a differentiable curve in $D$ given by $z(t)$ for $t \in (a,b)$. Consider the function $g(z) = z^{(4-d)/8}$ that maps $D$ to $\tilde{D}$, the sector of the upper right quadrant given by $\{w = \rho e^{i\varphi}\, |\, \rho \in (0,\rho_0 = r_0^{(4-d)/8}),\, \varphi \in (0,\frac{\pi}{2})\}$ (see ). Let $w(t) = g(z(t))$ for $t \in (a,b)$. Taking the derivative of both sides, we have $$w'(t) = g'(z(t))z'(t)$$

since $g'(z) \neq 0$ in $D$, we have $$\begin{aligned}
 \label{eq:conformal-relation}
    \arg{(g'(z(t))z'(t))} &= \arg{g'(z(t))} + \arg{z'(t)}\\
                          &= \left(\frac{4-d}{8} - 1\right)\theta + \arg(z'(t))
  \end{aligned}$$

In the case that $z(t)$ is a solution of equation \[eq:ODE\], we have $$\begin{aligned}
    \arg{w'(t)} &= \frac{d\theta}{4} + \left(\frac{4-d}{8} - 1\right)\theta\\ 
                &= -\frac{(4-d)\theta}{8} = -\varphi
  \end{aligned}$$

Thus $w'(t) = \alpha(t)e^{-i\varphi}$ for some function $\alpha(t)$. Writing $w(t) = x(t) + iy(t)$, we have $x'(t) = \alpha(t)\cos(\varphi)$, $y'(t) = -\alpha(t)\sin(\varphi)$. Thus $$\frac{dy}{dx} = -\tan(\varphi) = -\frac{y}{x} \implies y = \frac{A}{x}$$

for some constant $A$. This equation describes the family of hyperbolas in the first quadrant with asymptotes at $\varphi = 0$ and $\varphi = \frac{\pi}{2}$. The curve from this family passing through the point $g(q) = \rho_q e^{i\varphi_q}$ is given by $\{x + iy\, |\, xy = A_q\}$, where $A_q = \rho_q^2\sin{\varphi_q}\cos{\varphi_q}$. The curve 