While I tried to prove a problem,
Suppose that a linear fractional transformation carries one pair of concentric circles into another pair of concentric circles. Prove that the ratios of the radii must be the same.
Following is my try
Suppose concentric circles $z_1=\alpha+Re^{i\theta}$ and $z_2=\alpha+t Re^{i\theta}$, where $t>0,\ \theta\in [0,2\pi)$, through LFT $\frac{az+b}{cz+d}$ mapping to $$ \omega_1=\frac{a Re^{i\theta}+a\alpha+b}{cRe^{i\theta}+c\alpha +d};\\ \omega_2=\frac{a tRe^{i\theta}+a\alpha+b}{ctRe^{i\theta}+c\alpha +d}. $$
According to condition the mapping maps the concentric circles to concentric circles, so $\omega_1$ and $\omega_2$ are concentric. Suppose $\omega_1=\beta_1+R_1e^{i\theta}$ and $\omega_2=\beta_2+R_2e^{i\theta}$, solve the equations, we get the results as follows, $$ \beta_1\to \frac{(a \alpha+b) (\alpha c+d)-a c R^2}{(\alpha c+d)^2-c^2 R^2},R_1\to \frac{R (a d-b c)}{(\alpha c+d)^2-c^2 R^2}; $$ $$ \beta_2\to \frac{(a \alpha+b) (\alpha c+d)-a c R^2 t^2}{(\alpha c+d)^2-c^2 R^2 t^2},R_2\to \frac{R t (a d-b c)}{(\alpha c+d)^2-c^2 R^2 t^2}. $$ We simplify $\beta_1-\beta_2=0$, that is $$ c(c\alpha+ d)=0. $$ (i)$c=0$, $R_2/R_1=t$.
(ii)$c\alpha+d=0$, $R_2/R_1=1/t$.
Now my question is the first situation is normal and second one seems like the original proposition is not right, what's wrong?
Besides, I think my way to solve this question especially solving the equations are complicated, it there any other simpler solution? Thanks!
