I'm curious, in the Mandelbrot set, why is the escape radius $2$? I've seen few proofs of that on the internet, but i can't understand them enough.

Why is the bailout value of the Mandelbrot set 2?

Mandelbrot sets and radius of convergence


Some of the statements in them seem "out of the blue" for me.

For example, in the second in-site link I gave above: $ |c|≤2 \Rightarrow|z_n+1|≥|z_n|2−|c|>2|z_n|−2$

Where does $2|z_n|−2$ come from?

  • 1
    $\begingroup$ For what it's worth, I put a bounty on one of the questions you linked. Hopefully that gets a couple of answers that explain this is a way that doesn't require as much background knowledge. $\endgroup$
    – Mr. Brooks
    Commented Feb 12, 2020 at 20:57

1 Answer 1


From the third link I interpeted the proof like this :

What we want to proof is the criteria $|z_n| \le 2$.

Let $|z_n|>2$ and $|z_n|>|c|$. This is done to create a situation where the ratio from $|z_{n+1}|$ to $|z_n|$ is always greater 1 so the next term in the sequence always gets bigger than the one before (sequence is unbound) : $\frac{|z_{n+1}|}{|z_n|}>1$.

So we now have to prove $\frac{|z_{n+1}|}{|z_n|}>1$ :

In the linked article they first use the triangle inequality on the term (https://en.wikipedia.org/wiki/Triangle_inequality) :

$\frac{|z_{n+1}|}{|z_n|} = \frac{|z_n^2+c|}{|z_n|} \ge \frac{|z_n|^2-|c|}{|z_n|} = |z_n|-\frac{|c|}{|z_n|}$

From $|z_n|>|c|$ we get $\frac{|c|}{|z_n|}<1$ so :

$|z_n|-\frac{|c|}{|z_n|} > |z_n|-1$

From $|z_n|>2$ we get :

$|z_n|-1 > 1$ , so all in all we have proven $\frac{|z_{n+1}|}{|z_n|}>1$

This inequality is now true (and the sequence unbound) for all $|c| \le 2 < |z_n|$.

But what if $|c|>2$ (second case)? Then we can show that the sequence always "escapes" at least after 2 iterations : $z_{n=2}=|c^2+c| \ge |c|^2-|c| > 2$.

If we put this all togehter we get the criteria for the sequence to be unbound if $|z_n| > 2$ , so the criteria for the sequence to be bound (number in the mandelbrot set) is : $|z_n| \le 2$.

PS : I'm a student for myself and not an expert. As I mentioned above this is just my interpretation of the proof in the linked article. If some expert comes along this thread pls have a quick look over my explaination and if everything is correct.

PPS : Sorry for my English im from Germany xD

  • $\begingroup$ Maybe I’m missing something, but why does the triangle inequality step end with: |z|^2 - |c|, doesn’t the triangle inequality imply |z^2 + c| <= |z^2| + |c| $\endgroup$
    – user668074
    Commented Dec 6, 2021 at 11:33

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