2019 AMC 12A Problems/Problem 21

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

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.

~ Baolan

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

Video Solution by Richard Rusczyk

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

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

Page

Toolbox

Search