fractional linear maps what is the fractional linear map maps the circles $|z-5|=3$ and $|z+5|=3$ to concentric circles? What is the general method to find such maps? What is the image of $|z+2|=2$ under the map: $z \to 1/\bar{z}$? Thank you very much.
 A: EDIT: Sorry for all the confusion, but I believe I now have a way to solve your first two problems (and my solution to your third question is not correct seeing as how you meant z is mapped to the complex conjugate of 1/z and I read z is mapped to just 1/z).  Well, I'll give you the general method without giving the details, since I don't have the time to carry them out myself.  A linear fractional transformation must preserve order, but there is a sort of exception in that you can wrap points around infinity.  For example, in this problem you have the following points on your circles, -8 < -2 < 2 < 8.  This order must be preserved by a LFT, except that you can cyclically permute as well.  Think of stretching out everything to the left until -8 passes past infinity and ends up on the right, so that we get $f(-2) < f(2) < f(8) < f(-8)$.  You MUST do something like this in this problem because right now neither circle is inside the other and you want one to get inside the other.  The only way to do this, while preserving the order of those points, is to do something like I just described.  So, you might start with a LFT that takes -8 to 1, -2 to -1, and 8 to 0.  This should get you two circles, with one inside the other, but not necessarily with the same centers and thus not concentric.  Aha, but now the link I put earlier actually applies.
Conformal mapping of a doubly connected domain onto an annulus
A: The map $z \to 1/\bar{z}$ is geometrically an inversion over the unit circle.  That is, $z$ is mapped to the point $z'$ such that $z'$ is on the ray $\overrightarrow{0z}$ and $|z|\cdot|z'|=1$.
In general, inversion over a circle centered at $c$ with radius $r$ maps $z$ to the point $z'$ such that $z'$ is on the ray $\overrightarrow{cz}$ and $|c-z|\cdot|c-z'|=r^2$.
It appears to me that any inversion over a circle centered at -4 or 4 will take your two given circles to a pair of concentric circles, but I haven't done much algebra to justify it.  If that's true, then $z\to\frac{1}{z+4}-4$ or $z\to\frac{1}{z-4}+4$ ought to do it (they aren't quite just the inversions, since they don't use conjugation, but the circles that result are the same because of symmetry over the real axis).
A: Per Numth's comment, let's send the negative circle to the unit circle with the positive circle to be concentric inside it. So we need to send:
$$\begin{array}{rcl}
-5-3 = -8 &\to& +1\\
-5+3 = -2 &\to& -1\\
+5-3 = +2 &\to& -x\\
+5+3 = +8 &\to& +x\end{array}$$
With a rational transformation given by $f(z) = (Az+B)/(z+C)$ we have four equations in the four unknowns $x,A,B,C$:
$$\begin{array}{rcl}
(-8A+B)/(-8+C) &=& +1\\
(-2A+B)/(-2+C) &=& -1\\
(+2A+B)/(+2+C) &=& -x\\
(+8A+B)/(+8+C) &=& +x\end{array}$$
All the circles are on the real axis, so all the variables are real.
