Therefore, by a result of Betker [@Betker p.110], see also [@Ikkei-LM §5.1], $f$ can be extended to a $(1/\sqrt2)$-q.c. automorphism $\Phi:{\mathbb{C}}\to{\mathbb{C}}$ as follows.

The image $f({\mathbb{D}})$ is a starlike Jordan domain symmetric w.r.t. ${\mathbb{R}}$ and bounded by two segments of logarithmic spirals. Namely, $\partial f({\mathbb{D}})={\{2\exp(-\pi/2+|\theta|+i\theta)\colon\theta\in[-\pi,\pi]\}}$. It follows that for any $z\in{\mathbb{D}}\setminus\{0\}$ the intersection of $\partial f({\mathbb{D}})$ and $\{tf(z):t\ge0\}$ consists of one point $\zeta(z)$, with $r(z):=|\zeta(z)|=2\exp\!\big(|\!{\mathop{\operator@font Arg}\nolimits}f(z)|-\pi/2\big)$, where ${\mathop{\operator@font Arg}\nolimits}w$ stands for the value of $\arg w$ that belongs to $(-\pi, \pi]$.

Betker’s q.c.-extension of $f$, see e.g. [@Ikkei-LM eq.(5.6) with $\lambda:=0$], is given by $$\Phi(z):=\frac{~r(1/\bar z)^2}{\hphantom{m}\vphantom{\int_0^1}\overline{f(1/\bar z)}\hphantom{m}}=\frac{4e^{-\pi}}{f(1/z)}\left(\frac{f(1/\bar z)}{|f(1/\bar z)|}\right)^{\! -2i\,\eta(z)}\!\!=~\frac{4e^{-\pi}}{f(1/z)}\left(\frac{f(1/z)}{f(1/\bar z)}\right)^{\! i\,\eta(z)}$$ for all $z\in\C\setminus{\mathbb{D}}$, where $\eta(z):={\mathop{\mathrm{sgn}}}\Im z$. Simple calculations give $$\frac{\Phi'_z(z)}{\Phi(z)}=\frac{1-i\eta(z)}{z^2}\frac{f'(1/z)}{f(1/z)},\quad \frac{\Phi'_{\bar z}(z)}{\Phi(z)}=\frac{i\eta(z)}{\bar{z}^2}\frac{f'(1/\bar z)}{f(1/\bar z)},\qquad |z|>1,~z\not\in{\mathbb{R}}.$$ Using the above formulas we see that for any $r>1$ the boundary of $D_r:=\Phi(r{\mathbb{D}})$ consists of two real-analytic arcs with common end-points at $\Phi(\pm r)$, where they form angle of magnitude $2{\mathop{\operator@font arctg}\nolimits}(1/2)<\pi/2$. The angle at $\Phi(-r)$ is internal w.r.t. $D_r$. It follows that conformal mappings of ${\mathbb{D}}$ onto $D_r$ do not belong to the Hardy space $H^2({\mathbb{D}})$. Therefore, by the main result of [@Betker_Hardy], there is no Loewner chain with image domains $D_r$ that defines a q.c.-extension via . Therefore, the homeomorphism 