This is a problem from Bak-Newman's "Complex Analysis", #4 from Chapter 14 "The Riemann Mapping Theorem".
The question is this:
Verify directly that $F(z) = z + \frac{1}{z}$ is the unique conformal mapping (up to an additive constant) of the set $|z| > 1$ onto the exterior of a horizontal interval, such that $F(z) \sim{} z$ near $z = \infty$, and which takes the boundary of the unit disk onto the horizontal interval.
There is a hint which suggests that you look at the Laurent expansion of $F(z)$ on the region $|z| >1$.
Using the condition $F(z) \sim{} z$ near $z = \infty$ you get $F(z) = z + A_0 + \frac{A_1}{z} + \frac{A_2}{z^2} + ...$, and then can take $A_0 = 0$ since uniqueness is up to an additive constant.
The rest seems to boil down to showing $A_1 = 1$ and $A_k = 0$ for $k \ge 2$.
There is a solution in the back of the book which uses Fourier Series techniques.
I thought that maybe there could be another way to solve it, maybe using Schwarz reflection?
Here's my attempt at a different solution.
Since $F(z)$ is real when $|z| = 1$, use Schwarz reflection to write $F(\frac{1}{\overline{z}}) = \overline{F(z)}$. Then write $0 = F(\frac{1}{\overline{z}}) - \overline{F(z)}$ as a Laurent series in $w = \frac{1}{\overline{z}}$. You get $$0 = \dots - A_3 w^{-3} - A_2 w^{-2} + (1-A_1) w^{-1} + (\overline{A_1}-1) w + \overline{A_2} w^2 + \overline{A_3} w^3 + \dots$$
All the coefficients of the Laurent series will be $0$, by uniqueness of Laurent series, and so you're left with $A_1 = 1$ and so $F(z) = z + \displaystyle \frac{1}{z}$.
It feels pretty solid to me, but I'm not always sure about the conditions under which you can apply this kind of Schwarz reflection argument.
Thanks you very much for any input.
