Assume we have a linear operator $A: \mathscr{H}\supseteq D(A) \rightarrow \mathscr{H}$ which is symmetric hence densely defined but not self adjoint. Where $\mathscr{H}$ is a complex Hilbert space.
The numerical range is defined as $N(A):=\lbrace \langle Ax,x\rangle | x\in D(A), \|x\|=1\rbrace$
Then we know $cl(N(A))\subseteq \mathbb{R}$, because of the symmetry.
I wonder if $cl(N(A))=\mathbb{R}$. Because I came accross a statement that says that for a symmetric operators, which are not self adjoint, the spectrum $\sigma(A)$ is either $\mathbb{C}$, $\lbrace{z\in\mathbb{C}| \mathrm{Im}(z)\geq 0}\rbrace$ or $\lbrace{z\in\mathbb{C}| \mathrm{Im}(z)\leq 0}\rbrace$.
But if that would be the case and $cl(N(A)) \subsetneqq \mathbb{R}$ we could find $z_0$ in some open and connected domain $\Omega \subseteq \mathbb{C}\setminus cl(N(A))$ where $\Omega\cap \sigma(A)\neq \emptyset$ (when $\sigma(A)=\lbrace{z\in\mathbb{C}| \mathrm{Im}(z)\leq 0}\rbrace$ or $\lbrace{z\in\mathbb{C}| \mathrm{Im}(z)\geq 0}\rbrace$). Where $\mathrm{ran}(z_0-A)=\mathscr{H}$ and thus $\Omega\subseteq \rho(A)$.
But this is a contradiction.
I also attached a picture to clarify my thoughts:
