Let's look at another example.

f of z equals z squared minus 1, so the case where the constant c equals minus 1.

What is the filled-in Julia set in that case?

Let's go find out.

Again, we're looking at f of z equals z squared minus 1.

So, that's the case where this constant c is equal to -1.

We want to find the build-in Julia set, so

all those point z that remain bounded under iteration.

So let's check some orbits to get a feeling.

For example, when z is equal to 0,

then f of z is 0 squared minus 1 so that's minus 1.

What's f of f of z?

Well f of f of 0 is f of minus 1,

because we already calculated that f of 0 is minus 1.

So, we just need to apply f to minus 1.

What's f of minus 1?

That's minus 1 squared, which is 1 minus 1, so a 0.

If we apply another f to that, we get f of 0, which is minus 1.

If we apply another f, we get f4 of 0 which is 0.

And so the orbit just bounces back and forth between 0 and minus 1.

0 gets mapped to minus 1 under f, minus 1 gets mapped back to 0.

0 gets mapped to minus 1, minus 1 gets mapped to 0, 0 to minus 1, and so forth.

It just bounces back and forth between 0 and minus 1.

We call this a periodic orbit because it just keeps repeating zero,

minus one, minus one, zero, and so forth.

It's a periodic orbit and

the period is two because there's exactly two elements in that orbit.

And such a point clearly needs to belong to the filled-in Julia set

because the orbit never runs away to infinity.

It stays between 0 and negative 1.