Find a Fractional Linear Transformation that maps the region between $\{|z+1| = 1\}$ and $\{|z|=2\}$ to the region between $Im(z) = 1$ and $Im(z) = 2$

Question

I'm trying to find a Fractional Linear Transformation (if one exists) that maps the region between the circles $\{|z+1| = 1\}$ and $\{|z|=2\}$ to the region between the horizontal lines $\operatorname {Im}(z) = 1$ and $\operatorname{Im}(z) = 2$.

I know that since $-2$ is a point of both circles, I need to find a transformation which has a pole at $-2$, so the denominator of the transformation should be $z+2$ so that both circles get mapped to parallel lines.

From here, I'm getting really stuck. I know how to find a transformation between two chosen triples, but I'm having trouble figuring out what other two points I need for those triples besides $-2$ and $\infty$.

Any hint or link to a place where I could do some more reading would be really helpful! Thank you!

Answer 1

Lets pause for the second after the sentence (if one exists). Go find the statement of Riemann mapping theorem (more general result would be Uniformization theorem theorem). Make sure you convince yourself such mapping exists.

Next step is to find the actual map. That is actually a harder problem. Linear fractional transformation are so interesting that you one can write a whole book on them. For example this is a good one

http://www.amazon.com/Fuchsian-Groups-Chicago-Lectures-Mathematics/dp/0226425835

You problem is kind easy and playing with composition of the generators of the Fuchsian it will be easy to solve.

But you are raising an interesting question. Can I find algorithmic way to find actual open mapping between to regions. The answer is yes and relies on the discreat version of Riemann mapping which is proved by Denis Sullivan but the place to start learning that business is from is Ken Stephens

http://www.math.utk.edu/~kens/

So a good semester long project would be to write computer program which will solve the above HW question for you.

Answer 2

Taking two triples $(-2, 0, 2)$ and $(\infty, \alpha , \beta )$, where $\alpha $ and $\beta $ should be on the lines $\operatorname{Im}\,w=1$ and $\operatorname{Im}\,w=2$ respectively(, or $\operatorname{Im}\,w=2$ and $\operatorname{Im}\,w=1$ respectively), we try to determine $$f(z)=\frac{az+b}{z+2}$$ so that $f$ maps the region between the circles $|z+1|=1$ and $|z|=2$ to the region between the horizontal lines $\operatorname{Im}\,w=1$ and $\operatorname{Im}\,w=2$.

Example 1. Let $\alpha =2i, \beta =i$. Then $$f(z)=\frac{4i}{z+2}.$$ Succesful! See Fig.1 below.

Example 2. Let $\alpha =2+2i, \beta =1+i$. Then $f(z)=\frac{4(1+i)}{z+2}$.
Failure! See Fig.2 below. Why ?

Since the line $l$ from $z=0$ to $z=2$ is perpendicular to the circles $|z+1|=1$ and $|z|=2$ the image $f(l)$ should be perpendicular to the lines $\operatorname{Im}\,w=1$ and $\operatorname{Im}\,w=2$ by the conformality of $f$. Thus the image $f(l)$ should be a vertical line and hence we must take $\alpha $ and $\beta $ so that their real parts are identical. In example 2 we mistaked this.

Conversely every pair of $\alpha $ and $\beta $ having identical real parts produces good results. For example, $\alpha =3+i$ and $\beta =3+2i$ leads to $f(z)=\frac{(3+3i)z+6+2i}{z+2},$ which is an appropriate linear transformation.