|^{\frac{5}{48}}\end{aligned}$$ The correct definition of the square root is: $$\begin{aligned}
 \sqrt{x+i y}&=&\frac{1}{\sqrt{2}}(  {{\mathrm{sign}}}(y) \sqrt{r+x}  + i \sqrt{r-x}) \\
 r &=& \sqrt{x^2+y^2}\end{aligned}$$ Hence the true map $g_A(z)$ reads for $z=x+i y$, $y \geq 0$: $$\begin{aligned}
{\nonumber}g_A(z) &=& \frac{1}{\sqrt{2}}(  {{\mathrm{sign}}}(x-a) \sqrt{\rho+t}  + i \sqrt{\rho-t})  \\
&&  + {{{\mathrm{sign}}}}(a) \sqrt{a^2+h^2} \\
 \rho &=& \sqrt{((x-a)^2-y^2+h^2)^2+4 (x-a)^2 y^2}\qquad  \\
 t &=& (x-a)^2-y^2+h^2\end{aligned}$$ One finds: $$\begin{aligned}
 |g'_A(z)|^2 &=& \frac{(x-a)^2+y^2}{\sqrt{h^4 + 2 h^2 ((x-a)^2-y^2)+((x-a)^2+y^2)^2}}{\nonumber}\\
 & =& \frac{(x-a)^2+y^2}{\rho} \end{aligned}$$ $$\begin{aligned}
{\cal P} = \left|\frac{a^{5/8} z_0^{5/4}
   \Im(g_A(z_0))^{25/48}
   g_A'(z_0)^{5/48}}{\left(a^2+h^2\right)^{5/16}
   y_0^{25/48} g_A(z_0)^{5/4}}\right|\end{aligned}$$ This simplifies considerably for $x=a$ $$\begin{aligned}
p= \left(\frac{a^2}{a^2 + h^2} \right)^{{5}/{16}} \left(\frac{y^2-h^2}{y^2}\right)^{5/24} \ ,\end{aligned}$$ which is the result given in the main text.

Appendix B: Force on a disk
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Let us recall [@bauer] that the probability that a SAW from $1$ to $e^{i \phi}$ on the unit disk avoids a disk centered at $0$ of radius $\rho\ll 1$ is $${\cal P}(\rho,\phi) \approx 1 - c \rho^{2/3} \sin^2(\phi/2) \ ,$$ where $c$ is still elusive. Consider the map $z=z_0 \frac{1-w}{1 - \frac{z_0}{\bar z_0} w}$, equivalent to ${w=\frac{\bar z_0}{z_0} \frac{z_0-z}{\bar z_0-z}}$ from the upper half plane in $z$, to the unit disk in $w$, with $w(0)=1$, $w(z_0)=0$ and $w(\infty)=\bar z_0/z_0=e^{i \phi}$. We note $z_0=x_0+ i y_0=r_0 {{\mathrm{e}}}^{i \theta_0}$ with $\theta_0=-\phi/2$. 