At most one connected component of unbounded portion of entire function. Suppose $f$ is an entire complex analytic function and $R$ a positive real number. Define the set $E:= \{z\in\mathbb{C};|f(z)| < R\}$ to be the set of $z$ whose image is bounded by $R$. I want to show that the complement $E^c$ has at most one connected component.
$E^c$ has zero components $\iff E^c = \varnothing \iff f$ is bounded by $R \implies f$ is constant. So we can assume $E^c$ has at least one component. If $f$ is instead bounded by a larger constant $S>R$, then $E^c$ will have one component consisting of a single point. So assume $f$ is unbounded.
The boundary $\partial E^c = \{z\in\mathbb{C};|f(z)|=R\}$. So if a component $A$ is bounded, then since $|f| \geq R$ in $A$ and $|f|=R$ on $\partial A$, by maximum modulus principle, $f$ is constant in $A$. If $A$ has nonempty interior, this implies $f$ is a constant function. If $A$ has empty interior, then if there is an open neighborhood of $A$ contained in $E$, $f$ will again be constant by the maximum modulus principle. If $A$ has no such open neighborhood, there there must be a sequence in $E^c\setminus A$ converging to a point in $A$, which seems problematic but I'm not sure what else I can say about A.
Assuming that every component of $E^c$ must be unbounded and that $E^c$ has more than $1$ component, $E$ must also be unbounded, since if there were an open neighborhood of $\infty$ disjoint from $E$, then all components of $E^c$ would meet near $\infty$. So every open neighborhood of $\infty$ intersects both $E$ and $E^c$. This is enough to show that $\infty$ is an essential singularity of $f$, which implies that $f$ is not a polynomial.
Another way to see that $f$ is nonpolynomial is to observe that $E$ and $E^c$ both being unbounded implies that $\partial E^c$ is unbounded, hence noncompact, but $\partial E^c = f^{-1}(\{|z|=R\})$ is the preimage of a compact set. So $f$ is not a proper map and cannot be a polynomial.
[EDIT:] The condition on connected components is not something I've used too often, so I don't know many ways to make use of it. An open set is simply connected if and only if its complement in the extended plane is connected, but due to the unboundedness of each component of $E^c$, $E$ will be simply connected. But then we could define logarithms on $E$ if $0\notin E$ (or we could translate $f$ to achieve $0\notin E$).
We could take a pair of points $z_1, z_2$ in separate components of $E^c$, which could possibly have been helpful when working with winding numbers of paths in $E$, except here $E$ is simply connected. So this isn't useful because there is a natural bijection between paths $\gamma\subset E$, up to homotopy, and functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a Hölder continuity condition with constant $\alpha>1$, up to $\mathbb{R}$-scalar multiplication.
At this point I can't figure out anything else about $f$. What to do?
 A: I think this was resolved in the comments, but let me add some remarks and further background.
If $f$ is an entire function, and $R>0$, then the components of $\Omega:=\{z\in\mathbb{C}\colon |f(z)|>R\}$ are called the tracts of $f$ (over $\infty$). The number of tracts can be any positive finite number, or indeed infinite.
Indeed, for large $R$, there is one tract for $\exp$, namely the right half-plane. As noted in the comments, for $\sin$ there are two. For $z\mapsto \exp(z^d)$, there are $d$ tracts. For $z\mapsto \exp(\exp(z))$, there are infinitely many. So the statement is indeed false when the complement is taken in $\mathbb{C}$. It is, however, true if the order of growth $\rho(f)$ is strictly less than one. Indeed, by the Denjoy-Carleman-Ahlfors theorem, the number of tracts is at most $\max(1,2\rho(f))$.
On the other hand, every tract is unbounded (this is elementary), and hence $\Omega\cup\{\infty\}$ is connected. Thus also $\overline{\Omega}\cup\{\infty\}$ is connected, which - as noted in the comment - was likely the complement intended in the question.
