Does this set tend towards a disc? Let $p$ be a complex polynomial 
\begin{gather*}
p:\mathbb{C}\longrightarrow\mathbb{C},\\
\deg p = n,\quad n\in\mathbb{N}.
\end{gather*}
Define the set $\mathcal{R}=\{z\in\mathbb{C}:|p(z)|\leq R\}$, where $R>0$ is chosen large.
I believe there is some upper and lower bound on $\partial\mathcal{R}$ for large $R$ in the form of disks centred at the origin. That is
\begin{align*}
\mathcal{R}_{small}&=\{z\in\mathbb{C}:|z|<R_{small}\},\\
\mathcal{R}_{big}&=\{z\in\mathbb{C}:|z|<R_{big}\},
\end{align*} 
where $R_{big}>R_{small}>0$, such that $\mathcal{R}_{small}\cap\partial\mathcal{R}=\emptyset$ and $\mathcal{R}_{big}\cap\partial\mathcal{R}=\partial\mathcal{R}$. What is the biggest $\mathcal{R}_{small}$ and smallest $\mathcal{R}_{big}$ for large $R$?
My related post: Analysing a set in the complex plane
Example:
Let the polynomial $p$ be given by
$|p(x+iy)|^2=|(x+iy)^5+(x+iy)^2+1|^2=(1+x^2+x^5-y^2-10x^3y^2+5xy^4)^2+(2xy+5x^4y-10x^2y^3+y^5)^2$
try puting $|p(x+iy)|^2$ equal to 1, 2, 3 and so on in https://www.desmos.com/calculator
Also try this https://www.desmos.com/calculator/opbksj7j4t
 A: Let $$\begin{align}p(z)&=a_nz^n+a_{n-1}z^{n-1}+\ldots +a_1z+a_0 \\&=a_nz^n\cdot\left(1+\frac{a_{n-1}}{a_nz}+\ldots + \frac{a_{1}}{a_nz^{n-1}}+\frac{a_{0}}{a_nz^{n}}\right)\end{align}$$
The expression in parentheses tends to $1$ as $|z|\to\infty$.
Especially, there exists $r_0$ such that its absolute value is between $\let\epsilon\varepsilon1-\epsilon$ and $1+\epsilon$ for all $z$ with $|z|>r_0$.
Also, let $M=\max\{\,|p(z)|:|z|\le r_0\,\}$.
Then for $R$ with $R>\max\{r_0,M\}$ we conclude that $ \mathcal R$ contains the disk $\mathcal R_{\text{small}}$ with radius $\sqrt[n]{\frac{R}{(1+\epsilon)\left|a_n\right|}}$ and is contained in the disk $\mathcal R_{\text{big}}$ with radius $\sqrt[n]{\frac{R}{(1-\epsilon)\left|a_n\right|}}$.
As the quotient of these radii is $\sqrt[n]{\frac{1+\epsilon}{1-\epsilon}}$ one could indeed say that $\mathcal R$ approaches disk shape as $R\to\infty$.
Can we investigate the asymptotic behaviour more closely?
If $\epsilon$ is small enough, we will have $r_0\approx \left|\frac{a_{n-1}}{a_n\epsilon}\right|$ (if $a_{n-1}\ne 0$; more generally it is $r_0\approx \sqrt[k]{\left|\frac{a_{n-k}}{a_n\epsilon}\right|}$ where $k$ is minimal with $a_{n-k}\ne 0$; I shall consider only the case $k=1$) and $M\approx |a_n|r_0^n$, so for $R\approx \left|\frac{a_na_{n-1}^n}{a_n^{n}\epsilon^n}\right|$ the two bounding disks will have radii $$\approx \sqrt[n]{\left|\frac{a_na_{n-1}^n}{a_n^n\epsilon^n(1\pm \epsilon)a_n}\right|}=\left|\frac{a_{n-1}}{a_n\epsilon}\right|\frac1{\sqrt[n]{1\pm\epsilon}}\approx \left|\frac{a_{n-1}}{a_n\epsilon}\right|\left(1\mp\frac\epsilon n\right)=\left|\frac{a_{n-1}}{a_n\epsilon}\right|+\frac1n\left|\frac{a_{n-1}}{a_n}\right|.$$
If we express $\epsilon $ in terms of $R$, i.e., as $\epsilon\approx \left|\frac{a_{n-1}}{a_n}\right|\sqrt[n]{\left|\frac{a_n}{R}\right|}$, the expressions for the radii become
$$ \approx \sqrt[n]{\frac R{|a_n|}}\mp \frac1n\left|\frac{a_{n-1}}{a_n}\right|.$$
In other words, the absolute difference between $\mathcal R_{\text{small}}$ and $\mathcal R_{\text{big}}$ is more or less constant.
A: Suppose $p(z) = a_nz^n + \cdots.$ Then $|p(z)| \sim |a_n||z|^n$ as $z \to \infty.$ So given $\epsilon > 0,$ we will have
$$(1-\epsilon)|a_n|R^n < |p(Re^{it})| < (1+\epsilon)|a_n|R^n$$
for large $R.$ Not sure if you are looking for something more precise.
