Source: Oxford Exam $A2 \ 1999$
We want to construct a conformal map $F$ from the unit disc $\mathbb{D}=\{z:|z|<1\}$ to $\mathbb{C} \setminus S$ where $S$ is the half-line $\{x+i:x \in (-\infty,0] \}$ with the additional property that $F(0)=0$.
Here are my thoughts, can someone see if they are correct please:
- Use the Möbius transformation $f_1:z \mapsto \frac{1-z}{z+1}$ to send $\mathbb{D}$ to the right half plane.
- Use the map $f_2:z \mapsto z+ \alpha$ where $\alpha \in \mathbb{C}$ we will determine later. This simply takes the right half plane to the right half plane.
- Use the map $f_3:z \mapsto z^2$ to map the right half plane to $\mathbb{C}$ with a cut along the negative reals.
- Use the map $f_4:z \mapsto z+i$ to move the cut up and map to $\mathbb{C} \setminus S$.
Now the composition of these four maps is conformal so we certainly have a conformal map between the sets.
We now need $F(0)=f_4 \circ f_3 \circ f_2 \circ f_1(0)=0$.
$f_1(0)=1$, $f_2(1)=1+\alpha$, $f_3(1+\alpha)=\alpha^2+2\alpha+1$, $f_4(\alpha^2+2\alpha+1)=\alpha^2+2\alpha+1+i$
So we need to solve $\alpha^2+2\alpha+1+i=0$ and this is the $\alpha$ required to ensure $F(0)=0$.
We can find that $\alpha=\sqrt{-i}-1$.
Thus we are done.
Is this correct?
Best Answer
Your choice of $f_2$ is bad. We need a conformal map from the right-half plane to the right-half plane, that sends $f_1(0) = 1$ to $f_3^{-1}\circ f_4^{-1}(0) = f_3^{-1}(-i) = (1-i)/\sqrt 2$.
Your choice of a translation sends the right-half plane to one of its translate to the left, which is no longer the right-half plane, and then the square map messes up everything.
Since the automorphisms of the top-half plane are of the form $z \mapsto (az+b)/(cz+d)$ with $ad-bc > 0$, the conformal map we need is of the form $z \mapsto (az-ib)/(icz+d)$ with $ad-bc > 0$. For example, $f_2(z) = (z-i)/\sqrt 2 $ is one map that works, and it is very easy to see that it does map the right-half plane to itself.