Question 106, Let the complex number $\mathrm{z}$ be such that the real and imaginary parts of 1 are both greater than 1. If the set of points corresponding to $\mathrm{z}$ on the complex plane is the figure $C$, find the area of figure $C$.

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Question 106, Solution: Let $z=x+y i$, then $\frac{1}{z}=\frac{x}{x^{2}+y^{2}}-\frac{y}{x^{2}+y^{2}} i$. Therefore, $\left\{\begin{array}{l}\frac{x}{x^{2}+y^{2}}>1 \\ -\frac{y}{x^{2}+y^{2}}>1\end{array}\right.$, rearranging gives $\left\{\begin{array}{l}\left(x-\frac{1}{2}\right)^{2}+y^{2}=\frac{1}{4} \\ x^{2}+\left(y+\frac{1}{2}\right)^{2}=\frac{1}{4}\end{array}\right.$ Thus, the shaded area in figure C is as shown, with an area of $\frac{\pi-2}{8}$.