$A$ is the region of the complex plane $\{z: z/4$ and $4/z'$ have real and imaginary part in $(0, 1)\}$, then $[p]$(where $p$ is the area of the region $A$ and $[.]$ denotes the greatest integer function) is
My attempt: Taking $z=x+iy$ For first condition, I get $x \in (0,4)$ and $y \in (0,4)$ which is a square of side $4$. For the second condition I get $\displaystyle \frac{4x}{x^2+y^2} \in (0,1)$ and $\displaystyle \frac{4y}{x^2+y^2} \in (0,1)$. I dont know how to graph this and proceed.
It helps to draw the region out. The square part is easy.
Now, you have $\displaystyle 0<\frac{4x}{x^2+y^2}<1$, which you can reciprocate to get $\displaystyle 1<\frac{x^2+y^2}{4x}<\infty$, or just $\displaystyle 1<\frac{x^2+y^2}{4x}$.
This becomes $\displaystyle 4x<x^2+y^2$, which rearranges to $x^2+y^2-4x>0$, and $(x-2)^2+y^2>2^2$.
Similarly, $\displaystyle 0<\frac{4y}{x^2+y^2}<1$ becomes $x^2+(y-2)^2>2^2$.
I assume you know how to draw both of those circles, and the diagram looks as follows.
For the black area, we take the original $16$ squares, subtract $4$ in the bottom left, and then subtract $\displaystyle \pi$ for the top-left and bottom-right quarter-circles. The area is then $12-2\pi$, and so $[p]=5$.