# 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 $b=\pm1$ 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 $\boxed{\textbf{(C)}\ \sqrt{3}-1}$.

## Shortcut

We only need $Q$ in $f^4(z)=g^2(z)=\frac{Pz+\textcolor{red}{Q}}{Rz+S}$.

Set $Q=0$: $a(b+1)\left(b^2+2a+1\right)=0$. Since $|a|=1$, either $b+1=0$ or $b^2+2a+1=0$. $b+1=0\rightarrow b=-1$ so $|b|=1$. $b^2+2a+1=0\rightarrow b^2=-1-2a$. This is a circle in the complex plane centered at $(-1,0)$ with radius $2$ since $|a|=1$. The maximum distance from the origin is $3$ at $(-3,0)$ and similarly the minimum distance is $1$ at $(1,0)$. So $1\le |b^2|\le 3\rightarrow 1\le |b|\le \sqrt{3}$.

Both solutions give the same lower bound, $1$. So the range is $\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}$.

## Video Solution

~MathProblemSolvingSkills.com

## Note

This problem is kinda similar to 2002 AIME I Problems/Problem 12

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