I need calculate $f^ {-1}(\mathbb{R})$ and $f(\partial\mathbb{E})$ wherein $\mathbb{E}$ is the unit disk, and $f(z)=\frac{1}{2}(z+z^ {-1})$.
Idea: For $f^ {-1}(\mathbb{R})$:
Let $c\in f^ {-1}(\mathbb{R})$ be then $\frac{1}{2}(c+c^ {-1})=\frac{1}{2}(\bar c+\bar c^ {-1})$ and I suppose $c\notin\mathbb{R}\setminus{\{0}\}$ I think it should happen $|c|=1$
But I don't know how to prove it. Can you help me? I also don't know how to calculate $f(\partial\mathbb{E})$.
First note that $$ \frac{1}{z}=\frac{\bar z}{|z|^2} $$ Thus, $$ \textrm{Im}(z+z^{-1})=\textrm{Im}(z)\left(1-\frac{1}{|z|^2}\right) $$ which equals $0$ if and only if either $z\in\mathbb R\setminus \{0\}$ or $|z|=1$. Hence $f^{-1}(\mathbb R)=\mathbb R\setminus\{0\}\cup \partial \mathbb E$.
Secondly, note that $f(e^{i\theta})=\cos\theta$, and thus $f(\partial\mathbb E)=[-1,1]$.