Mapping circles using Möbius transformations. I need some help with the following problem from Ahlfors' Complex Analysis.
Problem: Find a single Möbius transformation $\phi$ (that is, a map of the form $\phi(z) = \dfrac{az + b}{cz + d}$, where $a,b,c,d$ are complex numbers) that maps the circles $|z| = 1$ and $\left|z - \frac{1}{4}\right| = \frac{1}{4}$ to concentric circles. Infinite values are fair game; I'm working on the Riemann sphere $\mathbb{C}_{\infty}$.
What I tried: I know that the family of maps $z \mapsto \dfrac{z - a}{1 - \bar{a}z}$, where $a \in \{z : |z| < 1 \}$, preserves the unit disk, and therefore also the circle $|z| = 1$. I then used the following facts:
(i) $\left|z - \frac{1}{4}\right| = \frac{1}{4}$ should be mapped to some circle inside the unit disk centered at $0$. Call this circle $C$.
(ii) $0$ is in the border of $\left|z - \frac{1}{4}\right| \leq \frac{1}{4}$, so $\phi(0) = a$ should go to the border of $C$. This means $C$ has radius $|\phi(0)| = |a|$. By rotating $C$, we can assume $a$ is positive and real.
(iii) $1/2$ is also in the border of $\left|z - \frac{1}{4}\right| \leq \frac{1}{4}$, and therefore it should also be mapped to the border of $C$. By solving the equation
$$\left|\phi(1/2)\right| = |a|$$
for $a$, where $a$ is real and positive because of the above reasoning, I should get the desired $a$, but instead I get functions that don't map $|z - 1/4| = 1/4$ to some circle centered at $0$. And now I'm stuck.
Any help appreciated!
 A: I kept making mistakes, so here is a pretty complete thing. Instead of guessing on the coefficients the way I intended, it is better to work with the cross ratio, pages 78-80. We start with the ordered triple $-1, 0, \frac{1}{2}$ and send these to $1,0,\infty$ by a Möbius transformation, namely $\frac{3z}{2z-1}.$ This takes the quadruple $-1, 0, \frac{1}{2},1$ to  $1,0,\infty,3,$ so the cross ratio of the quadruple is $3.$
Next, we want a quadruple with some real $\alpha > 0$ given by $1,0, - \alpha, -1 - \alpha$ that also has cross ratio $3.$ That is, the Möbius trans given by $\frac{(1 + \alpha)z}{z + \alpha}$ must take $ -1 - \alpha$ to $3.$ So we get $\alpha = \sqrt 3 - 1$ and $1 + \alpha = \sqrt 3.$ So, $ \frac{\sqrt 3 \; z}{z + \sqrt 3 - 1}  $ takes $1,0,1 - \sqrt 3, - \sqrt 3$ to $1,0,\infty,3.$ Which is good. 
However, what we really want is to take $1,0,\infty,3$ to $1,0,1 - \sqrt 3, - \sqrt 3,$ which requires the inverse item, namely $\frac{(\sqrt 3 - 1) \; z}{-z + \sqrt 3}.$ 
Taking the composition and dividing all entries by $\sqrt 3,$ we get
$$ f(z) = \frac{(3 - \sqrt 3) \; z \; }{(2 - \sqrt 3) \; z - 1}  $$
This gives us $$f(-1) = 1, \; \; f(0) = 0, \; \; f(\frac{1}{2}) = 1 - \sqrt 3, \; \;  f(1) = - \sqrt 3.$$
Let's see, the determinant is negative, real entries, so it does fix the real axis but takes positive imaginary part to negative imaginary part. 
The common center is the real number $$ \frac{1 - \sqrt 3}{2}.  $$ If you wish, you may subtract that off to move the center(s) to the origin, then multiply by a real number to adjust the radii. Thus it is possible to get the unit disk back to where it was.
