2019 AMC 12A Problem 21

Attempt Problem 21 of the 2019 AMC 12A below, then check your answer against the professionally curated solution 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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21.

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}

Answer: C

Concepts:roots of unity complex number modular arithmetic

Difficulty rating: 2160

Small hint:

z = e^{\frac{i\pi}{4}},

so

z^{k^2}

depends only on

k^2 \bmod 8

Big hint:

Over

k = 1, \ldots, 12,

the exponent

k^2 \bmod 8

is

1

six times,

4

three times, and

0

three times

Solution:

Since

z = e^{\frac{i\pi}{4}},

we have

z^{k^2} = e^{\frac{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.

Problem 21 in other years