11. If the real and imaginary parts of $\frac{z}{20}$ and $\frac{20}{z}$ are all between 0 and 1, then the area of the figure formed by the points corresponding to such complex numbers $z$ is $\qquad$ .
11. $300-50 \pi \quad$ Let $z=r(\cos \theta+i \sin \theta)$

So $\left\{\begin{array}{l}00,(x-10)^{2}+y^{2}>10^{2}\end{array}\right.$.
Similarly, $(y-10)^{2}+x^{2}>10^{2}$
$$
S=20^{2}-10^{2}-2 \times \frac{1}{4} \times \pi \times 10^{2}=300-50 \pi
$$