Thus the cut is the real positive axis with $g_A(x+i \epsilon)=- \sqrt{(x-a)^2+h^2}$ for $x<a$. In other words, one defines in the full plane minus the real positive axis $\sqrt{x+i y}:=\frac{1}{\sqrt{2}}\left(  {{{\mathrm{sign}}}}(y) \sqrt{\sqrt{x^2+y^2}+x} + i \sqrt{\sqrt{x^2+y^2}-x}\right)$.

To obtain the force when the polymer starts at $0$ and ends at $z_0=x_0+i y_0$ in the half plane, one has to use the map $w=\tilde g_A(z)$ which preserves $z_0$ rather than $\infty$. It can be constructed, by composing $g_A$ with a second Moebius map. The latter maps $H$ to $H$, $0$ to $0$ and $g(z_0)$ back to $z_0$. We claim that $w$ and $z$ are related by $$\frac{1}{w} - \Re \frac{1}{z_0} = \frac{\Im(1/z_0)}{\Im(1/g(z_0))} \left[\frac 1{g(z)} - \Re \frac{1}{g(z_0)}\right]\ ,$$ where $\Re$ and $\Im$ indicate real and imaginary parts. First, all $\Re$ and $\Im$ appearing above are real numbers. Thus, $w$ is a Moebius transform of $g(z)$ with real parameters. Second, $z=0$, i.e. $g(z)=0$ is mapped onto $w=0$. Third, $z_0$ is mapped to $w=z_0$. Using this, we find $$\begin{aligned}
\left| \tilde g_A'(0)\right| &=&\left| \frac{a z_0^2 \,\Im g_A(z_0)}{ \sqrt{a^2+h^2} y_0 g_A(z_0)^2} \right| \\
\left| \tilde g_A'(z_0)\right| &=& \left|\frac{y_0 g_A'(z_0)}{\Im  g_A(z_0)}\right|\end{aligned}$$ This leads to $$\begin{aligned}
{\cal P}=\left|\frac{a}{\sqrt{a^2 + h^2}} \frac{z_0^2}{g_A(z_0)^2}\right|^{\frac{5}{8}} \left| \frac{\Im g_A(z_0)}{\Im(z_0)} \right|^{\frac{5}{8}-\frac{5}{48}} \nonumber 
 \left|g'_A(z_0) \right 