Conformal mapping of the domain bounded by a line segment and a circular arc

Question

I am trying to construct a conformal map from the region $R$ which is the set of points in the complex plane bounded by the segment connecting $i$ and $1$ and the part of the unit circle in the first quadrant to the unit disk $\mathbb{D}$. I'm good with standard conformal map examples. So I just need a way to map $R$ to a familiar space. Thanks for any help.

Answer 1

So, starting with $f(z) = \frac{z-1}{z-i}$, where does the region go?

Near $1$ we have $f(z) \sim \frac{z-1}{1-i}$. Since the argument of $1-i$ is $-\pi/4$, ths map $f$ rotates the neighborhood of $1$ by $\pi/4$ counterclockwise. So, the circular arc (originally going straight up from $1$) is mapped to the line $\{(-1+i)t : t\ge 0\}$, as Mike Miller said. And the line segment, after rotation, goes to negative real axis. The image of the domain is between them.

Why is the map a surjection onto the region between these half-lines? Because it is a homeomorphism of the Riemann sphere onto itself, which implies that $\partial f(A) = f(\partial A)$ for every set $A$. So, the image of the domain is bounded by the image of the boundary.

Once you have the above, apply rotation (multiply by $e^{-3\pi/4}$), then power function, etc.