AMC12 2002 A

AMC12 2002 A · Q24

AMC12 2002 A · Q24. It mainly tests Complex numbers (rare), Powers & residues.

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$.

求满足 $(a+bi)^{2002} = a-bi$ 的实数有序对 $(a,b)$ 的个数。

(A) 1001 1001

(B) 1002 1002

(D) 2002 2002

(E) 2004 2004

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$. Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$, while the magnitude of $a-bi$ is $s$. We get that $s^{2002}=s$, hence either $s=0$ or $s=1$. For $s=0$ we get a single solution $(a,b)=(0,0)$. Let's now assume that $s=1$. Multiply both sides by $a+bi$. The left hand side becomes $(a+bi)^{2003}$, the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$. Hence the solutions for this case are precisely all the $2003$rd complex roots of unity, and there are $2003$ of those. The total number of solutions is therefore $1+2003 = \boxed{2004}$.

设 $s=\sqrt{a^2+b^2}$ 为 $a+bi$ 的模。则 $(a+bi)^{2002}$ 的模为 $s^{2002}$，而 $a-bi$ 的模为 $s$。因此有 $s^{2002}=s$，从而 $s=0$ 或 $s=1$。当 $s=0$ 时，得到唯一解 $(a,b)=(0,0)$。现在假设 $s=1$。两边同乘以 $a+bi$。左边变为 $(a+bi)^{2003}$，右边变为 $(a-bi)(a+bi)=a^2 + b^2 = 1$。因此此情形下的解恰好是所有 $2003$ 次单位根，共有 $2003$ 个。因此解的总数为 $1+2003 = \boxed{2004}$。

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