Let \begin{equation*} \begin{split} f \colon & \mathbb C \setminus \left\{0\right\} \to \mathbb C \\ & z \mapsto \frac{1}{2}\left( z+ \frac{1}{z}\right) \end{split} \end{equation*} I am asked to find the image of $\partial B_r := \{z \in \mathbb C : \vert z \vert = r\}$ (where $r >0$) under $f$.
Let me show you what I've done: hope it's correct.
Let $\vert z \vert = r$. Then \begin{equation*} \begin{split} \vert f(z) \vert & = \sqrt{f(z) \overline{f(z)}} = \sqrt{\frac{1}{2}\left( z+ \frac{1}{z}\right)\frac{1}{2}\left( \overline{z}+ \frac{1}{\overline{z}}\right)} = \\ & = \frac{1}{2}\sqrt{\left( z\overline{z} + \frac{z}{\overline{z}} + \frac{\overline{z}}{z} + \frac{1}{z\overline{z}}\right)} = \\ & = \frac{1}{2}\sqrt{\left( r^2 + 2\Re{\frac{z}{\overline{z}}} + \frac{1}{r^2} \right)} \end{split} \end{equation*}
Now we calculate $ \Re{\frac{z}{\overline{z}}} = \Re{\frac{x+iy}{x-iy}} = \Re{\frac{x^2+y^2 + 2ixy}{x^2+y^2}} = 1 $ hence we get \begin{equation*} \begin{split} \vert f(z) \vert & = \frac{1}{2}\sqrt{\left( r^2 + 2\Re{\frac{z}{\overline{z}}} + \frac{1}{r^2} \right)} = \\ & = \frac{1}{2}\sqrt{\left( r^2 + 2 + \frac{1}{r^2} \right)} = \frac{r^2+1}{2r} =: k. \end{split} \end{equation*}
So we can conclude that $ f(\partial B_{r}) = \partial B_{k} $
What do you think? Is it correct? Thanks for your help.