AMC12 2019 A

AMC12 2019 A · Q21

AMC12 2019 A · Q21. It mainly tests Complex numbers (rare), Remainders & modular arithmetic.

Let $$ z = \frac{1+i}{\sqrt{2}}. $$ What is $$ (z^{12} + z^{22} + z^{32} + \dots + z^{122}) \cdot \left( \frac{1}{z^{12}} + \frac{1}{z^{22}} + \frac{1}{z^{32}} + \dots + \frac{1}{z^{122}} \right) ? $$

设 $$ z = \frac{1+i}{\sqrt{2}}. $$ 何为 $$ (z^{12} + z^{22} + z^{32} + \dots + z^{122}) \cdot \left( \frac{1}{z^{12}} + \frac{1}{z^{22}} + \frac{1}{z^{32}} + \dots + \frac{1}{z^{122}} \right) ? $$

(A) 18 18

(B) $72 - 36\sqrt{2}$ $72 - 36\sqrt{2}$

(D) 72 72

(E) $72 + 36\sqrt{2}$ $72 + 36\sqrt{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Note that $z^2=i$, so $z^4=-1$. For $k=1,2,\ldots,6$, $$z^{(2k)^2}=(-1)^{k^2} \quad \text{and} \quad z^{(2k-1)^2}=z^{4k^2-4k+1}=(-1)^{k(k-1)}z=z.$$ Because $$\sum_{k=1}^{6}(-1)^{k^2}=0,$$ the first factor is $6z=3\sqrt{2}+3\sqrt{2}i$. Because $|z|=1$, the second factor is the conjugate, $3\sqrt{2}-3\sqrt{2}i$. The requested product is therefore the square of the modulus of $3\sqrt{2}+3\sqrt{2}i$, $$\left(3\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2=18+18=36.$$

答案（C）：注意 $z^2=i$，所以 $z^4=-1$。对 $k=1,2,\ldots,6$， $$z^{(2k)^2}=(-1)^{k^2} \quad \text{且} \quad z^{(2k-1)^2}=z^{4k^2-4k+1}=(-1)^{k(k-1)}z=z.$$ 因为 $$\sum_{k=1}^{6}(-1)^{k^2}=0,$$ 第一个因子为 $6z=3\sqrt{2}+3\sqrt{2}i$。由于 $|z|=1$，第二个因子为其共轭 $3\sqrt{2}-3\sqrt{2}i$。因此所求乘积等于 $3\sqrt{2}+3\sqrt{2}i$ 的模的平方： $$\left(3\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2=18+18=36.$$

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