Region of complex numbers!

Let us denote $z$ as $cos(\theta) + j\cdot sin(\theta)$ and $a$ as $x + jy$

$| z^2 + az + 1| = 1$

$| cos(2\theta) + jsin(2\theta) + (xcos(\theta) - ysin(\theta)) + j(xsin(\theta) +ycos(\theta)) + 1| = 1$

$(cos(2\theta) + (xcos(\theta) - ysin(\theta)) + 1)^2 + (xsin(\theta) +ycos(\theta) + sin(2\theta))^2 = 1$

$(2cos^2(\theta) + xcos(\theta) - ysin(\theta) )^2 + (xsin(\theta) +ycos(\theta) + 2sin(\theta)cos(\theta))^2 = 1$

$x^2 + y^2 + 4cos^2(\theta) + 4xcos(\theta) = 1$

$(x+2cos(\theta))^2 + y^2 = 1$

Since $2cos(\theta)$ varies from $-2$ to $2$ , the figure is equivalent to a circle with center with $y$ coordinate equal to $0$ and $x$ coordinate varying from $-2$ to $2$ , which means a figure like:

Which means that the area will be a rectangle of base $4$ and height $2$ plus 2 semi-circles (1 circle) of radius equal to $1$ . Or:

$A = 4\cdot 2 + \pi\cdot 1^2$

$A = 8 + \pi = 11.1416...$

Then $\color{#3D99F6} \large \lfloor \frac{A}{2} \rfloor = \fbox{5}$

It seems that this question would have $n$ solvers $n$ likes