complex-analysisconformal-geometrygeometry
I haven't found a similar question on here, though I suspect the question may be rather well-covered.
I want to find a conformal map from the vertical strip $\{z:-1<Re(z)<1\}$ onto the unit disc.
Under the exponential map the region is taken to the annulus with radii $e$ and $e^{-1}$, but I'm not sure how useful this will be.
Can anyone advise on what the conformal map may be?
Thanks.
Since you didn't show too many own thoughts, here are some hints only. By conformal I understand biholomorphic.
First take $f(z) = z^2$ to map the quadrant biholomorphically onto the upper half-plane, then compose with the Cayley transform $\kappa(z) = \frac{z-i}{z+i}$ to get $\kappa(f(z)) = \frac{z^2-i}{z^2+i}$.
Look at $\cos{(z)}$ and modify appropriately.
Impossible, since $G$ is not simply connected.
Map the region $G$ to the strip between two parallel lines using a Möbius transformation sending $1$ to infinity (e.g. using the inverse Cayley transformation). Then use the exponential function.
This should be enough to figure the solutions out.
For the precise relationship between "conformal" and "analytic", as well as for explanations on how to find such maps, I refer you to Ahlfors or (probably—I never really read it) Needham or any decent text on complex analysis treating conformal mapping.
The characterization of biholomorphisms between simply connected regions is essentially the content of the Riemann mapping theorem.
Sometimes biholomorhic mappings between polygonal regions and the unit disk can be computed via the Schwarz-Christoffel formula, but usually it leads to elliptic integrals that can't be solved explicitly in elementary terms.
Added:
Since the solution of 4. is a bit trickier, here's a rather detailed outline:
First note that $G$ is the region enclosed between the circles $\{|z| = 1\}$ and $\{|z - \frac{1}{2}| = \frac{1}{2}\}$. Applying the Möbius transformation (= the inverse Cayley transform) $\kappa^{-1}(z) = i\frac{1+z}{1-z}$ sends $G$ to the horizontal strip $\{0 \lt \operatorname{Im}{z} \lt 1\}$. To see this, look at this picture from Wikipedia illustrating the Cayley transform:
Finally, the exponential function $g(z) = \exp{(\pi z)}$ sends this strip to the upper half plane. Composing this with the Cayley transform we get the biholomorphic map $h = \kappa \circ g \circ \kappa^{-1}: G \to \mathbb{D}$.
Your map sending $(-1,0,1)$ to $(0,i,\infty)$ is given by $i = e^{i\pi/2}$ times the cross ratio $[z,0,-1,1] = \frac{z+1}{z - 1} : \frac{1}{-1}$, so it is $$\tilde{\phi}(z) = e^{i\pi/2}\cdot \frac{1+z}{1-z}.$$ It is easy to see geometrically (or by a direct calculation) that $\tilde\phi(L)$ is the quadrant $\{z:\,\operatorname{Im}{z} \gt |\operatorname{Re}{z}|\}$ since the circles $|z-i|=\sqrt{2}$ and $|z+i|=\sqrt{2}$ are sent to the lines $\{\operatorname{Im}{z} = \operatorname{Re}{z}\}$ and $\{\operatorname{Im}{z} = -\operatorname{Re}{z}\}$. Note that the angles at the vertices of the lune are equal to $\pi/2$.
It is more convenient to work with $\displaystyle\phi(z) = e^{i\pi/4}\cdot \frac{1+z}{1-z}$ which sends $L$ to the quadrant $\{z:\operatorname{Re}{z}, \, \operatorname{Im}{z}\gt 0\}$, then square to get to the upper half plane and apply the Cayley-transfom $\displaystyle\kappa(z) = \frac{z-i}{z+i}$ to get to the unit disk.
The solution to your problem then is $\kappa((\phi(z))^2)$, which you can compute yourself if needed.
The general procedure is succinctly explained in GEdgar's answer.
Best Answer
First apply $z \mapsto iz$ to map the strip onto the corresponding horizontal strip $-1 < \operatorname{Im} z < 1$.
Next step, apply $z \mapsto \exp(\frac{\pi z}2)$. This will give you the right half-plane $\operatorname{Re} z > 0$.
To finish off, the Möbius transformation $z \mapsto \dfrac{z-1}{z+1}$ takes the half-plane onto the unit disc.