Determining the number of critical points requires consideration of imaginary as well as real values.

The formula:

zn+1 = zn5 – 1.5zn + c

f(z) = z5 – 1.5z + c
f'(z) = 5z4 – 1.5

the critical points are the roots of

f'(z) = 5z4 – 1.5 = 0

which are the 4th roots of 0.3.

The value of the 4th root of 0.3 is 0.740082804492, its negative value is also a root. So that’s 2 roots.

Taking in account imaginary values there are two more roots 0.740082804492i and -0.740082804492i, this is because i to the power 4 is i squared multiplied by i squared and since i squared is -1 i to the fourth power of i is 1.

So the formula has 4 roots, not 2. I’m only going to show a picture of one of the roots because all roots produce the same shape but in different positions.

Critical point 0.740082804492

All four critical points combined

So far all the examples have been formulae containing only powers of z. Now for a formula with a trigonometric function:

zn+1 = sin(zn) + c

f(z) = sin(z) + c
f'(z) = cos(z)

the critical points are the roots of

cos(z) = 0

which are -π/2, π/2, and repeat at 2π intervals.

There are only two versions of the picture the ones ate -π/2 and its repeats and thes ones at π/2 and its repeats. The images are mirror images of each other. Note: the original post had critical points at 0 and π which was for z = cos(z) + c which produces the same pictures.

π/2 critical point

-π/2 critical point

Critical points combined