已知 $\mathrm{i}$ 为虚数单位, 设 $z_{1}=x+2 \mathrm{i}, z_{2}=3-\mathcal{y}(x, y \in \mathbf{R})$, 且 $z_{1}+z_{2}=5-6 \mathbf{i}$,则 $z_{1}-z_{2}=$ \$ \qquad \$
$ecause z_{1}+z_{2}=5-6 \mathrm{i}, 	herefore(x+2 \mathrm{i})+(3-y i)=5-6 \mathrm{i}$$\left\{egin{array}{l}x+3=5 \ 2-y=-6\end{array}
ight.$, 即 $\left\{egin{array}{l}x=2 \ y=8\end{array}, 	herefore z_{1}=2+2 i
ight.$$z_{2}=3-8 i, \quad 	herefore z_{1}-z_{2}=(2+2 i)-(3-8 i)=-1+10 i$