<s>[INST] Image of $\partial B_r $ under an holomorphic function 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. 
 [/INST] We have, writing $a+ib:=re^{it}$ that 
\begin{align}
f(re^{it})&=\frac 12\left(a+ib+\frac 1{a+ib}\frac{a-ib}{a-ib}\right)\\
&=\frac 12\left(a+ib+\frac{a-ib}{r^2}\right)\\
&=\frac 12\left(a\left(1+\frac 1{r^2}\right)+ib\left(1-\frac 1{r^2}\right)\right)\\
&=\frac 12\left(r\cos \theta\frac{r^2+1}{r^2}+ir\sin\theta\frac{r^2-1}{r^2}\right)\\
&=\frac 1{2r}(\cos\theta (r^2+1)+i\sin\theta(r^2-1)).
\end{align}
Do you recognize curve in the plane of the form $\gamma(t)=(A\cos t,B\sin t)$, where $A$ and $B$ are fixed?
</s>