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Difference between revisions of "2019 AMC 12A Problems/Problem 21"
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==Problem==
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== Problem ==
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Let <cmath>z=\frac{1+i}{\sqrt{2}}.</cmath>What is <cmath>\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)?</cmath>
 
Let <cmath>z=\frac{1+i}{\sqrt{2}}.</cmath>What is <cmath>\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)?</cmath>
 
<math>\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2</math>
 
<math>\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2</math>
  
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==Solution 1==
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=== Solution 1 ===
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Note that <math>z = \mathrm{cis }(45^{\circ})</math>.
 
Note that <math>z = \mathrm{cis }(45^{\circ})</math>.
  
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The term thus <math>\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)</math> simplifies to <math>6\mathrm{cis }(45^{\circ})</math>, while the term <math>\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)</math> simplifies to <math>\frac{6}{\mathrm{cis }(45^{\circ})}</math>. Upon multiplication, the <math>\mathrm{cis }(45^{\circ})</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>.
 
The term thus <math>\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)</math> simplifies to <math>6\mathrm{cis }(45^{\circ})</math>, while the term <math>\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)</math> simplifies to <math>\frac{6}{\mathrm{cis }(45^{\circ})}</math>. Upon multiplication, the <math>\mathrm{cis }(45^{\circ})</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>.
  
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==Solution 2==
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=== Solution 2 ===
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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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==Solution 3 (bashing)==
 
 
We first calculate that <math>z^4 = -1</math>. After a bit of calculation for the other even powers of <math>z</math>, we realize that they cancel out 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.
 
We first calculate that <math>z^4 = -1</math>. After a bit of calculation for the other even powers of <math>z</math>, we realize that they cancel out 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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=== Video Solution by Richard Rusczyk ===
 
=== Video Solution by Richard Rusczyk ===
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https://artofproblemsolving.com/videos/amc/2019amc12a/493
 
https://artofproblemsolving.com/videos/amc/2019amc12a/493
  
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~ dolphin7
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== See Also ==
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==See Also==
 
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{{AMC12 box|year=2019|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2019|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:18, 19 October 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 that $z^4 = -1$. After a bit of calculation for the other even powers of $z$, we realize that they cancel out add up to zero. Now we can simplify the expression to $(z^{1^2} + z^{3^2} + ... + z^{11^2})(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}})$. 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 $z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}$, and $\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}} = \frac{6}{z^{1^2}}$. When multiplied together, we get $6 * 6 = \boxed{\textbf{(C) } 36}$ as our answer.

~ Baolan

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2019amc12a/493

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AMC 12 Problems and Solutions

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