Does there exist a conformal map from the unbounded region outside the disks $\{z:|z-1|\le 1\}\cup \{z:|z+1|\le 1\}$ to the upper half plane?
In order to find the conformal map we need to find $f(z)$ which keeps the points in the unbounded region outside the disks $\{z:|z-1|\le 1\}\cup \{z:|z+1|\le 1\}$ which are in the upper half plane intact but moves the points in the lower half plane to upper half plane.
I thought of $f(z)=z^2$ but its not working.
Can I get some hint?
The complement of those two disk in the extended complex plane $$ D = \hat{\Bbb C} \setminus (\{z:|z-1|\le 1\}\cup \{z:|z+1|\le 1\}) $$ is simply-connected. The Riemann mapping theorem guarantees that there is a bijective conformal mapping from $D$ to the upper half-plane.
If you want an explicit map then I would start with $z \mapsto 1/z$, which maps $D$ to the region between two parallel lines. This strip is mapped to $\{ z: 0 < \operatorname{Im}(z) < \pi \}$ with a rotation and translation, and finally to the upper half-plane with the exponential function.
All used functions are conformal and bijective in the domain where they are used, so that the composition is conformal and bijective.