Conformal map taking quadrants to half-planes 
The function $f$ takes the values indicated in the picture. It has
  poles at $1$ and $-1$ and $f(i)=f(-i)=1$. Elsewhere it is analytic and
  maps quadrants one-to-one onto the indicated half-planes. What is it?

Ideas: If we can construct a map that agrees with this one on a neighborhood, we are done. 
I've thought of trying to look at $f^2$, which would take $\text{UHP}\to \text{UHP}$ and preserve $\mathbb{R}$. Perhaps we could construct an FLT with these properties mapping the given points where we want, but how would we know that this is $f$?

 A: We want a rational function of order $2$ (because nothing else can map the quadrants conformally to the half planes) that has poles in $1$ and $-1$, so we will have
$$f(z) = \frac{??}{z^2-1}.$$
We want $f(i) = f(-i) = 1$, so the denominator must take the value $-2$ in both $i$ and $-i$. One option is the constant $-2$, so
$$f_0(z) = \frac{2}{1-z^2},$$
and one sees that it meets the requirements.
If the numerator isn't constant, it must have the form $c(z^2+1)-2$ in order to attain the value $-2$ in $\pm i$ and be quadratic. The parameter $c$ must be real, since the numerator must be real-valued on $\mathbb{R}$, as the denominator is. We can write that in the form
$$f_c(z) = \frac{c(z^2+1)-2}{z^2-1} = \frac{c(z^2-1) + 2(c-1)}{z^2-1} = c + \frac{2(1-c)}{1-z^2} = c + (1-c)f_0(z).$$
From that, it is seen that for $c\neq 1$, $f_c(\pm 1) = \infty$, and $f_c(\pm i) = 1$, and $f_c$ conformally maps the quadrants to the upper resp. lower half-plane. For $c < 1$, it maps the first quadrant to the upper half-plane, as required, and for $c > 1$, it maps the first quadrant to the lower half-plane, hence $f$ can be any of the $f_c$ for $c < 1$.
