Let $\omega(k)=\alpha_{n}k^{n}+\alpha_{n-1}k^{n-1}+\dots+\alpha_{0}$ be a polynomial of degree $n\in\mathbb{N}$ on $\mathbb{C}$.
Define $D=\{k\in\mathbb{C}:\text{Re}(\omega(k))<0 \}$.
How do i show that $\mathbb{C}-\partial D$ is a union of disjoint unbounded simply connected open sets?
There is a hint to use the fact that the real part of $\omega(k)$ is a harmonic function.
Note: I have limited experiences with topology hence i would like an answer that is as elementary as possible.
The key lemma is that there can be no closed curve $S$ such that $\forall s \in S: \text{Re}(\omega(s)) = 0$$.
Here is a sketch of a proof of that lemma: Note that any closed curve on $\Bbb{C}$ has an interior. If $\text{Re}(\omega(z)) = 0$ for all $z$ in the interior, then by analyticity $\omega(k)$ is a constant, which must be pure imaginary, and $D = \emptyset$, which vacuously satisfies the requirement that it be a union of sets of any description -- it is the empty union. So we are free to choose $z$ in the interior such that $\text{Re}(\omega(z)) \neq 0$. Assume $\text{Re}(\omega(z))> 0$; the other case can be tackled in an analogous way.
For our given point $z$ and for $\text{Re}(\omega(k)) > \epsilon > 0$ define $r_z(\theta, \epsilon)$ as the smallest $r > 0$ such that $\text{Re}(\omega(z + r e^{i\omega})) = \epsilon$. Such a value exists for all $\theta$ by the intermediate value theorem: A continuous function of $r$ is going from $\text{Re}(\omega(z)) > \epsilon$ to $0 < \epsilon$. And by continuity of $\omega(k)$ it is easy to show that the curve $w = z + r_z(\theta, \epsilon)e^{i\omega}$ is a closed continuous curve.
But on that curve, $\text{Re}(\omega(k)) = \epsilon$ so the integral along that curve of $\omega(k)$ is positive (or rather, has a positive real part). But that integral is proportional to the simple pole value at $z$, which for a polynomial is zero. So no closed curve at real part $0$ can exist.
Whith that lemma, the main proof becomes easy.
First, a lemma in general topology related to this problem:
Lemma: Let $X$, $Y$ topological space, $X$ locally connected, $f\colon X \to Y$ a continuous, open and closed map. Let $B$ a closed subset of $Y$ and $A \colon = f^{-1}(B)$. Then the image of any connected component of $X\backslash A$ is a connected component of $Y \backslash B$.
Proof: The connected components of $X\backslash A$ are open because $X$ is locally connected. Let $C$ be one. Then $f(C)$, being connected, is contained in a connected component $C'$ of $Y \backslash B$. Now the union of the other connected components of $X\backslash A$ is open, hence $C \cup A$ is closed. Therefore $\bar C \subset C \cup A$, and so $f(\bar C)$ equals $f(C)$ with some elements of $B$ added, hence $f(C) = f(\bar C) \cap C'$. We see that $f(C)$ is an open ($f$ open so $f(C)$ open) and closed ($f(\bar C)$ closed) subset of $C'$, so it must be the full $C'$.
Applying this to the continuous open and closed map $f \colon \mathbb{C} \to \mathbb{C}$ polynomial of degree $n$ and $B \colon = \{ \mathcal{Re}\ z = 0\}$ we see that each component of $\{z \ | \mathcal{Re}\ f(z) = 0\}$ maps surjectively onto $\{ \mathcal{Re}\ z > 0\}$ or $\{ \mathcal{Re}\ z < 0\}$, so in particular it cannot be bounded. From here we conclude that each of these connected components is simply connected, using this fundamental fact: an open subset of $\mathbb{C}$ is simply connected if and only if the complement of its closure has no bounded components.
It was tempting to conclude that since we have a branched covering $f \colon C \to \{ \mathcal{Re}\ z > 0\}$ we can right away conclude that $C$ is simply connected. However, this seems to be not correct, one can map an annulus to a disk by a $2-1$ branched covering.
$\bf{Added:}$ The polynomial map $f$ should be seen a a ramified $n-1$ covering from the Riemann sphere $S$ to $S$. The preimage of the imaginary axis $\cup \{\infty\}$ (a circle), is a graph on the sphere $S$. This has to be understood better.
