Mapping of unit circle under $f(z) = \frac{1}{2}(z + \frac{1}{z})$ I'm trying to find the mapping of the unit circle centered at the origin $A=\{ z\in \mathbb{C} : |z|=1\}$ and a straight line $B=\{ \arg (z) = \theta_0 \}$, under the mapping $ f(z) = \frac{1}{2}(z + \frac{1}{z})$.  
For the first set i got that, writing the set in polar form, $A =\{ re^{i\theta} : r=1, \theta \in [0, 2\pi]\}=\{ e^{i\theta} :\theta \in [0, 2\pi]\}$, so:
$$f(e^{i\theta}) = \frac{1}{2}(e^{i\theta} + \frac{1}{e^{i\theta}}) = \frac{1}{2}(e^{i\theta} + {e^{-i\theta}})= \{\cos \theta : \theta\in [0,2\pi]\} = [-1,1]$$
I tried to apply the same logic to the second set:
$B =\{ re^{i\theta_0} , r\in \mathbb{R}\} $ so:
$$f( re^{i\theta_0}) = \frac{1}{2}( re^{i\theta_0} + \frac{1}{ re^{i\theta_0}}) = \frac{1}{2} re^{i\theta_0} + \frac{1}{2 re^{i\theta_0}}  =  \frac{1}{2} re^{i\theta_0} + {\frac{1}{2r} re^{-i\theta_0}}  $$
But haven't been able to rewrite $f$ as a known set in the complex plane.
 A: Starting from where you ended off, mapping $r$ to  $$\frac{1}{2} re^{i\theta_0} + {\frac{1}{2r} re^{-i\theta_0}}
= \left( \frac{r}2 + \frac1{2r} \right) \cos \theta_0 + i  \left( \frac{r}2 - \frac1{2r} \right) \sin \theta_0 
$$
we can then let $z = x+iy$ and this becomes
$$
\left( \begin{array}{c} x =  \frac{r}2 + \frac1{2r} \\y =  \frac{r}2 - \frac1{2r} \end{array}\right)
$$
then $$\left( \begin{array}{c} 
x+y = r\\ x-y = \frac1r
\end{array}\right) \implies (x+y)(x-y) = x^2 - y^2 = 1$$
and so the line maps to a hyperbola.  
The tricky part is that since you don't map the part of the line on the "other" side of zero, the ray maps to just one branch of that hyperbola.  You can try mapping the point with $r=1$ to see which branch.
A: The mapping is:
$$ (x,y) \mapsto \left( \frac{x}{2} + \frac{x}{2 (x^2 + y^2)}, \frac{y}{2} - \frac{y}{2 (x^2 + y^2)} \right) $$
The unit circle will map to the line segment joining the points -1 and 1.  The unit circle and its interior will map to the whole plane.  Each circle centered at zero  interior to the unit circle maps to an ellipse.
A line through the origin maps to an hyperbola (as posted in another answer) and the particular hyperbola will depend on $\theta_0$.
