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I would like you to follow my logic, confirming it if correct, suggesting change when flawed or suboptimal.

Let $p$ be a complex polynomial \begin{gather*} p:\mathbb{C}\longrightarrow\mathbb{C},\\ \deg p = n,\quad n\in\mathbb{N}. \end{gather*} Let $z=x+iy$, and write $p(z)=u(x,y)+iv(x,y)$. Note that $u$ and $v$ are harmonic functions. Define the set $\mathcal{R}=\{z\in\mathbb{C}:|p(z)|\leq R\}$, where $R>0$ is chosen sufficiently large.

  • $\partial\mathcal{R}=\{z\in\mathbb{C}:|p(z)|=R\}$ by continuity and the maximum modulus principle.

  • Hence $\mathcal{R}$ is closed. Since $|p(z)|\rightarrow\infty$ as $|z|\rightarrow\infty$, the set $\mathcal{R}$ is also bounded.

  • $\mathbb{C}\setminus\mathcal{R}$ is open and by the maximum modulus principle necessarily unbounded.

  • $\operatorname{int}\mathcal{R}=\mathcal{R}\setminus\partial\mathcal{R}$ is open and bounded. It's also simply connected by the maximum modulus principle.

  • $\mathcal{R}_{1/2}=\left\lbrace z\in\mathbb{C}: |p(z)|\leq\frac{R}{2}\right\rbrace\subset\operatorname{int}\mathcal{R}$ by continuity of $|p(z)|$.

  • Hence $\partial\mathcal{R}\subset\mathbb{C}\setminus\mathcal{R}_{1/2}=\left\lbrace k\in\mathbb{C}: |p(k)|>\frac{R}{2}\right\rbrace$.

Now suppose $R$ is so large that \begin{align*} p^{\prime}(z)\neq0\text{ when }z\in\mathbb{C}\setminus\mathcal{R}_{1/2}. \end{align*}

  • Then the boundary $\partial\mathcal{R}$ is given by a smooth curve: Suppose $x_{0}+iy_{0}=z_{0}\in\partial\mathcal{R}$. The condition on the boundary is $|p(z_{0})|=R$ and $p^{\prime}(z_{0})\neq0$. Let $F(x,y)=(u(x,y))^{2}+(v(x,y))^{2}-R^{2}$. The zero level set of $F$ is precisely the set $\partial\mathcal{R}$. Note that \begin{align*} \nabla F = 2\, \underbrace{\begin{pmatrix} \partial_{x}u & \partial_{x}v \\ \partial_{y}u & \partial_{y}v \end{pmatrix}}_{A}\, \underbrace{\begin{pmatrix} u\\ v \end{pmatrix}}_{\bar{u}}, \end{align*} where $\det A=p^{\prime}\neq0$ at the point $z_{0}$. Since the matrix $A$ is invertible at the point $z_{0}$, the only element in it's nullspace is the zero vector. However $\left\vert\bar{u}\right\vert^{2}=R^{2}$ at $z_{0}$. That meas $\left\vert\nabla F\right\vert\neq0$ at $z_{0}$, hence atleast one of the partial derivatives $\partial_{x}F$ or $\partial_{y}F$ is non zero at the point $z_{0}$. So the implicit function theorem gives that there exists a neighbourhood of the point $z_{0}$ such that the set of $z$ such that $F(z)=0$ in that neighbourhood consists of a smooth curve passing through $z_{0}$.
  • The above smooth curve is bounded.

Here are some things I'm wondering about

  • Is the smooth curve giving the boundary $\partial\mathcal{R}$ simple (for big enough $R$)? I also think it's closed but not sure how to show it.

  • 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 $\mathcal{R}_{big}$?

  • Are there some more interesting properties to discuss?
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