Bounded, non-constant, analytic function on C\[-1,1]? Find an explicit analytic function on $\mathbb{C}\setminus[-1,1]$ which is bounded and non-constant.
Suggestions on how to approach this problem?
 A: One approach: If $\phi$ is any function integrable on $[-1.1]$ and you define $f$ to be tha Cauchy  integral $$f(z)=\int_{-1}^1\frac{\phi(t)\,dt}{z-t}$$then $f$ will be holomorphic in $\Bbb C\setminus[-1,1]$. If $\phi$ is smooth enough you can show $f$ is bounded...
A: Try to find a holomorphic function from your domain into the unit
disk. Proceed in steps and start with
$$
T(z) = \frac{z-1}{z+1}
$$
which maps $\mathbb{C}\setminus[-1,1]$ into the complex plane without
the negative real axis. Continue with a mapping into the right half-plane,
and you are almost done.
Spoiler:

 $$ f(z) = \dfrac{\sqrt{\frac{z-1}{z+1}} - 1}{\sqrt{\frac{z-1}{z+1}} + 1}$$

A: $y = \sqrt {z^2-1}$ defines a $2$-fold cover over of $\Bbb C$ with two branch points at $\pm 1$. When you consider one of its branches $y(z)$ on $\Bbb C \setminus [-1;1]$, you only get half of the Riemann surface associated with it. Then you simply want to find a function defined on that Riemann surface whose poles are not seen on that branch.
Here, the Riemann surface has genus $0$ so there are functions with a single pole anywhere you want.
For example, look at $z/ (2 y + \sqrt 3 z)$.
Its poles are included in the roots of $(2y+\sqrt 3z)(2y-\sqrt 3z) = 4y^2-3z^2 = z^2-4 = (z-2)(z+2)$, which are the $4$ points $(z=2, y= \pm \sqrt 3)$ and $(z=-2, y = \pm \sqrt 3)$.
So  $z/ (2 y + \sqrt 3 z)$ has two poles on the Riemann surface, located at $(z=2, y = - \sqrt 3)$ and $(z=-2, y = \sqrt 3)$.  
Then you simply have to choose the branch where you only see the points  $(z=2, y = \sqrt 3)$ and  $(z= -2, y = - \sqrt 3)$ to view $z/(2y(z)+ \sqrt 3 z)$ as a bounded holomorphic function on $\Bbb C \setminus [-1;1]$
