Can I really get a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$? I should end up with $h(u) = \frac{iR + u + i \sqrt{R^2 − u^2}}{2}$, with $u = x_1 +ix_2$, but I cannot understand where this solution could come from.
I had the following ideas.
- Mapping the region outside the semicircle to the upper half plane. The region outside the semicircle, ($\mathbb{C} \setminus S_R$), can be mapped to the upper half plane using the function ($f(z) = \sqrt{R^2 – z^2}$). This function takes the points outside the semicircle and maps them to the upper half plane.
- Mapping the upper half plane to the unit disk. The upper half plane could be mapped to the unit disk using the Möbius transformation ($g(w) = \frac{w – i}{w + i}$). This function takes the points in the upper half plane and maps them to the inside of the unit disk.
- Scaling the unit disk to a disk of radius ($\frac{R}{\sqrt 2}$). The unit disk can be scaled to a disk of radius ($\frac{R}{\sqrt 2}$) using the function ($s(z) = \frac{R}{\sqrt{2}} z$). This function scales the points inside the unit disk to the inside of a disk of radius ($\frac{R}{\sqrt 2}$).
This does not give me the desired result, but I cannot see any other possible solution. What steps should I take? Any hints for arriving at the correct solution are appreciated.
Best Answer
The problem asks to show explicitly a conformal isomorphism between a semicircle exterior and a disc exterior. The main issue here is that while both are "one hole domains" bounded by circle arcs in the extended Mobius sense (so they are conformally isomorphic by Riemann mapping theorem extended version to finite hole domains since both are isomorphic to punctured discs), the angles do not match, so we need to do a boundary straightening which kills conformality there (conformal isomorphisms preserve angles inside the domain where they are so), hence such an isomorphism will not extend conformally over the boundary and we have to be careful.
For the easier problem of doing the semicircle inside to disc, the angles are $90$ vs $180$ so a squaring function will do, but here the angles are $270$ vs $180$ due to orientation so we need a power $2/3$ at some point to match the boundary angles.
First, we use a Mobius isomorphism (which is conformal in the extended plane) to send the semicircle (which has angle $90$ between the arc and the segment on the boundary) to two perpendicular rays and since we need to deal with the exterior, we need to keep track where $\infty$ goes.
As such a map we can use $e^{it}\frac{z+R}{z-R}$ which sends the intersection of the boundary components to $0, \infty$, hence sending the two to perpendicular rays through the origin and $\infty$ to $e^{it}$. For example, if $t=0$, checking where $0, -iR$ are sent we see that the segment is sent to the negative ray and the lower half circle to the upper imaginary ray, while infinity gets sent to $1$ so the inside of the semicircle goes to the third quadrant, and the outside to the rest, so the three quadrants remaining including the positive ray and the lower imaginary ray.
Now since we need to get to a disc, we need to straighten the angle and for this we use the conformal map $z \to z^{2/3}$ which is well defined on our three quadrant region; we can think of it as $re^{i\theta} \to r^{2/3}e^{2i \theta/3}$ where here $-\pi < \theta < \pi/2$; again for $t=0$ this sends the two boundary rays to the rays through angles $\pi/3, -2\pi/3$ (up and down respectively, so they continue each other forming a line), hence sends our three quadrant region to the right-hand plane bounded by this and keeps infinity at $1$ of course.
Now we can use a Mobius transform to send that to the unit disc and $1$ to $0$ and invert $z \to c/z$, or directly send the half plane to the unit disc exterior and $1 \to \infty$ and then dilate to get the radius we want, but an explicit form of this can be complicated so we can proceed in stages.
First we rotate the half plane to become the right half or the upper one, then use the standard Mobius transforms (eg $\frac{z-1}{z+1}$ for the right half plane $\Re z >0$) to send those to the unit disc and then use a disc automorphism $\frac{z-a}{1-\bar a z}$ to send $a$ the image of $1$ to $0$ and finally invert $z \to c/z$. To avoid this, one can try to see if there is a $t$ st when we do the straightening and rotating, the final image of $\infty$ is either $1$ or $i$ in the right half plane or upper half plane so then we can dispense with the disc automorphisms and use the standard Mobius maps.
So as a summation - the required map will be a composition of Mobius transform with $z^{2/3}$ with another Mobius transform and the reasons mentioned above, make this unavoidable for matching the boundary angles of the original domain to the ones of the target domain