已知 $z_{1}, z_{2} \in C,\left|z_{1}+z_{2}\right|=2 \sqrt{2},\left|z_{1}\right|=\sqrt{3},\left|z_{2}\right|=\sqrt{2}$, 则 $\left|z_{1}-z_{2}\right|=$ ( )
本题可用特值法, 不妨设 $z_{1}=\sqrt{3}, z_{2}=\sqrt{2}(\cos 	heta+i \sin 	heta), 	heta \in[0,2 \pi)$,由于 $\left|z_{1}+z_{2}
ight|=2 \sqrt{2}$, 可得 $\sqrt{(\sqrt{3}+\sqrt{2} \cos 	heta)^{2}+(\sqrt{2} \sin 	heta)^{2}}=2 \sqrt{2}$,解得 $\cos 	heta=rac{\sqrt{6}}{4}$,则 $\left|z_{1}-z_{2}
ight|=\sqrt{(\sqrt{3}-\sqrt{2} \cos 	heta)^{2}+(\sqrt{2} \sin 	heta)^{2}}=\sqrt{5-6 \cos 	heta}=\sqrt{2}$, 故选 D.