Let $z$ be a complex number. The number $1$ is written on a board. You perform a series of moves. When can you make terms tend to $0$?

The region $A$ looks like a 4-leaf clover, and has area $$2 + \pi = 8\int_0^{\pi/4} \cos^2(\pi/4 - \theta) d\theta.$$ A picture of the region is at the bottom of the answer. A nice idea pointed out in a comment below is that one can also compute the area without an integral by dissecting the region into a square of side length $\sqrt{2}$ and two circles of radius $r = 1/\sqrt{2}$.

Here's an outline of how to determine the region $A$. Polar coordinates are well-suited for getting our hands on the complex multiplication option in the game, and we work mostly in them. Writing every detail would be rather lengthy — I'm happy to explain any of the steps that are unclear.

• $z$ is in $A$ if and only if there is a point $w$ on the boundary of the unit square so $\max(|\rm{Re}(wz)|, |\rm{Im}(wz)|) < 1.$ To see why this suffices, adjust $1$ to $w$ with move 2, apply move $1$, and adjust back to a positive real number $c < 1$. Repeating $n$ times yields the real number $c^n \to 0$.
• Now try to find all $z$ in the region of the form $re^{i\theta}$ for a fixed $\theta$, using the condition above.
• We may as well work with $0 \leq \theta \leq \pi/4$ and $w$ with $\rm{Re}(w) = 1, |\rm{Im}(w)| \leq 1$. (That is $w$ along the right side of the unit square). The rest follows by symmetry, as $\max(|\rm{Re}(\zeta)|, |\rm{Im}(\zeta)|)$ is invariant under conjugation of $\zeta$ and multiplication by $i$.
• The relevant side of the square is parametrized as $\sec(\nu) e^{i\nu}$ for $-\pi/4 \leq \nu \leq \pi/4$, so that the relevant points $zw$ are given by $$r \sec(\nu)e^{i(\theta+\nu)} = r\sec(\nu) \cos(\theta+\nu) + ir\sec(\nu) \sin(\theta + \nu).$$
• Analysis of critical points (where derivatives of either are $0$, or the two are equal, or endpoints) of $\max(|\rm{Re}(r \sec(\nu)e^{i(\theta+\nu)})|, |\rm{Im}(r \sec(\nu)e^{i(\theta+\nu)})|)$ shows that the minimum occurs at $\nu = \pi/4 - \theta$. There, we have $e^{i\pi/4} = \cos(\pi/4) + i \sin(\pi/4) = 1/\sqrt{2}+i/\sqrt{2}$ so $$ \max(|\rm{Re}(r \sec(\pi/4 - \theta)e^{i\pi/4})|, |\rm{Im}(r \sec(\pi/4 - \theta)e^{i\pi/4})|) = \frac{1}{\sqrt 2}r \sec(\pi/4 -\theta) .$$
• So $re^{i\theta}$ is in the region if and only if $\sqrt{2}\cos(\pi/4 - \theta) > r$ for $0 \leq \theta \leq \pi/4$. Symmetry considerations above together with this yield the polar integral for area above.

The region is bounded by the below curve, plotted in Desmos.