Showing existences of biholomorphic maps. Define a complex polynomial $p:\mathbb{C}\longrightarrow\mathbb{C}$ where $\deg p=n\in\mathbb{N}\setminus\{2\}$. Define the set $\mathcal{S}=\{z\in\mathbb{C}:\text{Re }p(z) < 0\}\setminus\{z\in\mathbb{C}:|p(z)|\leq R\}$ where $R>0$ is so large that $p(z)$ and $p^{\prime}(z)$ are none-zero in $\mathcal{S}$, essentially make $R$ as big as necessary. Let
\begin{equation} 
p(z) = \alpha_{n}z^{n}+\alpha_{n-1}z^{n-1}+\dots+\alpha_{1}z+\alpha_{0},\quad \alpha_{n}\neq0.
\end{equation}
Hence $\mathcal{S}$ approaches the $n$ sectors
\begin{align*}
\arg\alpha_{n}+n\arg{k}\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)+2m\pi,\,m=0,\dots,n-1.
\end{align*} 
 Moreover, denote the components of $\mathcal{S}$ by $\mathcal{S}_{m}$ for $m\in\{0,\ldots,n-1\}$, where $m$ is increasing in a counter-clockwise direction, starting from the positive real line.
Now there is a claim:
Suppose $l,k\in\{0,\ldots,n-1\}$ and $l\neq k$. Then there exists biholomorphic maps 
\begin{align*}
\lambda_{k,l}:\overline{\mathcal{S}_{l}}\rightarrow\overline{\mathcal{S}_{k}}
\end{align*}
such that
\begin{align*}
p(\lambda_{k,l}(z))=p(z),\quad\forall z\in\mathcal{S}_{l},
\end{align*} 
for all such possible $l$ and $k$.
Is this true and if so how does one go about proving this fact?
Edit:
Look here, eq (1.16), (2.17) and (2.18)
http://imi.cas.sc.edu/django/site_media/media/papers/1999/1999_11.pdf
 A: This (and the assertion that $\mathcal{S}$ has exactly $m$ components) follows from Rouche's theorem.  Assume WLOG that $p(z)$ is monic.  If $|a|$ is sufficiently large, then Rouche's theorem (comparing $p(z)-a$ to $z^n-a$ in a disk around each $n$th root of $a$) implies that $p(z)=a$ has $n$ distinct roots, and arguments of these roots are very close to the arguments of the $n$th roots of $a$.  It follows that if $R$ is sufficiently large and you let $\mathcal{S}_m=\mathcal{S}\cap \{z:\arg z\in(\pi/2n+2\pi m/n-\epsilon,3\pi/2n+2\pi m/n+\epsilon)\}$ for some appropriate $\epsilon>0$, then $p$ maps each $\mathcal{S}_m$ bijectively onto the set $H=\{a:\operatorname{Re}(a)<0, |a|>R\}$.  The inverse $p|_{\mathcal{S}_m}^{-1}:H\to\mathcal{S}_m$ is then holomorphic, and you can define $\lambda_{k,l}=p|_{\mathcal{S}_k}^{-1}\circ p|_{\mathcal{S}_l}$.
(To get the extension to the boundary, you can just observe that everything would work the same if you slightly enlarged $H$ and correspondingly enlarged $\mathcal{S}$, so $p$ actually gives a biholomorphism between a neighborhood of $\overline{\mathcal{S}_m}$ and a neighborhood of $\overline{H}$.)

The following answers an earlier version of the question which had the wrong definition of $\mathcal{S}$.
This is not true.  For instance, consider $p(z)=(z-1)^n$; fix some $r$ that is much larger than $R$.  Then for $\epsilon$ very small, $z_m=re^{i\pi/2n+2\pi im/n+\epsilon}$ is very close to the boundary of $\mathcal{S}_m$, so $z_m+1$ will be in $\mathcal{S}_m$ for some values of $m$ and not in $\mathcal{S}_m$ for other values of $m$.  But $a=p(z_m+1)$ is the same for all $m$, and the values $z=z_m+1$ are all of the roots of $p(z)=a$.  Thus for some $m$ there is a root of $p(z)=a$ in $\mathcal{S}_m$, and for other $m$ there is not.  But if your desired biholomorphisms $\lambda_{k,l}$ existed, then the image $p(\mathcal{S}_m)$ would have to be the same for all $m$.
More generally, for any $p$, if you choose $z_0$ very near the boundary of $\mathcal{S}_0$, the other roots of $p(z)=p(z_0)$ will be very near the boundary of each $\mathcal{S}_m$, and in all likelihood some of them will be inside $\mathcal{S}_m$ and some of them will be outside $\mathcal{S}_m$.  
