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2013 AMC 12A 第 25 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

25.

定义函数 f:C→Cf : \mathbb{C} \to \mathbb{C} 为 f(z)=z2+iz+1f(z) = z^2 + iz + 1。有多少个复数 zz 满足 Im⁡(z)>0\operatorname{Im}(z) \gt 0，并且 f(z)f(z) 的实部和虚部都是绝对值至多为 1010 的整数？

Let f:C→Cf : \mathbb{C} \to \mathbb{C} be defined by f(z)=z2+iz+1.f(z) = z^2 + iz + 1. How many complex numbers zz are there such that Im⁡(z)>0\operatorname{Im}(z) \gt 0 and both the real and the imaginary parts of f(z)f(z) are integers with absolute value at most 10?10?

399399

401401

413413

431431

441441

答案：A
知识点：复数格点
难度评级：2790
小提示：

证明 ff 在上半平面上一一对应到其像，所以计数 zz 等价于计数有效的 f(z)f(z)

Show ff is one-to-one on the upper half-plane, so counting zz equals counting valid f(z)f(z)

大提示：

像集为 {w:Re⁡(w)<(Im⁡(w))2+1}\{w : \operatorname{Re}(w) \lt (\operatorname{Im}(w))^2 + 1\}；统计满足 ∣Re⁡∣,∣Im⁡∣≤10|\operatorname{Re}|, |\operatorname{Im}| \le 10 的格点

The image is {w:Re⁡(w)<(Im⁡(w))2+1};\{w : \operatorname{Re}(w) \lt (\operatorname{Im}(w))^2 + 1\}; count lattice points with ∣Re⁡∣,∣Im⁡∣≤10|\operatorname{Re}|, |\operatorname{Im}| \le 10

解答：

在上半平面 HH 上，若 f(z1)=f(z2)f(z_1) = f(z_2)，则 (z1−z2)(z1+z2+i)=0(z_1 - z_2)(z_1 + z_2 + i) = 0；因为 Im⁡(z1),Im⁡(z2)>0\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0，因子 z1+z2+i≠0z_1 + z_2 + i \ne 0，所以 ff 在 HH 上是一一的。

当 rr 取实数时，边界上的值 f(r)=r2+1+irf(r)=r^2+1+ir 描出抛物线 Re⁡(w)=(Im⁡(w))2+1\operatorname{Re}(w)=(\operatorname{Im}(w))^2+1。因为 f(i)=−1f(i)=-1 位于它的左侧，而 ff 在 HH 上连续且一一对应，所以它的像恰好由满足 Re⁡(w)<(Im⁡(w))2+1\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1 的所有 ww 组成。因此我们计数满足 a,b∈Za, b \in \mathbb{Z}、∣a∣,∣b∣≤10|a|, |b| \le 10 且 a<b2+1a \lt b^2 + 1 的 w=a+ibw = a + ib： ∣S∣=212−∑b=−33(10−b2)=441−42=399。 \begin{gathered} |S| = 21^2 \\ {}- \sum_{b=-3}^{3}(10 - b^2) \\ = 441 - 42 = 399 \end{gathered}\text{。}

因此，正确答案是 A。

On the upper half-plane H,H, if f(z1)=f(z2)f(z_1) = f(z_2) then (z1−z2)(z1+z2+i)=0;(z_1 - z_2)(z_1 + z_2 + i) = 0; since Im⁡(z1),Im⁡(z2)>0,\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0, the factor z1+z2+i≠0,z_1 + z_2 + i \ne 0, so ff is one-to-one on H.H.

For real r,r, the boundary values f(r)=r2+1+irf(r)=r^2+1+ir trace the parabola Re⁡(w)=(Im⁡(w))2+1.\operatorname{Re}(w)=(\operatorname{Im}(w))^2+1. Since f(i)=−1f(i)=-1 lies to its left and ff is continuous and one-to-one on H,H, its image consists precisely of the values ww satisfying Re⁡(w)<(Im⁡(w))2+1.\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1. Thus we count w=a+ibw = a + ib with a,b∈Z,a, b \in \mathbb{Z}, ∣a∣,∣b∣≤10,|a|, |b| \le 10, and a<b2+1:a \lt b^2 + 1: ∣S∣=212−∑b=−33(10−b2)=441−42=399. \begin{gathered} |S| = 21^2 \\ {}- \sum_{b=-3}^{3}(10 - b^2) \\ = 441 - 42 = 399. \end{gathered}

Thus, the correct answer is A.

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