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

Revision as of 14:12, 22 September 2013

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

Solution 1

Answer: (C) $\sqrt{3} - 1$

Lemma) if $f(z) = \frac{z + a}{z + b}$ , then $f(f(f(f(\frac{-a(1+b)}{a+1})))) = \frac{a ( 1 + b)}{ a+b^2}$

$f(f(f(f(\frac{-a(1+b)}{a+1})))) = f(f(f(-a))) = f(f(0)) = f(\frac{a}{b}) = \frac{a + ab}{a + b^2}$

The tedious algebra is left to the reader. (it is not bad at all)

Well, let us consider the cases where each of those step is definite ( $f(-b)$ is never evaluate).

So, we have $\frac{-a(1+b)}{a+1} \neq - b$ ,

$-a - ab \neq - ab - b$

$a \neq b$ --- (exception -> case 2)

$-a \neq -b$ --- (exception -> case 3)

$0 \neq -b$ --- (exception -> case 4)

$\frac{a}{b} \neq -b$

$a \neq -b ^2$ --- (exception -> case 5)

If it is not any of the above 5 cases, then

$\frac{-a(1+b)}{a+1} = \frac{a + ab}{a + b^2}$

if $a(1+b) \neq 0$ (--- exception -> case 6), then $-(a+b^2) = a+1$ , $b^2 = -2a - 1$ , $1 \le |b^2| \le 3$

Hence, it is possible maximum of $|b| = \sqrt{3}$ and minimum is 1.

2 possible combination of $(a,b)$ are $(1, 3i)$ and $(-1, 1)$ . Verification is left upto the reader. Right now, (C) is the most possible answer out of those 5.

Case 2) $a = b$ , then $|b| = 1$

Case 3) $|b|$ = 1, which is in the range.

Case 5) $b^2 = -a$ , hence $|b| = 1$

Case 6) Since $a \neq 0$ , $(1+b) = 0$ , $|b| = 1$

Case 4) $b = 0$ , this is quite an annoying special case. In this case, $f(z) = \frac{z+a}{z}$ , $f(0)$ is not define.

In this case, $f(f(f(f(\frac{a}{a-1})))) = f(f(f(a)))= f(f(2)) = f( \frac{2+a}{2}) = \frac{2+3a}{2+a}$ and $f(f(f(f(\frac{a}{a-2})))) = f(f(f(2a)))= f(f(\frac{3}{2})) = f( \frac{3+2a}{3}) = \frac{3+5a}{3+2a}$

Hence, $(2+a)a = (a-1)(2+3a)$ and $(a)(3+2a) = (a-2)(3+5a)$ . Once, you work out this system, you will get no solution with $|a| = 1$ .

Thus, answer is (C).

Solution 1

After a bit of tedius algebra (that isn't too bad, but a little lengthy) we obtain $h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}$ where $P=(a+1)^2+a(b+1)^2$ , $Q=a(b+1)(b^2+2a+1)$ , $R=(b+1)(b^2+2a+1)$ , and $S=a(b+1)^2+(a+b^2)^2$ . In order for $h(z)=z$ , we must have $R=0$ , $Q=0$ , and $P=S$ . The first implies $b=-1$ or $b^2+2a+1=0$ , the second implies $a=0$ , $b=-1$ , or $b^2+2a+1=0$ , and the third 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 $|b^2+1|=|-2a|=2$ so that $2=|b^2+1|\leq |b^2|+1$ and hence $1\leq|b|^2\Rightarrow1\leq |b|$ . On the other hand $2=|b^2+1|\geq|b^2|-1$ so that $|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}$ . Thus $1\leq |b|\leq \sqrt{3}$ . Hence in any case 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). Hence the answer is $\sqrt{3}-1$ .\\ \

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 == Solution ==
+===Solution 1===
 Answer: (C) <math>\sqrt{3} - 1</math>
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-SOLUTION 2:
+===Solution 1===

2011 AMC 12A (Problems • Answer Key • Resources)
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Difference between revisions of "2011 AMC 12A Problems/Problem 23"

Revision as of 14:12, 22 September 2013

Contents

Problem

Solution

Solution 1

Solution 1

See also