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2019 AMC 12A Problems/Problem 21

Revision as of 21:03, 17 February 2019 by Sevenoptimus (talk | contribs) (Fixed formatting (and added degree symbols))

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 1

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

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

$1^2, 5^2,$ 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,

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

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

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

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

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

Solution 2

It is well known that if $|z|=1$ then $\bar{z}=\frac{1}{z}$. Therefore, we have that the desired expression is equal to \[(z^1+z^4+z^9+...+z^{144})(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144})\] We know that $z=e^{\frac{i\pi}{4}}$ so $\bar{z}=e^{\frac{i7\pi}{4}}$. Then, by De Moivre's Theorem, we have \[(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi})(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi})\] which can easily be computed as $\boxed{36}$.

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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