Question

Let f be the polynomial f()25. Show that f has a parabolic fixed point at the origin, and that f2 has a multiple fixed point at the origin. By calculating fo2, show that f has 4 attracting petals.

Answer 1

Given that

$f(z)=-z+z^5 \\ \implies f'(z)= 5z^4-1$.

Hence clearly

f(z) = 0 when z E {0, 1,-1, i,-i} & viceversa ,1,-1, i, -i, f(f(z))0 moreover wheneverz E 10

Hence f has a periodic fixed point at origin 0 and the Julia set is given below

Now $f'(z)=-1+5z^4=5(z^4-1/5)=5(z^2-1/√5)(z^2+1/√5)\\ \implies f'(z)$ $f'(z)=-1+5z^4=5(z^4-1/5)=5(z^2-1/√5)(z^2+1/√5)\\ \implies f'(z)$√5)(z^2+1/√5)

Hence f^2 has multiple fixed points at origin namely ± 1/5^4