AMC12 2008 B

AMC12 2008 B · Q19

AMC12 2008 B · Q19. It mainly tests Absolute value, Inequalities (AM-GM etc. basic).

A function $f$ is defined by $f(z) = (4 + i) z^2 + \alpha z + \gamma$ for all complex numbers $z$, where $\alpha$ and $\gamma$ are complex numbers and $i^2 = - 1$. Suppose that $f(1)$ and $f(i)$ are both real. What is the smallest possible value of $| \alpha | + |\gamma |$ ?

函数 $f$ 定义为对所有复数 $z$，$f(z) = (4 + i) z^2 + \alpha z + \gamma$，其中 $\alpha$ 和 $\gamma$ 为复数，且 $i^2 = - 1$。已知 $f(1)$ 和 $f(i)$ 都是实数。$| \alpha | + |\gamma |$ 的最小可能值是多少？

(A) 1 1

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

(D) 2\sqrt{2} 2\sqrt{2}

(E) 4 4

Answer

Correct choice: (B)

正确答案：（B）

Solution

We need only concern ourselves with the imaginary portions of $f(1)$ and $f(i)$ (both of which must be 0). These are: \begin{align*} \text{Im}(f(1)) & = i+i\text{Im}(\alpha)+i\text{Im}(\gamma) \\ \text{Im}(f(i)) & = -i+i\text{Re}(\alpha)+i\text{Im}(\gamma) \end{align*} Let $p=\text{Im}(\gamma)$ and $q=\text{Re}{(\gamma)},$ then we know $\text{Im}(\alpha)=-p-1$ and $\text{Re}(\alpha)=1-p.$ Therefore \[|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},\] which reaches its minimum $\sqrt 2$ when $p=q=0$ by the Trivial Inequality. Thus, the answer is $\boxed B.$

我们只需关注 $f(1)$ 和 $f(i)$ 的虚部（都必须为 0）。它们为： \begin{align*} \text{Im}(f(1)) & = i+i\text{Im}(\alpha)+i\text{Im}(\gamma) \\ \text{Im}(f(i)) & = -i+i\text{Re}(\alpha)+i\text{Im}(\gamma) \end{align*} 令 $p=\text{Im}(\gamma)$ 且 $q=\text{Re}{(\gamma)},$ 则有 $\text{Im}(\alpha)=-p-1$ 且 $\text{Re}(\alpha)=1-p.$ 因此 \[|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},\] 由 Trivial Inequality 可知当 $p=q=0$ 时取到最小值 $\sqrt 2$。因此答案为 $\boxed B.$

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