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Difference between revisions of "2019 AMC 12A Problems/Problem 21"

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It is well known that if <math>|z|=1</math> then <math>\bar{z}=\frac{1}{z}</math>. Therefore, we have that the desired expression is equal to <cmath>\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)</cmath> We know that <math>z=e^{\frac{i\pi}{4}}</math> so <math>\bar{z}=e^{\frac{i7\pi}{4}}</math>. Then, by De Moivre's Theorem, we have <cmath>\left(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)</cmath> which can easily be computed as <math>\boxed{36}</math>.
 
It is well known that if <math>|z|=1</math> then <math>\bar{z}=\frac{1}{z}</math>. Therefore, we have that the desired expression is equal to <cmath>\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)</cmath> We know that <math>z=e^{\frac{i\pi}{4}}</math> so <math>\bar{z}=e^{\frac{i7\pi}{4}}</math>. Then, by De Moivre's Theorem, we have <cmath>\left(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)</cmath> which can easily be computed as <math>\boxed{36}</math>.
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==Solution 3 (bashing)==
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We first calculate <math>z^4 = -1</math>. After a bit of calculation for the other even powers of <math>z</math>, we realize that they add up to zero. Now we can simplify the expression to <math>(z^1^2 + z^3^2 + ... + z^11^2)(\frac{1}{z^1^2} + \frac{1}{z^3^2} + ... + \frac{1}{z^11^2})</math>. Then, we calculate the first few odd powers of <math>z</math>. We notice that <math>z^1 = z^9</math>, so the values cycle after every 8th power. Since all of the odd squares are a multiple of <math>8</math> away from each other, <math>z^1 = z^9 = z^25 = ... = z^121</math>, so <math>z^1^2 + z^3^2 + ... + z^11^2 = 6z^1^2</math>, and <math>\frac{1}{z^1^2} + \frac{1}{z^3^2} + ... + \frac{1}{z^11^2} = \frac{6}{z^1^2}</math>. When multiplied together, we get <math>6 * 6 = \boxed{\textbf{(C) } 36}</math> as our answer.
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~ Baolan
  
 
==See Also==
 
==See Also==

Revision as of 12:31, 3 February 2020

Problem

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

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 $\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)$ simplifies to $6\mathrm{cis }(45^{\circ})$, while the term $\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)$ 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 \[\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)\] 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 \[\left(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)\] which can easily be computed as $\boxed{36}$.


Solution 3 (bashing)

We first calculate $z^4 = -1$. After a bit of calculation for the other even powers of $z$, we realize that they add up to zero. Now we can simplify the expression to $(z^1^2 + z^3^2 + ... + z^11^2)(\frac{1}{z^1^2} + \frac{1}{z^3^2} + ... + \frac{1}{z^11^2})$ (Error compiling LaTeX. Unknown error_msg). Then, we calculate the first few odd powers of $z$. We notice that $z^1 = z^9$, so the values cycle after every 8th power. Since all of the odd squares are a multiple of $8$ away from each other, $z^1 = z^9 = z^25 = ... = z^121$, so $z^1^2 + z^3^2 + ... + z^11^2 = 6z^1^2$ (Error compiling LaTeX. Unknown error_msg), and $\frac{1}{z^1^2} + \frac{1}{z^3^2} + ... + \frac{1}{z^11^2} = \frac{6}{z^1^2}$ (Error compiling LaTeX. Unknown error_msg). When multiplied together, we get $6 * 6 = \boxed{\textbf{(C) } 36}$ as our answer.

~ Baolan

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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