2019 AMC 12A Problems/Problem 21

Problem

Let $z=\frac{1+i}{\sqrt{2}}.$ What is $(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}) \cdot (\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}})?$ $\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

Solution

Note that $z = \mathrm{cis }(45)$ .

Also note that $z^{k} = z^{k + 8}$ for all positive integers $k$ because of DeMoivre's Theorem. Therefore, we want to look at the exponents of each term mod 8.

$1^2, 5^2,$ and $9^2$ are all $1 \pmod{8}$

$2^2, 6^2,$ and $10^2$ are all $4 \pmod{8}$

$3^2, 7^2,$ and $11^2$ are all $1 \pmod{8}$

$4^2, 8^2,$ and $12^2$ are all $0 \pmod{8}$

Therefore,

$z^{1^2} = z^{5^2} = z^{9^2} = \mathrm{cis }(45)$

$z^{2^2} = z^{6^2} = z^{10^2} = \mathrm{cis }(180) = -1$

$z^{3^2} = z^{7^2} = z^{11^2} = \mathrm{cis }(45)$

$z^{4^2} = z^{8^2} = z^{12^2} = \mathrm{cis }(0) = 1$

The term $(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2})$ simplifies to $6\mathrm{cis }(45)$ , while the term $(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}})$ simplifies to $\frac{6}{\mathrm{cis }(45)}$ . Upon multiplication, the $\mathrm{cis }(45)$ cancels out and leaves us with $\boxed{\textbf{(C) }36}$ .

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2019 AMC 12A Problems/Problem 21

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