The Mandelbrot set is often rendered by calculating whether a point, a complex number $c$, diverges under the iterated function $z_{n+1}=z^2_n+c$, where $z_0=0$ and $c$ is the point being tested.
Those points which diverge under that iteration are designated as being outside the Mandelbrot set.
As a test for divergence, many simple algorithms check to see if $|z_n|$ grows larger than 2.
I tried to find the origin of the very common computational test, and finally a derivation in Chaos and Fractals: New Frontiers of Science by Peitgen, Jurgens and Saupe, 2004 [ISBN-13 : 978-0387202297].
Question: My interpretation is that two, not one, conditions need to be met for divergence. Am I correct? Most software implementations only use of the two conditions.
See below for derivation, based on that source.
Start with triangle inequality $|a+b|\leq|a|+|b|$, with $a=z^{2}+c$ and $b=- c$, noting also $|-c|=|c|$,
$$|z^{2}+c-c|\leq|z^{2}+c|+|c|$$
We're interested in $|z^2+c|$, so isolate it
$$|z^{2}+c|\geq|z^{2}|-|c|$$
Also, $|z^{2}|$ is the same as $|z|^{2}$ so we have
$$|z^{2}+c|\geq|z|^{2}-|c|$$
Condition 1: if we say that $|z|$ is bigger than $|c|$, that would mean
$$|z|^{2}-|c|>|z|^{2}-|z|$$
Which means we can also say,
$$|z^{2}+c|>|z|^{2}-|z|$$
Which can be factorised as
$$|z^{2}+c|>|z|\left(|z|-1\right)$$
If $z_{n}$ is the $n^{th}$ point in an orbit, then the next point in the orbit is $z_{n+1}=z_{n}^{2}+c$. We can rewrite the above expression as a ratio,
$$\frac{|z_{n+1}|}{|z_{n}|}>|z_{n}|-1$$
Condition 2: If we now also say that $|z|$ is greater than 2, we have $|z|-1>1$. So we finally have
$$\frac{|z_{n+1}|}{|z_{n}|}>1$$
This is telling us that $|z_{n+1}|$ is always greater than $|z_{n}|$, that is we have divergence, as long as:
$|z|>|c|$, and
$|z|>2$
So we've shown that if the above two conditions are true, then the orbit of a point will escape.
Thoughts
My thinking is that every software-implemented algorithm I have seen only implements the test $|z|>2$ because $|c|$ for most of the region of interest is less than 2. Is this correct?
The region of interest is the rectangle with corners $-2.25+1.5i$ and $0.75-1.5i$, and the largest $|c|$ here is about 2.7.
The Mandelbrot set only contains points $c$ with $|c| \le 2$.
To see this, let's go from $z_0 = 0$ to $z_1 = c$ to $z_2 = c^2+c$. If $|c| > 2$, then $|c^2| = |c|^2 > 2|c|$, which means that $$|z_2| = |c^2+c| \ge |c^2| - |c| > |c|.$$ So for $z_2$, both of your conditions for divergence are satisfied, which means that $c$ is not in the Mandelbrot set.
As a result, though the proof of divergence in the question logically requires both $|z_n| > 2$ and $|z_n| > |c|$ to work, we can dispense with the second condition: for all points $c$ which possibly could be in the Mandelbrot set, the first condition implies the second.