复数

$\displaystyle 75-\frac{25 \pi}{2}$ ． 由复数 $z=x+y \mathrm{i}$ 使得 $\displaystyle \frac{10}{\bar{z}}$ 及 $\displaystyle \frac{z}{10}$ 的实部和虚部都是小于 1 的正数，可得 $$ \begin{aligned} \left\{\begin{array}{l} \frac{10}{\bar{z}}=\frac{10 x}{x^{2}+y^{2}}+\frac{10 y}{x^{2}+y^{2}} \mathrm{i}, \\ \frac{z}{10}=\frac{x}{10}+\frac{y}{10} \mathrm{i} \end{array}\right. & \Rightarrow\left\{\begin{array} { l } { 0 < \frac { 1 0 x } { x ^ { 2 } + y ^ { 2 } } < 1 , } \\ { 0 < \frac { 1 0 y } { x ^ { 2 } + y ^ { 2 } } < 1 } \end{array} \text { 且 } \left\{\begin{array}{l} 0<\frac{x}{10}<1, \\ 0<\frac{y}{10}<1 \end{array}\right.\right. \\ & \Rightarrow\left\{\begin{array} { l } { ( x - 5 ) ^ { 2 } + y ^ { 2 } > 5 ^ { 2 } , } \\ { x ^ { 2 } + ( y - 5 ) ^ { 2 } > 5 ^ { 2 } } \end{array} \text { 且 } \left\{\begin{array}{l} 0<x<10, \\ 0<y<10 . \end{array}\right.\right. \end{aligned} $$ 由图可求．

$2 \sqrt{3}$ ． 由题意，$\displaystyle z_{1}=\frac{1 \pm \sqrt{3} \mathrm{i}}{4} z_{2}$ ，则 $\displaystyle \left|z_{1}\right|=\left|\frac{1 \pm \sqrt{3} \mathrm{i}}{4}\right|\left|z_{2}\right|=2$ ，故 $\displaystyle \arg \left(\frac{z_{1}}{z_{2}}\right)=\arg \frac{1 \pm \sqrt{3} \mathrm{i}}{4}= \pm \frac{\pi}{3}$ ，则 $\displaystyle \angle P O Q=\frac{\pi}{3}, S_{\triangle O P Q}=2 \sqrt{3}$ ．