function $\rho$, the family $(f^\Phi_t)_{t\ge0}$ is a Loewner chain. Using Courant’s Theorem, see e.g. [@Tsuji Theorem IX.14], it is possible to show that formula  defines a homeomorphism $F$ of $\C$. 1ex [**Construction 2.**]{} Let $f\in{\mathcal{S}}_k$. Denote by $g$ the conformal map of ${\mathbb{C}}\setminus\overline{\mathbb{D}}$ onto ${\mathbb{C}}\setminus\overline{f({\mathbb{D}})}$. For ${t\ge0}$, consider the conformal map $f_t^g$ of ${\mathbb{D}}$, ${f^g_t(0)=0}$, ${\omega(t)\big(f^g_t\big)'(0)>0}$, onto the Jordan domain bounded by ${g\big(\{z:|z|=\rho(t)\}\big)}$. For a suitable choice of the function $\rho$, the family $(f^g_t)_{t\ge0}$ is a Loewner chain and the map $F$ that it generates via  is a homeomorphism of $\C$. 1ex We will say that the function $\rho$ is *admissible* in Construction 1 or, respectively, in Construction 2, if the family $(f_t^\Phi)$, or respectively, the family $(f^g_t)$ is a Loewner chain. Note that admissibility of $\rho$ does not depend on the choice of $\omega$.

There exists a $(1/\sqrt{2})$-q.c. map $\Phi:{\mathbb{C}}\to{\mathbb{C}}$ with $f:=\Phi|_{\mathbb{D}}\in{\mathcal{S}}$ such that the homemorphisms $F$ defined in Constructions 1 and 2 are not quasiconformal for any admissible ${\rho:[0,+\infty)\to[1,+\infty)}$ and any locally absolutely continuous ${\omega:[0,+\infty)\to\partial{\mathbb{D}}}$.

Consider the function $$\label{EQ_f-example}
f(z):=\frac{2z(iz+\sqrt{1-z^2})^i}{1+\sqrt{1-z^2}}=\frac{2ze^{-\arcsin z}}{1+\sqrt{1-z^2}},\quad z\in{\mathbb{D}},$$ choosing the unique single-valued branch in ${\mathbb{D}}$ that belongs to ${\mathcal{S}}$. It is not difficult to check that $$\label{EQ_p}
p(z):=\frac{f(z)}{zf'(z)}=\sqrt{\frac{1+z}{1-z}}\quad\text{for all $z\in{\mathbb{D}}$}.$$ In particular, $\big|\arg p(z)\big|\le\pi/4$. 