To find the position of the ant after 2020 moves is $(p, q)$, An ant is moving on the coordinate plane. Initially it was at (6, 0). Each move
of the ant consists of a counter-clockwise rotation of 60◦ about the origin fol-
lowed by a translation of 7 units in the positive x-direction. If the position of
the ant after 2020 moves is $(p, q)$, find the value of $p^2$ + $q^2$
For this problem,I just drew some and tried to figure out how this is going to go... But I have absolutely no idea what is the trajectory of the ant after 2020 moves.
I need help.
 A: Let $e^{i\pi/3}=:\omega$. Then we have to iterate the map
$$z\mapsto T(z):=\omega z+7\ .$$
This $T$ has a fixed point $a:={7\over1-\omega}$, and is in fact a $60^\circ$  rotation of the plane around $a$. We can therefore write
$$T(z)-a=\omega(z-a)\ ,$$
and this implies
$$T^n(z)-a=\omega^n(z-a)\ .$$
As $2020=6\cdot336+4$ and $\omega^4=-\omega$ we obtain the equation
$$T^{2020}(6)-a=-\omega(6-a)\ .$$
Solving this leads to
$$\bigl|T^{2020}(6)\bigr|^2=57\ .$$
A: In the complex plane, a rotation of $60^\circ$ is achieved by multiplying by $e^{i\frac{\pi}{3}} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$ For ease of notation, let $\alpha = e^{i\frac{\pi}{3}}$.
Let $z_0 = 6$ be the initial position of the ant and $z_k$ be the location of the ant after $k$ moves. Then
$$\begin{align}
z_1 &= z_0 \alpha + 7 \\
z_2 &= z_1 \alpha + 7 &&= z_0 \alpha^2 + 7 \alpha + 7 \\
z_3 &= z_2 \alpha + 7 &&= z_0 \alpha^3 + 7 \alpha^2 + 7 \alpha + 7 \\
& \;\; \vdots \\
z_n &= z_{n-1} \alpha + 7 &&= z_0 \alpha^n + 7 \alpha^{n-1} + 7 \alpha^{n-2} + 7 \alpha^{n-3} + \ldots + 7 \alpha + 7 \\
\\
& &&= z_0 \alpha + 7 \left[ \alpha^{n-1} + \alpha^{n-2} + \alpha^{n-3} + \ldots + \alpha + 1 \right] \\
\\
& &&= z_0 \alpha + 7 \left[ \frac{\alpha^n - 1}{\alpha - 1} \right]
\end{align}$$
In particular, noting that $\alpha^6 = 1$, so $\alpha^7 = \alpha$, we have that
$$\begin{align}
z_7 &= z_0 \alpha^7 + 7 \left[ \frac{\alpha^7 - 1}{\alpha - 1} \right] \\
\\
&= z_0 \alpha + 7 \left[ \frac{\alpha - 1}{\alpha - 1} \right] \\
\\
&= z_0 \alpha + 7 = z_1
\end{align}$$
This means that every six moves returns the ant to its starting position. Thus, since $2020 \equiv 4 \bmod{6}$, we have that $z_{2020} = z_4$.
We now use the fact that $\alpha^n = \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}$, and in particular, $\alpha^3 = -1$. Then
$$\begin{align}
z_{2020} = z_4 &= z_0 \alpha^4 + 7 ( \alpha^3 + \alpha^2 + \alpha+ 1 ) \\
\\
&= 6 \alpha^4 + 7 ( \alpha^2 + \alpha ) \\
\\
&= 6 ( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} ) + 7 \left[ ( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} ) + ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} ) \right] \\
\\
&= 6 ( \frac{-1}{2} - i \frac{\sqrt{3}}{2}) + 7 \left[ ( \frac{-1}{2} + i \frac{\sqrt{3}}{2} ) + ( \frac{1}{2} + i \frac{\sqrt{3}}{2} ) \right] \\
\\
&= -3 -3 \sqrt{3}i + 7 \sqrt{3}i \\
\\
&= -3 + 4 \sqrt{3}i
\end{align}$$
Thus $p = -3$ and $q = 4 \sqrt{3}$, so the answer is
$$ p^2 + q^2 = 9 + 48 = \boxed{57} $$
