By F. v. Haeseler, H.-O. Peitgen (auth.), Heinz-Otto Peitgen (eds.)

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If1'o each M > 1 there exists kEN such that I(R n ) I (z) 12M > 1 for all z E J and n 2 nJ = 0. then for k. 2 applies as well. 2 . 6 Let Zo be an attractivejzxed point of R. Suppose A*(zo} is simply connected and 8A*(zo} n1'o = 0. Then. if 8A*(zo} is not a straight line or a circle 8A*( zo} does not have a tangent at any point. F. -O. ria " j=1 with lajl < z - aj 1 - a,'z 1 for j = 1, ... , n. As the example R(z) = (z + 1)2 4z shows, the Julia set can be a straight line, although Q n J =10. Indeed, the critical points of R have finite orbits: 1 ~ 1 and 1 is a superattractive fixed point, -1 ~ 0 ~ 00 and 00 is a repulsive fixed point The Julia set of R is [ -00, 0].

But it turns out that this condition would be much to strong. : 4> is a quasiconformal mapping (see [2,50] ) with fz4> = 0 on K f. Figure 27 : The basin of attraction and the Julia set for). 35689684451. If J f is of measure zero then the condition :-;4> = 0 on K f just means that 4> is analytic on the interior of K f. 4 (Straightening Theorem) Every polynomial-like mapping I : U' --. U of degree d is hybrid equivalent to a polynomial of degree d. If K f is connected, then p is unique up to a conjugation by an affine map.

Then for k. 2 applies as well. 2 . 6 Let Zo be an attractivejzxed point of R. Suppose A*(zo} is simply connected and 8A*(zo} n1'o = 0. Then. if 8A*(zo} is not a straight line or a circle 8A*( zo} does not have a tangent at any point. F. -O. ria " j=1 with lajl < z - aj 1 - a,'z 1 for j = 1, ... , n. As the example R(z) = (z + 1)2 4z shows, the Julia set can be a straight line, although Q n J =10. Indeed, the critical points of R have finite orbits: 1 ~ 1 and 1 is a superattractive fixed point, -1 ~ 0 ~ 00 and 00 is a repulsive fixed point The Julia set of R is [ -00, 0].