Find Conformal Mapping Between Circles

Question

How would I go about finding the conformal mapping bewtween $|z| = 2$ and $|z-1| = 1$? I have done the following so far: Step 1: map regions between circles onto a strip: $$\frac{az + b}{cz +d} = \frac{iz}{z-2}$$ Step 2: I assume this is the step you rotate it into a horizontal strip, although I am not sure how. Step 3: Transform from strip to upper half plane using $e^z$ (should I use $e^z$ or $e^{iz}$).
Step 4: Transform from upper half plane onto unit disc using Möbius transformation: $$\frac{z-1}{z+1}$$ Then do $f_4 \circ f_3 \circ f_2 \circ f_1$. Is this correct? Am I missing anything? Please let me know! Thank you!!

Answer 1

In general, when dealing with circles/planes use Mobius Transforms. You simply need to carry three points on the starting circle to three points on the end circle, and you are guaranteed a unique transformation. Any good complex analysis text will outline the details and provide basic formulas to use.

In this case, we can save some time and simply consult a list of known transforms, e.g. here, and find that:

$f_1(z)=\frac{z-1}{z+1}$ carries the right half plane to the unit disk

$f_2(z) = z+1/2\,$ carries the right half plane to $\{z : \operatorname{Re}(z) > 1/2\}$

$f_3(z) = 1/z\,$ carries $\{z : \operatorname{Re}(z) > 1/2\}$ to the disk $|z-1| = 1$

In other words, $$\text{unit disk} \stackrel{f_1^{-1}}{\longrightarrow} \text{right half plane} \stackrel{f_2}{\longrightarrow} \{z : \operatorname{Re}(z) > 1/2\} \stackrel{f_3}{\longrightarrow} \{z : |z-1| = 1\}$$