The range of complex -valued functions How can I find the range of $\phi(z)=\frac{1}{2}(z+1/z)$ for $z$ in the upper plane $Im z>0$, or for $z$ outside the unit sphere.
It is really difficult to do it...as I have seen Finding the Range of a Complex Function.
 A: Let $z = re^{i\theta}$. We have
\begin{align}
\phi(z) &= \phi(re^{i\theta}) = \frac{1}{2}\left(re^{i\theta} + r^{-1}e^{-i\theta}\right) \\
&= \frac{1}{2}\left((r + r^{-1})\cos\theta + i(r - r^{-1})\sin\theta\right).
\end{align}
Thus
\begin{align}
x &= \Re \phi(z) = \frac{1}{2}\left((r + r^{-1})\cos\theta\right),\\
y &= \Im \phi(z) = \frac{1}{2}\left((r - r^{-1})\sin\theta\right).
\end{align}
For a fixed $r > 1$, the image of $z = re^{i\theta}$ is
$$
\frac{x^2}{(r + r^{-1})^2/4} + \frac{y^2}{(r - r^{-1})^2/4} = 1.
$$
Hence the image of a circle centered at the origin with radius $r > 1$ is an ellipse centered at the origin with major axis $(r + r^{-1})/2$ and minor axis $(r - r^{-1})/2$.
Similarly for a fixed $\theta \in (0, \pi) - \{\pi/2\}$, the image of $z = re^{i\theta}$ is
$$
\frac{x^2}{\cos^2\theta} - \frac{y^2}{\sin^2\theta} = 1.
$$
Hence the image of a ray in the upper half-plane with initial point at the origin is a branch of a hyperbola.
I'll let you study the extreme cases and find the images you're looking for. This should be straightforward given the above. You'll find that the image of $\{z \in \Bbb C : |z| > 1\}$ is $\Bbb C - [-1, 1]$ and the image of $\{z \in \Bbb C : \Im z > 0\}$ is $\Bbb C - ((-\infty, -1] \cup [1, \infty))$.
A: Best to do this on the Riemann sphere. The map $z\mapsto \frac{1}{2}\left(z+\frac{1}{z}\right)$ decomposes into the product of three maps: 
$$z\mapsto \frac{z+1}{z-1}=t,\; t\mapsto t^2=w,\; w\mapsto \frac{w+1}{w-1} = \frac{1}{2}\left(z+\frac{1}{z}\right)$$
Now the fractional linear transformation $z\mapsto \frac{z+1}{z-1}$ is a rotation of the Riemann sphere about an axis at 45 degrees from the vertical (i.e., the axis from 0 to $\infty$). So it's of order 2, and your function is the squaring function conjugated by this rotation.
Note that the rotation interchanges $\infty$ with 1, and 0 with $-1$. And your original function leaves 1 and $-1$ fixed.
So to look at the image of the upper half plane: the rotation interchanges the upper and lower half planes. The squaring map then wraps the lower (closed) half plane around the entire sphere, and the second rotation doesn't change that. However, if you only want the image of the open upper half plane, then the squaring map applied to the lower half plane omits the non-negative reals, and after the rotation this omitted part is $[-\infty,-1]\cup[1,+\infty]$.
For the image of the (closed) exterior of the unit circle: the rotation interchanges the imaginary axis with the unit circle, and the right half plane ($\Im z\geq 0$) with the exterior of the unit circle. The squaring map again maps this to the whole sphere, so the image is the whole sphere again. If you want only the image of open exterior, then the squaring map applied to the right (open) half plane omits the non-positive reals, and so the image of your original function omits $[-1,1]$.
Motivation: two principles suggested the decomposition. (1) For rational functions, always work over the Riemann sphere. (2) Use FLTs to move the "significant points" (fixed points, poles, etc.) to a "standard position". Now for the example in question, the function $(z^2+1)/2z$ is quadratic over linear, suggesting a comparison with the simplest quadratic rational function, $z^2$. That has fixed points $\infty,0,1$, while $(z^2+1)/2z$ has fixed points $\infty,\pm 1$. From there it's a short step to the FLT $(z+1)/(z-1)$. It's just a piece of luck that this turned out to be a rotation on the sphere. (However, an FLT can move any three distinct points to any three distinct points, so we have a fair amount of flexibility.)
