AMC12 2011 12A, Problem 23

Let $f(z)= \frac{z+a}{z+b}$ and $g(z)=f(f(z))$, where $a$ and $b$ are complex numbers. Suppose that $\left| a \right| = 1$ 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 $\left| b \right|$?

Solution: By algebraic manipulations, we obtain \[h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}\] where \[P=(a+1)^2+a(b+1)^2\] \[Q=a(b+1)(b^2+2a+1)\] \[R=(b+1)(b^2+2a+1)\] \[S=a(b+1)^2+(a+b^2)^2\] In order for $h(z)=z$, we must have $R=0$, $Q=0$, and $P=S$. $R=0$ implies $b=-1$ or $b^2+2a+1=0$. $Q=0$ implies $a=0$, $b=-1$, or $b^2+2a+1=0$. $P=S$ implies $b=\pm1$ or $b^2+2a+1=0$. Since $|a|=1\neq 0$, in order to satisfy all 3 conditions we must have either $b=\pm1$ or $b^2+2a+1=0$. In the first case $|b|=1$. For the latter case note that \[|b^2+1|=|-2a|=2\] \[2=|b^2+1|\leq |b^2|+1\] and hence, \[1\leq|b|^2\Rightarrow1\leq |b|\]. On the other hand, \[2=|b^2+1|\geq|b^2|-1\] so, \[|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}\]. Thus $1\leq |b|\leq \sqrt{3}$. Hence the maximum value for $|b|$ is $\sqrt{3}$ while the minimum is $1$ (which can be achieved for instance when $|a|=1,|b|=\sqrt{3}$ or $|a|=1,|b|=1$ respectively). Therefore the answer is $\fbox{\sqrt{3}-1}$.

Answer: \sqrt{3}-1