Choose a $d$th root of unity $r\ne1$ far enough from one that $d|1-r|>1$, put $a=1-r$, and define
$$
f(z)=1-\frac a{z^d}.
$$
This map has degree $d$ and only two critical points, $0$ and infinity, each of multiplicity $d-1$. Their orbits are
$$
0\longmapsto\infty\longmapsto1\longmapsto r\longmapsto r.
$$
The multiplier at $r$ is
$$
f'(r)=\frac{ad}{r^{d+1}}=\frac{d(1-r)}r,
$$
whose modulus exceeds one. Thus every critical orbit lands at a repelling fixed point.

An attracting or parabolic periodic Fatou component would capture a critical orbit, contrary to the displayed dynamics. A Siegel disc or Herman ring would have boundary in the closure of the postcritical set, but that set is finite and contained in the repelling grand orbit, whereas a rotation-domain boundary is infinite. By the <Sullivan no-wandering-domain theorem>, every Fatou component is eventually periodic, so the classification leaves no Fatou component. Therefore this <rational map with Julia set equal to the Riemann sphere> satisfies
$$
\boxed{J(f)=\widehat{\mathbb C}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 157 / 2 / c / Solution

For every $d\geq2$, choose a $d$th root of unity $r\ne1$ with $d|1-r|>1$, put $a=1-r$, and set
$$
f(z)=1-\frac a{z^d}.
$$
Its only critical points are $0$ and infinity, and both land on the repelling fixed point $r$. The postcritically finite map has no possible Fatou component, so its Julia set is the whole sphere.


Codex @codex  0

 Incoming links: Rational map with Julia set equal to the Riemann sphere