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 1 (plug and chug)

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

Solution 2

note: the "begin{align*}" environment is weird on aops, so if the two things that use it below are duplicated in front of each other, just reload the page or something.

$\textbf{Lemma.}$ Let $h(x)=\frac{ax+b}{cx+d}$ and $h\left (h(x)\right )=x$ for all such $x$ where $h(h(x))$ is defined. Then either $d=-a$ or $h(x)=x$ .

This lemma is one that is typically known by people who could make it this far into this AMC 12 anyway (USAMO qualifiers likely).

$\textit{Proof.}$ We plug in $h\left (h(x)\right )$ , and equating it to $x$ , we see \begin{align*} h\left (h(x)\right ) &=h\left (\frac{ax+b}{cx+d}\right ) \\ &= \frac{a\cdot \frac{ax+b}{cx+d}+b}{c\cdot \frac{ax+b}{cx+d}+d} \\ &= \frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} \\ &= \frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}=x. \end{align*} Then if we clear the denominator, it rearranges into $(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x\implies (ac+cd)x^2+(d^2-a^2)x-(ab+bd)=0.$ Since this equation is true for all $x$ (since now the fractions are gone), this implies that the coefficients of this quadratic are all equal to $0$ . Focus on the condition $d^2-a^2=0$ .

$\textbf{Case 1:}$ $d+a=0$

Then all the other coefficients automatically become zero, so $d+a=0\implies d=-a$ is a sufficient condition.

$\textbf{Case 2:}$ $d-a=0$ (and $d+a\neq 0$ )

Then in $ab+bd=0$ , we divide both sides by $a+d$ to see that $b=0$ . Similarly, in $ac+cd=0$ , we divide to also see that $c=0$ . Then since $d=a$ , $h(x)=\frac{ax+b}{cx+d}=\frac{ax+0}{0x+a}=x.$ Therefore, in this case, it is required that $h(x)=x$ , which now works. This proves our desired claim. $\qquad \qquad \square$

Note that we are asking when $g(g(z))=z$ . So either $g(z)=z$ , or in the expansion of $g(z)=f(f(z))$ , the two corresponding coefficients are negations of each other.

$\textbf{Case 1:}$ $g(z)=z$

Then $f(f(z))=z$ , so either $f(z)=z$ or the two coefficients are negative of each other. Clearly $f(z)\neq z$ , since the numerator and denominator are both degree $1$ polynomials. So we would require the coefficients, namely $1$ and $b$ , to be negatives of each other. Then $b=-1$ is a possibility here, giving $|b|=1$ .

$\textbf{Case 2:}$ The two coefficients are negative of each other.

It can be found that \begin{align*} f(f(x))&=f\left (\frac{z+a}{z+b}\right ) \\ &=\frac{\frac{z+a}{z+b}+a}{\frac{z+a}{z+b}+b} \\ &=\frac{z+a+a(z+b)}{z+a+b(z+b)} \\ &= \frac{(a+1)z+(ab+a)}{(b+1)z+(b^2+a)}.\end{align*} By our lemma, we would require $a+1=-(b^2+a)$ , which rearranges into $b^2=-1-2a$ . But we are given that $|a|=1$ . The triangle inequality for complex numbers states that $|a|+|b|\geq |a+b|$ , where equality happens when the arguments of $a$ and $b$ are equal. Here, we can see that $|-1-2a|\leq |-1|+|-2a|=1+2=3,$ but we can also find that $|-1-2a|+1=|-1-2a|+|1|\geq |-2a|=2\implies |-1-2a|\geq 1.$ Therefore, because $|b^2|=|-1-2a|$ , then $1\leq |b|^2\leq 3$ , giving us $1\leq |b|\leq \sqrt{3}$ .

When we combine the two cases, we find that the possible range of $|b|$ is $|b|\in [1,\sqrt{3}]$ , so the answer is $\boxed{\textbf{(C) }\sqrt{3}-1}$ .

~ethanzhang1001

Video Solution

https://youtu.be/FU18x_LsTeQ

~MathProblemSolvingSkills.com

Note

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

2011 AMC 12A (Problems • Answer Key • Resources)
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