Proving that $f(z) = \frac{1}{2} \left(z + \frac{1}{z}\right)$ is biholomorph on a certain set Let $f: \mathbb{C} \backslash \{0\} \to \mathbb{C}$ be given by 
$$f(z) := \frac{1}{2} \left(z + \frac{1}{z}\right)$$
I first want to find the image of the set $H = \{z \in \mathbb{C}: |z| < 1, Im(z) > 0\}$ (the open upper half of the unit circle) regarding $f$.
Next, I want to show that $f$ sends $H$ biholomorph onto $f(H)$. Finally, I want to figure out the inverse mapping $f^{-1}$, as defined on $f(H)$.
My attempt so far: I've noticed that $f(z) = f(\frac{1}{z})$. As $lim_{t \downarrow 0} |f(t i)| = \infty $, the values $f$ takes on $H$ can become "very large" regarding their absolute value, hence, $f(H)$ is unbounded. On the other hand, we have that $i$ is a root of $f$ (and, besides $-i$, the only root of it). I therefore suspect that $f(H)$ might be something like the lower half of the complex plane $\{z \in \mathbb{C}: Im(z) < 0\}$ or an unbounded subset of it? (Because $Im(f(z)) < 0 \space \forall z: 1 > Im(z) > 0$ if I'm not mistaken; therefore, it can only be a subset of the lower half of $\mathbb{C}$.) I don't really know how to cleverly continue from there?
I know that in order to show biholomorphy, I would need to show that $f^{-1}$ is also holomorph; hence, it would be useful to first figure out $f^{-1}$ regarding $f(H)$. Maybe, it's already visible then on why $f^{-1}$ is holomorph. So my main problem is to figure out $f(H)$, and determine $f^{-1}$.
EDIT: Actually, since $f(z) = f(\frac{1}{z})$, does that mean that $f(\{z \in \mathbb{C}: Im(z) < 0\} \backslash H) = f(H)$, or am I mistaken there? I'm not sure though how that might help me.
 A: Hint: Write 
$$ 
 f(z) = \frac{1}{2} \left(z + \frac{1}{z}\right) = \frac{\left( \frac{1+z}{1-z}\right)^2 + 1}{\left( \frac{1+z}{1-z}\right)^2 - 1}
 = T_2(S(T_1(z))
$$
as the concatenation of three "simple" mappings:
$$
 T_1(z) = \frac{1+z}{1-z}, S(z) = z^2, T_2(z) = \frac{z+1}{z-1}
$$
Use the basic properties of Möbiustransformations
to conclude
that $T_1$ is a biholomorphic mapping from $H$ to the first quadrant.
Now consider $S$ restricted to the first quadrant. Is it one-to-one?
What is the image? Finally determine the image under $T_2$ which
is again a Möbiustransformation.
Each of the three mappings can be inverted easily, that allows you
to compute $f^{-1}$.

I therefore suspect that $f(H)$
   might be something like the lower half of the complex plane ...

Correct!

Actually, since $f(z) = f(\frac{1}{z})$, does that mean that ...

$z \to 1/z$ maps $H$ to $G := \{ z \in \mathbb{C}: |z| > 1, \text{Im}(z) < 0\}$,
therefore $f(H) = f(G)$.
A: This is the Muskhelishvili transform, for the special case $R=1/2$. It maps circles into circles and ellipses and vice versa.
To see examples of it in action, parametrize an open sub-cover of the domain $H$ in question, as:
$$H_1=\{z\in\mathbb{C}\colon z=(\rho-\epsilon)\cdot\exp(\theta i), 0\lt \theta\lt\pi\}$$
$$H_2=\{z\in\mathbb{C}\colon z=r\cdot\exp(0 i)+\epsilon\cdot i, -\rho+\epsilon\lt r\lt\rho-\epsilon\}$$
It's holomorphic,so just look at the subcover under the map, for various values of $\rho$, below in the graphs.
Its inverse is gotten by solving the following equation for $z$:
$$w=\frac{1}{2}\left(z+\frac{1}{z}\right)$$
which gives for $f^{-1}$:
$$z=w\pm\sqrt{w^2-1}$$
Therefore the inverse has two branches and two branch points at $f^{-1}(z_c)$, with:
$$z_c=\{z\in\mathbb{C}\colon \frac{df(z)}{dz}=0\}=\{-1,1\}$$
Hence the two branches of $f^{-1}$ can be joined together through only a branch cut between -1 and 1, making the inverse holomorphic everywhere away from this branch cut.
Some Maple code to help a bit:
restart;
with(plots);
eps := 0.5e-1;
f := proc (z) options operator, arrow; (1/2)*z+(1/2)/z end proc;
z := proc (rho, theta) options operator, arrow; rho*exp(I*theta) end proc;
rho := 1;
p1 := complexplot(z(rho-eps, theta), theta = 0 .. Pi, scaling = constrained, color = red);
p2 := complexplot(z(r, 0)+I*eps, r = -rho+eps .. rho-eps, scaling = constrained, color = red);
p3 := complexplot(f(z(rho-eps, theta)), theta = 0 .. Pi, scaling = constrained, color = blue);
p4 := complexplot(f(z(r, 0)+I*eps), r = -rho+eps .. rho-eps, scaling = constrained, color = blue);
display(p1, p2, p3, p4);

Graphs for $\rho=1$, $\rho=1/2$ and $\rho=2$ follow:



