Question

Let $f(z) = \dfrac{z + a}{z + b}$ and $g(z) = f(f(z)),$ where $a$ and $b$ are complex numbers. Suppose that $|a| = 1$ and $g(g(z)) = z$ for all $z$ for which $g(g(z))$ is defined. What is the difference between the largest and smallest possible values of $|b|?$

$0$

$\sqrt{2} - 1$

$\sqrt{3} - 1$

$1$

$2$

Solution:

Represent $f$ by $M = \begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix}.$ Then $g(g(z)) = z$ says $f$ composed with itself four times is the identity, so $M^4$ is a scalar matrix.

This happens when the ratio of eigenvalues is a fourth root of unity. The order- $4$ case gives $(\operatorname{tr} M)^2 = 2\det M,$ i.e. $(1 + b)^2 = 2(b - a),$ which simplifies to $b^2 = -(1 + 2a).$ The order- $2$ case gives $b = -1.$

Then $|b|^2 = |1 + 2a|,$ and as $a$ runs over $|a| = 1,$ $|1 + 2a|$ ranges over $[1, 3],$ so $|b|$ ranges over $[1, \sqrt3]$ (the value $b = -1$ is included). The difference is $\sqrt3 - 1.$

Thus, the correct answer is C.

2011 AMC 12A Problem 23

Solution:

Problem 23 in Other Years