Difference between revisions of "2011 AMC 12A Problems/Problem 23"

Latest revision as of 14:49, 5 November 2023

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

https://youtu.be/FU18x_LsTeQ

~MathProblemSolvingSkills.com

Note

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

@@ Line 10: / Line 10: @@
 == Solution ==
-===Solution 1===
+By algebraic manipulations, we obtain
+<cmath>h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}</cmath> where
+<cmath>P=(a+1)^2+a(b+1)^2</cmath>
+<cmath>Q=a(b+1)(b^2+2a+1)</cmath>
+<cmath>R=(b+1)(b^2+2a+1)</cmath>
+<cmath>S=a(b+1)^2+(a+b^2)^2</cmath>
+In order for <math>h(z)=z</math>, we must have <math>R=0</math>, <math>Q=0</math>, and <math>P=S</math>.
-Lemma) if <math>f(z) = \frac{z + a}{z + b}</math>, then <math>f(f(f(f(\frac{-a(1+b)}{a+1})))) = \frac{a ( 1 + b)}{ a+b^2}</math>
+<math>R=0</math> implies <math>b=-1</math> or <math>b^2+2a+1=0</math>.
-<math>f(f(f(f(\frac{-a(1+b)}{a+1})))) = f(f(f(-a))) = f(f(0)) = f(\frac{a}{b}) = \frac{a + ab}{a + b^2}</math>
+<math>Q=0</math> implies <math>a=0</math>, <math>b=-1</math>, or <math>b^2+2a+1=0</math>.
+<math>P=S</math> implies <math>b=\pm1</math> or <math>b^2+2a+1=0</math>.
-Well, let us consider the cases where each of those step is definite (<math>f(-b)</math> is never evaluate).
+Since <math>|a|=1\neq 0</math>, in order to satisfy all 3 conditions we must have either <math>b=\pm1</math> or <math>b^2+2a+1=0</math>. In the first case <math>|b|=1</math>.
-So, we have <math> \frac{-a(1+b)}{a+1} \neq - b</math>,
+For the latter case note that
+<cmath>|b^2+1|=|-2a|=2</cmath>
-<math> -a - ab \neq - ab - b</math>
+<cmath>2=|b^2+1|\leq |b^2|+1</cmath>
+and hence,
-<math> a \neq b</math> --- (exception -> case 2)
+<cmath>1\leq|b|^2\Rightarrow1\leq |b|</cmath>.
+On the other hand,
-<math>-a \neq -b</math> --- (exception -> case 3)
+<cmath>2=|b^2+1|\geq|b^2|-1</cmath>
+so,
-<math>0 \neq -b</math> --- (exception -> case 4)
+<cmath>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</cmath>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence the maximum value for <math>|b|</math> is <math>\sqrt{3}</math> while the minimum is <math>1</math> (which can be achieved for instance when <math>|a|=1,|b|=\sqrt{3}</math> or <math>|a|=1,|b|=1</math> respectively). Therefore the answer is <math>\boxed{\textbf{(C)}\ \sqrt{3}-1}</math>.
-<math>\frac{a}{b} \neq -b</math>
-<math>a \neq -b ^2</math> --- (exception -> case 5)
-If it is not any of the above 5 cases, then
+==Shortcut==
+We only need <math>Q</math> in <math>f^4(z)=g^2(z)=\frac{Pz+\textcolor{red}{Q}}{Rz+S}</math>.
-<math> \frac{-a(1+b)}{a+1} = \frac{a + ab}{a + b^2}</math>
+Set <math>Q=0</math>: <math>a(b+1)\left(b^2+2a+1\right)=0</math>. Since <math>|a|=1</math>, either <math>b+1=0</math> or <math>b^2+2a+1=0</math>.
-if <math>a(1+b) \neq 0</math> (--- exception -> case 6), then <math>-(a+b^2) = a+1</math>, <math>b^2 = -2a - 1</math>, <math>1 \le |b^2| \le 3</math>
-Hence, it is possible maximum of <math>|b| = \sqrt{3}</math> and minimum is 1.
+<math>b+1=0\rightarrow b=-1</math> so <math>|b|=1</math>.
-possible combination of <math>(a,b)</math> are <math>(1, 3i)</math> and <math>(-1, 1)</math>.
+<math>b^2+2a+1=0\rightarrow b^2=-1-2a</math>. This is a circle in the complex plane centered at <math>(-1,0)</math> with radius <math>2</math> since <math>|a|=1</math>. The maximum distance from the origin is <math>3</math> at <math>(-3,0)</math> and similarly the minimum distance is <math>1</math> at <math>(1,0)</math>. So <math>1\le |b^2|\le 3\rightarrow 1\le |b|\le \sqrt{3}</math>.
-Case 2) <math>a = b</math>, then <math>|b| = 1</math>
-Case 3) <math>|b|</math> = 1, which is in the range.
+Both solutions give the same lower bound, <math>1</math>. So the range is <math>\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}</math>.
-Case 5) <math>b^2 = -a</math>, hence <math>|b| = 1</math>
-Case 6) Since <math>a \neq 0</math>, <math>(1+b) = 0</math>, <math>|b| = 1</math>
+== Video Solution ==
+https://youtu.be/FU18x_LsTeQ
-Case 4) If <math>b = 0</math>, <math>f(z) = \frac{z+a}{z}</math>, <math>f(0)</math> is not define.
+~MathProblemSolvingSkills.com
-In this case, <math>f(f(f(f(\frac{a}{a-1})))) = f(f(f(a)))= f(f(2)) = f( \frac{2+a}{2}) = \frac{2+3a}{2+a}</math> and <math>f(f(f(f(\frac{a}{a-2})))) = f(f(f(2a)))= f(f(\frac{3}{2})) = f( \frac{3+2a}{3}) = \frac{3+5a}{3+2a}</math>
+==Note==
+This problem is kinda similar to [[2002 AIME I Problems/Problem 12]]
-Hence, <math>(2+a)a = (a-1)(2+3a)</math> and <math>(a)(3+2a) = (a-2)(3+5a)</math>. Once, you work out this system, you will get no solution with <math>|a| = 1</math>.
-<math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>
-===Solution 2===
-By algebraic manipulations, we obtain
-<cmath>h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}</cmath> where <math>P=(a+1)^2+a(b+1)^2</math>, <math>Q=a(b+1)(b^2+2a+1)</math>, <math>R=(b+1)(b^2+2a+1)</math>, and <math>S=a(b+1)^2+(a+b^2)^2</math>. In order for <math>h(z)=z</math>, we must have <math>R=0</math>, <math>Q=0</math>, and <math>P=S</math>. The first implies <math>b=-1</math> or <math>b^2+2a+1=0</math>, the second implies <math>a=0</math>, <math>b=-1</math>, or <math>b^2+2a+1=0</math>, and the third implies <math>b=\pm1</math> or <math>b^2+2a+1=0</math>. Since <math>|a|=1\neq 0</math>, in order to satisfy all 3 conditions we must have either <math>b=1</math> or <math>b^2+2a+1=0</math>. In the first case <math>|b|=1</math>.
-For the latter case note that
-<cmath>|b^2+1|=|-2a|=2</cmath>
-<cmath>2=|b^2+1|\leq |b^2|+1</cmath>
-and hence,
-<cmath>1\leq|b|^2\Rightarrow1\leq |b|</cmath>.
-On the other hand,
-<cmath>2=|b^2+1|\geq|b^2|-1</cmath>
-so,
-<cmath>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</cmath>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence the maximum value for <math>|b|</math> is <math>\sqrt{3}</math> while the minimum is <math>1</math> (which can be achieved for instance when <math>|a|=1,|b|=\sqrt{3}</math> or <math>|a|=1,|b|=1</math> respectively). Therefore the answer is <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
 == See also ==

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

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