Determine the image of the set $G = \{ z \mid |z|<1,\; Im(z)>0\}$ under $f(z) = \frac{z+\frac{1}{z}}{2}$ I wish to describe the image of $G = \{ z \mid |z|<1,\; Im(z)>0\}$ under $f(z) = \frac{z+\frac{1}{z}}{2}$. $G$ is the open upper unit=semicircular region.
$f(z) = \frac{z^2 + 1}{2z}$. I thought this was a mobius transformation, but apparently not. Does anyone have ideas on how to proceed?
My attempt:
$z \to z^2$ forms a parabolic region
$z \to z^2 + 1$ translates the parabola up by $1$
$z^2 +1 \to \frac{z^2 +1}{2z}$ translates the parabola by $\frac{1}{2z}$ downward.
Is this correct?
 A: Let $\Gamma$ be a complex contour for the boundary of the set $G$ with a notch of radius $h$ around (and above) the pole of $f(z)$ at $0+0i$. We will let $h \to 0^+$ to get a better intuitive idea of the image of $G$ under $f$. Define the four points $A(-1,0),B(1,0),C(h,0),D(-h,0)$ that specify $\Gamma$.

It is clear that the points map as follows:
$$\begin{align}
A&\mapsto(-1,0) \\
B&\mapsto(1,0) \\
C&\mapsto(h+\tfrac{1}{h},0) \\
D&\mapsto(-h-\tfrac{1}{h},0)
\end{align}$$ 
For $z\in AB,|z|=1$ so 
$$f(z)=\frac{1}{2}\left(z+\frac{1}{z}\right)=\frac{1}{2}\left(z+\frac{\overline{z}}{|z|^2}\right)=\Re[z]$$
$\begin{array}{lc}
\text{Therefore }&AB\mapsto \{x+iy:-1\le x\le 1,y=0\} \\[2ex]  
\text{Similarly }&BC\mapsto \{x+iy:1\le x\le h+\tfrac{1}{h},y=0\} \\[2ex]
\text{finally }  &DA\mapsto \{x+iy:-h-\tfrac{1}{h}\le x\le -1 ,y=0\}
\end{array}$
Now $CD$ can be parametrised as $\{z=re^{i\theta}:r=h,0\le\theta\le\pi\}$.
So for $z\in CD$, we have
$$f(z)=\frac{1}{2}\left(z+\frac{1}{z}\right) \implies \frac{1}{|z|}-|z|\le|f(z)|\le \frac{1}{|z|}+|z| \qquad (\text{since }|z|<1)$$
and
$$f(z)=\frac{1}{2}\left(z+\frac{\overline{z}}{|z|^2}\right) \implies \Im[z]<0\text{ for }|z|<1$$
So the image of $CD$ lies between circular arcs of radius $\frac{1}{h}-h$ and $\frac{1}{h}+h$ in the lower half-plane.
Now if we let $h\to0^+$, then $C'$ tends to $(+\infty,0)$ and $D'$ tends to $(-\infty,0)$, and $CD'$ becomes an arc of radius $\infty$ clockwise from $C'$ to $D'$ in the lower half-plane.
So by the open mapping theorem, $G$ is sent to the lower half-plane (note: the open interval of the real axis between $-1$ and $1$ is not included, because $|z|<1$ for $G$). That is
$$f(G)=\{x+iy:x\in(-\infty,\infty),y\in(-\infty,0)\}$$
A: HINT:
Being $f(z)=\frac{z+\frac1z}{2}$, you can reduce the problem to study, the image of $G$ thru $h(z)=\frac1z$.
Call $\Delta=\{|z|<1\}$ and $\Bbb H:=\{\Im z>0\}$.
With some elementary computations you can deduce that
$$
h(\Bbb H)=-\Bbb H\\
h(\Delta\setminus\{0\})=\Bbb C\setminus\Delta\;\;.
$$
