Without loss of generality, we consider $\gamma$, the streamline crossing $s_1$ orthogonally. This streamline crosses $s_1$ into $S'$, the open sector adjacent to $S$ that is also bounded by $s_1$, and then exits $B(a,r_0)$ (see ).

![Streamlines in a neighborhood of a singularity become hyperbolas under a conformal map $g$.[]{data-label="fig:hyperbolic"}](hyperbolic.pdf){width="\linewidth"}

By [@viertel_approach_2019], the cross field in $B(a,r_0)$ can be written as $$f(z) = e^{i(\frac{d\theta}{4} + \frac{2k\pi}{4})} + o(r)$$ where $d/4$ is the index of the singularity, $z=re^{i\theta}$, $k \in \{0,1,2,3\}$, and $\theta = 0$ corresponds to $s_0$. For $r < r_0$ where $r_0$ is sufficiently small, we make the approximation $$f(z) = e^{i(\frac{d\theta}{4} + \frac{2k\pi}{4})}.$$ Streamlines of the cross field are given by $$z' = f(z).$$ Since we are looking for the streamline crossing through $s_1$, we consider $k = 0$. Thus, we are looking for the set $C = \{z(t) \in B(a,r_0)\, |\, t \in (t_a,t_b)\}$ where $z(t)$ on $(t_a,t_b)$ is the solution to the problem $$\begin{aligned}
\label{eq:ODE}
  z' &= e^{i\frac{d\theta}{4}}\\
  z(0) &= (r_q,\theta_q)\end{aligned}$$

in $D = \{z = re^{i\theta}\, |\, r \in (0,r_0),\, \theta \in (0,\frac{4\pi}{4-d})\}$.

\[prop:hyperbolic-trajectory\] $C = \{(x + iy)^{-(4-d)/8}\, |\, xy = A,\, x \in I_x\}$ for some constant $A$ on some interval $I_x$.

Consider 