Can the derivative of a conformal homeomorphism onto a bounded domain be unbounded? Consider $f$ being a holomorphic homeomorphism of the closed unit disk into the complex plane (ie. $f$ must be a homeomorphism - which also imply bijectivity to its image - of the closed unit disk, it must be holomorphic in the interior of the disk, and the image of the closed disk cannot contains infinity).


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*Must $f^{\prime}$ in the interior of the disk be bounded?

*(if the answer to the above is no) Can there be an infinite number of point on the unit circle such that $f^{\prime}$ is unbounded in every neighbourhood of that point?
 A: The derivative $f'$ can be unbounded. Consider a conformal map from the disk to the square. Four points of the boundary, say $\zeta_1,\dots,\zeta_4$, are mapped to vertices, where the interior angle is $\pi/2$. This means that near $\zeta_k$ the map behaves like $c(z-\zeta_k)^{1/2}$ and its derivative like $\frac{c}{2}(z-\zeta_k)^{-1/2}$. The Schwarz–Christoffel mapping makes this intuition precise. 
The map is a homeomorphism of closures, since the square is a Jordan domain.
To make this happen at infinitely many boundary points, we need a Jordan domain with infinitely many vertices of interior angle less than $\pi$. There are many ways to do this. One construction is to take the convex hull of the points $\{\exp(i/n):  n=0,1,2,3,4,\dots \}$; these points will be vertices. Another is to   let the domain be $\{x+iy:0<x<1, |x\sin(1/x)|<y<1\}$: the bottom boundary is shown below.

For maximal insanity, imagine conformally mapping the disk onto the domain bounded by Koch snowflake. And this example isn't even really bad: the interior of snowflake has   good geometry from the conformal viewpoint, but the boundary behavior is weird.
