Conformal map from $\{z \in \mathbb{C}: |z|>1\}\setminus (-\infty,-1)$ onto $\mathbb{C}\setminus(-\infty,0]$ Find a conformal map from the set 
$\{z \in \mathbb{C}: |z|>1\}\setminus (-\infty,-1)$ onto the set $\mathbb{C}\setminus(-\infty,0]$.
Here is my thought, but I'm not sure if it is correct, can anyone help me to verify my answer?
1: $\frac{1}{z}$ maps the set to $\mathbb{D}\setminus(-1,0]$.
2: $z^{1/2}$ maps the disk with slit to the right half disk.
3: rotate the right half disk to lower half disk.
4: the map $z+1/z$ maps the lower half disk to upper half plane.
5: rotate the upper half plane to right half plane.
6: $z^2$ maps the right half plane to $\mathbb{C}\setminus(-\infty,0]$.
Also, if anyone can think of an alternate route, it would be nice to see! Thanks!
 A: Your solution looks right to me.
I did this first: $z\to \sqrt z \to 1/\sqrt z.$ That leaves us in the right half disc. I'd prefer the upper half disc, so multiply by $i.$ Then we apply the map $(1+z)/(1-z),$ which takes us to the first quadrant. Rotate clockwise by $\pi/4,$ i.e., multiply by $e^{-i\pi /4}.$ Finally the map $z^4$ puts us in the desired domain. The formula I got is 
$$z \to -\left(\frac{1+i/\sqrt z}{1-i/\sqrt z}\right)^4.$$
I think we are doing basically the same thing.
A: The first thing that worked for me was to use the open 1st quadrant $Q_\mathrm{I} = \{ z \in \mathbb{C} : 0 < \mathrm{arg}(z) < \frac{\pi}{2} \}$ as an intermediate domain. 
You can produce a conformal bijection $f : Q_\mathrm{I} \to \mathbb{C} \setminus (-\infty,0]$ as follows.

  
*
  
*$z^4$ maps $Q_\mathrm{I}$ to $\mathbb{C} \setminus [0,\infty)$. 
  
*$-z$ maps $\mathbb{C} \setminus [0,\infty)$ to $\mathbb{C} \setminus (-\infty,0]$.
  

You can produce a conformal bijection $g: Q_\mathrm{I}  \to \{z \in \mathbb{C}: |z|>1\}\setminus (-\infty,-1)$ as follows.

  
*
  
*The Cayley transform $\frac{z-i}{z+i}$ maps the upper half plane to the unit disk and, restricting, maps $Q_\mathrm{I}$ to the lower half disk.
  
*$z^2$ maps the lower half disk to the disk minus $[0,1)$. 
  
*$\frac{1}{z}$ maps the disk  minus $[0,1)$ to $\{z \in \mathbb{C}: |z|>1\}\setminus (1,\infty)$.
  
*$-z$ maps $\{z \in \mathbb{C}: |z|>1\}\setminus (1,\infty)$ to $\{z \in \mathbb{C}: |z|>1\}\setminus (-\infty, -1)$.
  

Then you can use $g \circ f^{-1}$ to get a conformal bijection between the desired domains.
By the way, your approach seems good, although I didn't check that $z + \frac{1}{z}$ does what you claim.
