1. Express z from the first equation and substitute into the second:

$x^{2}-2 x+y^{2}-2 \sqrt{3} y=-4 \rightarrow(x-1)^{2}+(y-\sqrt{3})^{2}=0 \rightarrow\left\{\begin{array}{c}x=1 \\ y=\sqrt{3}\end{array} \rightarrow z=x^{2}+y^{2}+2 x=6\right.$
Answer: $x=1, y=\sqrt{3}, z=6$