# 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(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

~ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 