To simplify the notation, we use f(x) instead of fc(x), f(n)(x) as n th composition of f.
The idea is similar to http://www.mathlinks.ro/viewtopic.php?t=256478.
a)c) are trivial.
b) For c=1, we can easily show A1=B1. Suppose |z0|>1.7. Obviously p(z)=z will not hold. Let p(z)=f(n)(z).
We can show by induction in n that |p(z0)|≥1.7. Contradiction!
For d)e)f)g), it suffices to show the sequence 0,f(c),⋯,f(n)(c) has infinitely many distinct terms.
Suppose it is not true, then there exists n≠m such that f(n)(c)=f(m)(c).
If we define g1(x)=x2+x,gn(x)=gn−1(x)2+x. It is equivalent to say G(x)=gn(x)−gm(x)=0 has root c.
Now f) is straightforward.
e) is also trivial because G(x) is a monic polynomial, then rational root has to be integer.
For d), we define z0=c2+c,zn=f(zn−1).
We can prove by induction that |zn|=|z2n−1+c|>|zn−1| as follows:
Suppose by induction assumption |zn−1|>|c|, then
|zn|≥|zn−1|2−|c|>|c|2−|c|>|c|.
Then |zn|−|zn−1|≥|zn−1|2−|zn−1|−|c|>|c|2−2|c|>0.
For g), c<−2 is covered by d). Now assume c>14.
Still by induction, zn−zn−1=z2n−1−zn−1+c>(zn−1−12)2>0.
Q.E.D.