$$\label{19}
f_x = \frac{5}{8} k T \frac{h^2}{a(a^2+h^2)}\ , \quad f_y = -  \frac{5}{8} k T  \frac{h}{(a^2+h^2)} \ .$$ 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 uses the map $v=\tilde g_A(z)$ which preserves $z_0$ rather than $\infty$. Composing $g_A$ with a Moebius map which maps $H$ to $H$, $0$ to $0$ and $g(z_0)$ back to $z_0$, one finds a complicated formula which simplifies for $x_0=a$ to ${\cal P}= (\frac{a^2}{a^2 + h^2})^{\frac{5}{16}} y_0^{-5/12} (y_0^2-h^2)^{5/24}$. This gives for the force on the wall $f_y=-\frac{5k T }{24}(\frac{3h}{a^2+h^2}+\frac{2h}{y_0^2-h^2})$ which diverges as $y_0 \to h^+$.

Another example is a half disk of radius $r$ centered at $x=a>0$. The uniformizing map is $g(z)=z+\frac{r^2}{a}+\frac{r^2}{z-a}$. Hence the-no hit probability is ${\cal P}=(1 - \frac{r^2}{a^2})^{5/8}$, and the object is repelled with a force $f_x=\frac{5 kT}{4 a} r^2/(a^2-r^2)$.

The [*polymer piston*]{} is interesting for [*extreme-value statistics*]{}. Consider the strip geometry ${\cal D}$ on Fig. \[f:3\] and add an impenetrable region ${\sf P}$ (the piston) for $x>a$. The map $h_a(z)= [\cosh(\frac{\pi}{L}a)-\cosh(\frac{\pi}{L}(z-a))]/[\cosh(\frac{\pi}{L}(z-a))+\cosh(\frac{\pi}{L}a)]$ maps the strip minus ${\sf P}$ to the upper half plane, and both axes $y=0$ and $y=L$ to $0$. Hence the map which removes the piston is $g_A(z)=h_{\infty}^{-1}(h_a(z)) 