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

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

@@ Line 4: / Line 4: @@
 <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>
-== Solutions 1 ==
+== Solutions 1(Using Modular Functions) ==
 Note that <math>z = \mathrm{cis }(45^{\circ})</math>.
@@ Line 29: / Line 29: @@
 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>.
-== Solution 2 ==
+== Solution 2(Using Magnitudes and Conjugates to our Advantage) ==
 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>.
-== Solution 3 (bashing) ==
+== 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>\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)</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 \cdot 6 = \boxed{\textbf{(C) } 36}</math> as our answer.
-~ Baolan
 == Solution 4 (this is what people would write down on their scratch paper) ==
@@ Line 53: / Line 51: @@
 ~ Education, the Study of Everything
+== Solution 5 ==
+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 ===

2019 AMC 12A (Problems • Answer Key • Resources)
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