AChaoticEnd? 259
In the casen=4, 2n− 1 =15;z=exp( 2 kπ/ 15 ).Ifk=0,zis of period 1 (fixed), while
ifk=5, 10,zis of period 2 (using 2^2 − 1 =3). The other twelve fifteenth roots of 1 give
three 4-cycles.
Already with these facts, sensitive dependence looks likely. It is easy toproveif we write
z=eiθ; then what we have calledd(z,z′)is just|θ−θ′|.Nowftakesθto 2θ.Givenzandε,
letnbe such that( 1 / 2 n+^1 )<ε, and letz′=expi(θ+( 1 / 2 n+^1 )); thenfn(z)=expi( 2 nθ)and
fn(z′)=expi( 2 nθ+^12 ).
(b) Fixed points are given byz^2 +c=z,orz=^12 ( 1 +
√
1 − 4 c).Nowg′(z)= 2 z, and so at
a fixed point|g′(z)|=| 1 +
√
1 − 4 c|.If|g′(z)|=1, write| 2 z|=1,z=^12 exp(iθ). Then
c=
1
2
exp(iθ)−
1
4
exp( 2 iθ)
from the original equation. Sochas coordinates(^12 cosθ−^14 cos 2θ,^12 sinθ−^14 sin 2θ). This
describes a cardioid asθgoes from 0 to 2π.