2011 AMC 12A Problems/Problem 23

Problem

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 $\left| b \right|$ ?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \sqrt{2}-1 \qquad \textbf{(C)}\ \sqrt{3}-1 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

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 $b=1$ 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 $\boxed{\textbf{(C)}\ 2\sqrt{6}}$ .

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

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2011 AMC 12A Problems/Problem 23

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