5. Given in the Cartesian coordinate system $x O y$, the vertices of square $A B C D$ are $A(-1,1)$, vertex $C(1,1+2 \sqrt{3})$. Then, the coordinates of vertices $B$ and $D$ are $\qquad$
$\qquad$
5. $(\sqrt{3}, \sqrt{3})$ and $(-\sqrt{3}, 2+\sqrt{3})$.

It is easy to know that the midpoint $T(0,1+\sqrt{3})$ of $A C$.
Translate the square $A B C D$ so that its center $T$ moves to the origin $O$, then we have $A^{\prime}(-1,-\sqrt{3}), C^{\prime}(1, \sqrt{3})$.
Rotate $A^{\prime} \leqslant C^{\prime}$ by $90^{\circ}$ around point $O$, to get $B^{\prime}(\sqrt{3},-1), D^{\prime}(-\sqrt{3}, 1)$.
Therefore, $B(\sqrt{3}, \sqrt{3}), D(-\sqrt{3}, 2+\sqrt{3})$.