Assume $J(f)$ is connected. Then the filled Julia set $K(f)$ is connected and full, so its complement $A_\infty(f)$ is simply connected. The local Bottcher coordinate therefore extends by the functional equation to a conformal isomorphism
$$
\phi_f:A_\infty(f)\longrightarrow\{w:|w|>1\}.
$$
If a finite critical point $c$ lay in $A_\infty(f)$, differentiating
$$
\phi_f(f(z))=\phi_f(z)^d
$$
at $c$ would give
$$
0=d\phi_f(c)^{d-1}\phi_f'(c).
$$
Neither factor on the right vanishes on the exterior disc under a conformal coordinate, a contradiction. Hence every finite critical point lies in $K(f)$. This is one direction of the <connected Julia set criterion for a polynomial>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 157 / 3 / b / Solution

The Julia set of a polynomial is connected exactly when every finite critical point belongs to the filled Julia set. In this case the Böttcher coordinate extends conformally over the entire basin of infinity; an escaping critical point is precisely an obstruction to that continuation.


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 Incoming links: Connected Julia set criterion for a polynomial