jmc

Let $z=x+yi.$ Then $\frac{z}{40}=\frac{x}{40}+\frac{y}{40}\cdot i,$ so $$0\le \frac{x}{40}\le 1$$and $$0\le \frac{y}{40}\le 1.$$In other words $0\le x\le 40$ and $0\le y\le 40.$

Also, $$\frac{40}{\overline{z}}=\frac{40}{x-yi}=\frac{40(x+yi)}{x^2+y^2}=\frac{40x}{x^2+y^2}+\frac{40y}{x^2+y^2}\cdot i,$$so $$0\le \frac{40x}{x^2+y^2}\le 1$$and $$0\le \frac{40y}{x^2+y^2}\le 1.$$Since $x\ge 0,$ the first inequality is equivalent to $40x\le x^2+y^2.$ Completing the square, we get $$(x-20{)}^2+y^2\ge 20^2.$$Since $y\ge 0,$ the second inequality is equivalent to $40y\le x^2+y^2.$ Completing the square, we get $$x^2+(y-20{)}^2\ge 20^2.$$Thus, $A$ is the region inside the square with vertices $0,$ $40,$ $40+40i,$ and $40i,$ but outside the circle centered at $20$ with radius $20,$ and outside the circle centered at $20i$ with radius $20.$



To find the area of $A,$ we divide the square into four quadrants. The shaded area in the upper-left quadrant is $$20^2-\frac{1}{4}\cdot \pi \cdot 20^2=400-100\pi .$$The shaded area in the lower-right quadrant is also $400-100\pi .$ Thus, the area of $A$ is $$2(400-100\pi )+400=\boxed{1200-200\pi }.$$