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While watching this Numberphile video on Mandelbrot sets, it's more or less stated that the fractal will "blow up" if it's radius of convergence is greater than 2. What is the mathematical basis for this? I'm currently a master's student in math so I'd like a rigourous reason for this if possible.

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the fractal will "blow up" if it's radius of convergence is greater than 2

That's an inaccurate description of what was said. Indeed, we consider iteration of the function $f(z)=z^2+c$, starting from $0$, where $c$ is a fixed complex number. That is, $z_0=0$ and after that $z_{n+1}=f(z)$.

But the Mandelbrot set does not consist of the numbers $z_n$. It consists of the numbers $c$ such that the sequence $z_n$ is bounded. So, if $z_n\to \infty$, this only means that the number $c$ we used for this sequence is not in the Mandelbrot set.

The claim is that if $|z_n|>2$ for some $n$, then the sequence $z_n$ tends to infinity ("blows up" or "escapes to infinity"), and consequently $c$ is not in the set. Instead of calling $2$ the "radius of convergence", I would call it "radius of escape".

The proof consists of two cases.

Case 1: $|c|\le 2$. Then $|z_{n+1}|\ge |z_n|^2 - |c| > 2|z_n|-2$. Rewriting this as $$|z_{n+1}|-2 > 2(|z_n|-2)$$ we see that $|z_n|-2$ tends to infinity.

Case 2. $|c|>2$. What helps here is that $|z_1|=|c^2+c|\ge |c|^2-|c|>|c|$. For subsequent iterations, $$|z_{n+1}|\ge |z_n|^2 - |c| \ge |c||z_n|-|c| \ge |z_n|+(|c|-1)|z_n|-|c|> |z_n|+(|c|-1)|c|-|c|$$ Here $(|c|-1)|c|-|c|$ is a positive constant. By induction, the inequality $|z_n|>|c|$ continues to hold, and the differences $|z_{n+1}|-|z_n|$ are bounded from below by a positive constant. Hence, $|z_n|\to \infty$.

The answer benefited from this post by Mike Hurley.

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could you tell me why my statement is inaccurate? If the sequence tends to infinity won't the fractal be unbounded and hence "blow up"? –  Millardo Peacecraft Aug 7 at 21:36
@MillardoPeacecraft See my edit. –  Weapon of Choice Aug 7 at 21:44

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