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

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~ Education, the Study of Everything
 
~ Education, the Study of Everything
  
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== Solution 5 ==
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We notice that <math>z = e^{\pi i/4}</math> and <math>\frac1z = \overline{z} = e^{-\pi i/4}</math>. We then see that: <cmath>z^4 = e^{\pi i} = -1.</cmath> This means that: <cmath>z^{(2n)^2} = z^{4n^2} = (-1)^{n^2}.</cmath> In the first summation, there are <math>6</math> even exponents, and the <math>-1</math>'s will cancel among those. This means that: <cmath>\sum_{k=1}^{12} z^{k^2} = \sum_{m=0}^5 z^{(2m+1)^2}.</cmath> We can simplify <math>z^{(2m+1)^2}</math> to get: <cmath>z^{(2m+1)^2} = z^{4m^2} \cdot z^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot z.</cmath> We know that <math>m^2</math> and <math>m</math> will have the same parity so the <math>-1</math>'s multiply into a <math>1</math>, so what we get left is: <cmath>\sum_{m=0}^5 z^{(2m+1)^2} = \sum_{m=0}^5 z = 6z.</cmath> Now, for the conjugates, we notice that: <cmath>\overline{z}^4 = \overline{z^4} = -1.</cmath> This means that: <cmath>\overline{z}^{(2n)^2} = (-1)^{n^2}.</cmath> Therefore: <cmath>\sum_{k=1}^{12}\overline{z}^{k^2} = \sum_{m=0}^5 \overline{z}^{(2m+1)^2}.</cmath> Now, we see that: <cmath>\overline{z}^{(2m+1)^2} = \overline{z}^{4m^2} \cdot \overline{z}^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot \overline{z}.</cmath> Again, <math>m^2</math> and <math>m</math> have the same parity, so the <math>-1</math>'s multiply into a <math>1</math>, leaving us with <math>\overline{z}</math>. Therefore: <cmath>\sum_{m=0}^5 \overline{z}^{(2m+1)^2} = \sum_{m=0}^5 \overline{z} = 6\overline{z}.</cmath> Now, what we have is: <cmath>6z \cdot 6\overline{z} = 6e^{\pi i/4} \cdot 6e^{-\pi i/4} = 6 \cdot 6 = \boxed{\textbf{(C) }36}</cmath>
  
 +
~ ap246
  
 
=== Video Solution by Richard Rusczyk ===
 
=== Video Solution by Richard Rusczyk ===

Latest revision as of 11:36, 3 December 2023

Problem

Let \[z=\frac{1+i}{\sqrt{2}}.\]What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]

$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$

Solutions 1(Using Modular Functions)

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(Using Magnitudes and Conjugates to our Advantage)

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 $\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)$. 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 $6 \cdot 6 = \boxed{\textbf{(C) } 36}$ as our answer.

Solution 4 (this is what people would write down on their scratch paper)

$z=\mathrm{cis }(\pi/4)$

Perfect squares mod 8: $1,4,1,0,1,4,1,0,1,4,1,0$

$1/z=\overline{z}=\mathrm{cis }(7\pi/4)$

$6\mathrm{cis }(\pi/4)\cdot 6\mathrm{cis }(7\pi/4)=\boxed{36}$

~ MathIsFun286

Video Solution1

https://youtu.be/58pxV5Lkiks

~ Education, the Study of Everything

Solution 5

We notice that $z = e^{\pi i/4}$ and $\frac1z = \overline{z} = e^{-\pi i/4}$. We then see that: \[z^4 = e^{\pi i} = -1.\] This means that: \[z^{(2n)^2} = z^{4n^2} = (-1)^{n^2}.\] In the first summation, there are $6$ even exponents, and the $-1$'s will cancel among those. This means that: \[\sum_{k=1}^{12} z^{k^2} = \sum_{m=0}^5 z^{(2m+1)^2}.\] We can simplify $z^{(2m+1)^2}$ to get: \[z^{(2m+1)^2} = z^{4m^2} \cdot z^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot z.\] We know that $m^2$ and $m$ will have the same parity so the $-1$'s multiply into a $1$, so what we get left is: \[\sum_{m=0}^5 z^{(2m+1)^2} = \sum_{m=0}^5 z = 6z.\] Now, for the conjugates, we notice that: \[\overline{z}^4 = \overline{z^4} = -1.\] This means that: \[\overline{z}^{(2n)^2} = (-1)^{n^2}.\] Therefore: \[\sum_{k=1}^{12}\overline{z}^{k^2} = \sum_{m=0}^5 \overline{z}^{(2m+1)^2}.\] Now, we see that: \[\overline{z}^{(2m+1)^2} = \overline{z}^{4m^2} \cdot \overline{z}^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot \overline{z}.\] Again, $m^2$ and $m$ have the same parity, so the $-1$'s multiply into a $1$, leaving us with $\overline{z}$. Therefore: \[\sum_{m=0}^5 \overline{z}^{(2m+1)^2} = \sum_{m=0}^5 \overline{z} = 6\overline{z}.\] Now, what we have is: \[6z \cdot 6\overline{z} = 6e^{\pi i/4} \cdot 6e^{-\pi i/4} = 6 \cdot 6 = \boxed{\textbf{(C) }36}\]

~ ap246

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
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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