2011 AMC 12A Problems/Problem 23

Revision as of 14:49, 5 November 2023 by Ryanjwang (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $\left| b \right|$?

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


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 $\boxed{\textbf{(C)}\ \sqrt{3}-1}$.


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




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

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png