AMC12 2011 A

AMC12 2011 A · Q23

AMC12 2011 A · Q23. It mainly tests Rational expressions, Complex numbers (rare).

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 $\left| b \right|$?

设 $f(z)= \frac{z+a}{z+b}$ 且 $g(z)=f(f(z))$，其中 $a$ 和 $b$ 为复数。已知 $\left| a \right| = 1$ 且对所有使 $g(g(z))$ 有定义的 $z$ 都有 $g(g(z))=z$。求 $\left| b \right|$ 的最大可能值与最小可能值之差。

(A) 0 0

(B) \sqrt{2}-1 \sqrt{2}-1

(D) 1 1

(E) 2 2

Answer

Correct choice: (C)

正确答案：（C）

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

通过代数运算可得 \[h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}\] 其中 \[P=(a+1)^2+a(b+1)^2\] \[Q=a(b+1)(b^2+2a+1)\] \[R=(b+1)(b^2+2a+1)\] \[S=a(b+1)^2+(a+b^2)^2\] 要使 $h(z)=z$，必须有 $R=0$、$Q=0$ 且 $P=S$。 $R=0$ 推出 $b=-1$ 或 $b^2+2a+1=0$。 $Q=0$ 推出 $a=0$、$b=-1$ 或 $b^2+2a+1=0$。 $P=S$ 推出 $b=\pm1$ 或 $b^2+2a+1=0$。由于 $|a|=1\neq 0$，要同时满足这 3 个条件，必须有 $b=\pm1$ 或 $b^2+2a+1=0$。第一种情况下 $|b|=1$。对第二种情况，注意 \[|b^2+1|=|-2a|=2\] \[2=|b^2+1|\leq |b^2|+1\] 因此 \[1\leq|b|^2\Rightarrow1\leq |b|\]. 另一方面， \[2=|b^2+1|\geq|b^2|-1\] 所以 \[|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}\]. 因此 $1\leq |b|\leq \sqrt{3}$。故 $|b|$ 的最大值为 $\sqrt{3}$，最小值为 $1$（例如分别可在 $|a|=1,|b|=\sqrt{3}$ 或 $|a|=1,|b|=1$ 时取得）。因此答案为 $\boxed{\textbf{(C)}\ \sqrt{3}-1}$。

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