Question

Sea $z = \dfrac{1 + i}{\sqrt{2}}.$ ¿Cuánto vale

$\begin{aligned} &\left(z^{1^2} + z^{2^2} + z^{3^2} + \cdots + z^{12^2}\right) \\ &\quad {}\cdot \scriptsize \left(\dfrac{1}{z^{1^2}} + \dfrac{1}{z^{2^2}} + \dfrac{1}{z^{3^2}} + \cdots + \dfrac{1}{z^{12^2}}\right)? \end{aligned}$

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

$\begin{aligned} &\left(z^{1^2} + z^{2^2} + z^{3^2} + \cdots + z^{12^2}\right) \\ &\quad {}\cdot \scriptsize \left(\dfrac{1}{z^{1^2}} + \dfrac{1}{z^{2^2}} + \dfrac{1}{z^{3^2}} + \cdots + \dfrac{1}{z^{12^2}}\right)? \end{aligned}$

$18$

$72 - 36\sqrt{2}$

$36$

$72$

$72 + 36\sqrt{2}$

Solución:

Como $z = e^{i\pi/4},$ tenemos $z^{k^2} = e^{i\pi k^2/4},$ que depende solo de $k^2 \bmod 8.$

Para $k = 1$ hasta $12,$ el residuo $k^2 \bmod 8$ es $1$ (dando $z$ ) seis veces, $4$ (dando $-1$ ) tres veces, y $0$ (dando $1$ ) tres veces. Así que la primera suma es $6z - 3 + 3 = 6z.$

La segunda suma es análogamente $\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}.$ Su producto es $6z \cdot \dfrac{6}{z} = 36.$

Por lo tanto, la respuesta correcta es C.

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.

2019 AMC 12A Problema 21

Solución:

El Problema 21 en otros años