3.11. $\left\{\begin{array}{l}x+y=2, \\ x y-z^{2}=1 .\end{array}\right.$
3.11. From the second equation, it follows that $x y \geqslant 1$. The numbers $x$ and $y$ cannot both be negative, since their sum is 2. Therefore, the numbers $x$ and $y$ are positive and $x+y \geqslant 2 \sqrt{x y} \geqslant 2$, and the equality $x+y=2$ is possible only in the case when $x=y=1$. In this case, $z=0$.