Let $D$ be a circular quadrilateral (that is a Jordan region whose boundary consists of 4 arcs of circles) whose interior angles are all equal to 0, the vertices lie on the unit circle, and $D$ is inside the unit disc. Suppose also that $D$ is symmetric with respect to the real and imaginary axes, and has one vertex $z_1=\exp(i\theta)$ where $\theta\in (0,\pi/2)$. Then this number $\theta$ determines such a $D$ completely.

Let $f$ be the inverse of the Riemann map, so that $f$ is the conformal map from the unit disc onto $D$, $f(0)=0$ and $f'(0)>0$.

Is it true that maximum $f'(0)$ is achieved when $\theta=\pi/4$ ?

There is a strong computer evidence for this, as well as the general considerations (where else can the maximum be?).

Let $D$ be a circular quadrilateral (that is a Jordan region whose boundary consists of 4 arcs of circles all orthogonal to the unit circle) whose interior angles are all equal to 0, the vertices lie on the unit circle, and $D$ is inside the unit disc. Suppose also that $D$ is symmetric with respect to the real and imaginary axes, and has one vertex $z_1=\exp(i\theta)$ where $\theta\in (0,\pi/2)$. Then this number $\theta$ determines such a $D$ completely.

Let $f$ be the inverse of the Riemann map, so that $f$ is the conformal map from the unit disc onto $D$, $f(0)=0$ and $f'(0)>0$.

Is it true that maximum $f'(0)$ is achieved when $\theta=\pi/4$ ?

There is a strong computer evidence for this, as well as the general considerations (where else can the maximum be?).