{n+1} = |z_n^2| + c$) which have no symmetry. Usually I can define a line somewhere on the graph intersecting the origin and it will be a mirror image on both sides, but not for everything I can define.
Consider a basic example like $z_{n+1} = z_n^0 + c$:

This has symmetry but it's just a circle. This has a finite area as well as a finite perimeter. Or am I wrong? Only $\{0\} \in F$ for this example, so it's not a fractal.
I'll disclaim I've only learned about fractals in the context of $z_{n+1} = z_n + c$, with $z_n$ to the power of an integer, or adding extra terms. Is it safe to assume this is not a particularly good means of defining fractals?
 A: Your last picture, actually, answers your title question!
What the "thing" you are asking about that the Mandelbrot set is an example of is called a bifurcation locus of a self-map on the complex plane. To put this in some cleaner notation, if you have a complex self-map $f_c: \mathbb{C} \mapsto \mathbb{C}$ with one parameter $c$, then the orbits of the map with a starting point $z_0$ are the sequences $O(f_c, z_0)$ defined by
$$[O(f_c, z_0)](n) := f_c^n(z_0)$$
where the exponent represents repeated application of $f_c$. At each $n$ the orbit gives the result of $n$ self-applications of $f_c$ to the starting point $z_0$. The bifurcation locus developed with the starting point $z_0$ is
$$BL[f, z_0] := \{\ c \in \mathbb{C} : \mbox{$O(f_c, z_0)$ is a bounded sequence}\ \}$$
The reason for the name "bifurcation locus" is that this is related to the notion of a bifurcation diagram like that commonly used to represent a real self-map, most popularly seen in the real logistic (Verhulst) map. The difference is the bifurcation locus is simply the points where you get a bounded orbit subject to bifurcation, while the bifurcation diagram is a plot of the whole orbit at every single parameter value, or at least the limiting behaviors of the orbits. For complex functions, the latter actually can't be done so faithfully because it would require a four-dimensional space.
In the last picture, your map $f_c(z) := z + c$ has a bifurcation locus consisting only of the point $0$. This can be understood by physical reasoning: repeated action of it is like a discretized motion with steady velocity, which here is $c$ - one step of $c$ for each "unit" of time. If you keep up a steady motion with nonzero speed long enough, eventually it will cross any boundary you put in its way. As a result, unless $c = 0$, all starting points $z_0$ go to infinity.
For the mathematical proof, it is easy to see that
$$f_c^n(z_0) = \underbrace{(\cdots((z_0 + c) + c) + c)}_{\mbox{$n$ nestings}} = z_0 + nc$$
and thus $\lim_{n \rightarrow \infty} f_c^n(z_0) = \infty$ unless $c = 0$, hence $BL[f, z_0] = \{0\}$ no matter what $z_0$ is. A single point set is not a fractal.
Your picture almost shows this, except that you didn't iterate enough times. As you mentioned, it gets very slow to plot because, effectively, the "speed" of the point gets arbitrarily slow as you get close to the origin. Thus for drawing near there, the computer has to do many, many, many iterations to reach your approximated escape boundary. You could get an accurate picture if you increased the iterations more and left it to cook, but the above should illustrate to you that you don't need to. Nonetheless, you can imagine what the final plot would look like: it will just be a kaleidoscope of colors converging at that one point.
