Let $f$ be a polynomial with a super attracting fixed point at $x=0$. Can the immediate basin of attraction of the fixed point contain other roots? If so, please provide a specific example with the immediate basin of attraction. If not, why?

  • $\begingroup$ How about $z^3-z^2$? $\endgroup$ Feb 24, 2020 at 20:20
  • $\begingroup$ @Mark is zero an attracting fixed point? Is, say, $z=1/2$ attracted to zero? $\endgroup$ Feb 24, 2020 at 21:17
  • $\begingroup$ @Mark is 1 in the immediate basin of attraction of 0? It is not obvious $\endgroup$
    – user152801
    Feb 24, 2020 at 21:35
  • $\begingroup$ what does super attracting mean ? $\endgroup$ Feb 24, 2020 at 21:39
  • $\begingroup$ @Piyush A fixed point is super-attracting if the derivative is 0. It means the points in the basin of attraction approach specifically at an exponential rate. In this case, for $z$ in the basin of attraction,$ |f(z)-0|\approx b|z-0|^p$ for some constant p. $\endgroup$
    – user152801
    Feb 24, 2020 at 21:48

1 Answer 1


Let $f(z)=az^2(z-1)$. Zero is superattracting. Now choose $a$ so small that $|f(z)|<|z|/2$ for $|z|<2$. Then the root $z_0=1$ is in the immediate domain of attraction.


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