Let z = \dfrac1 + i√2. What is (z^1^2 + z^2^2 + z^3^2 + \cdots + z^12^2) × (\dfrac1z^1^2 + \dfrac1z^2^2 + \dfrac1z^3^2 + \cdots + \dfrac1z^12^2)?

2019 AMC 12A Problem 21 — Solution

2019 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2019 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

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Concepts:roots of unity complex number modular arithmetic

Difficulty rating: 2160

21.

Let $z = \dfrac{1 + i}{\sqrt{2}}.$ What is

$\left(z^{1^2} + z^{2^2} + z^{3^2} + \cdots + z^{12^2}\right) \cdot \left(\dfrac{1}{z^{1^2}} + \dfrac{1}{z^{2^2}} + \dfrac{1}{z^{3^2}} + \cdots + \dfrac{1}{z^{12^2}}\right)?$

$18$

$72 - 36\sqrt{2}$

$36$

$72$

$72 + 36\sqrt{2}$

Solution:

Since $z = e^{i\pi/4},$ we have $z^{k^2} = e^{i\pi k^2/4},$ depending only on $k^2 \bmod 8.$

For $k = 1$ to $12,$ the residue $k^2 \bmod 8$ is $1$ (giving $z$ ) six times, $4$ (giving $-1$ ) three times, and $0$ (giving $1$ ) three times. So the first sum is $6z - 3 + 3 = 6z.$

The second sum is likewise $\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}.$ Their product is $6z \cdot \dfrac{6}{z} = 36.$

Thus, the correct answer is C.

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