设复数 $z_{1}, \quad z_{2}$ 满足 $\left|z_{1}\right|=\left|z_{2}\right|=2, \quad z_{1}+z_{2}=\sqrt{3}+i$, 则 $\left|z_{1}-z_{2}\right|=$
答案: $2 \sqrt{3}$

【解析】 $\because\left|z_{1}\right|=\left|z_{2}\right|=2$, 可设 $z_{1}=2 \cos \theta+2 \sin \theta \cdot i, \quad z_{2}=2 \cos \alpha+2 \sin \alpha \cdot i$,

$\therefore z_{1}+z_{2}=2(\cos \theta+\cos \alpha)+2(\sin \theta+\sin \alpha) \cdot i=\sqrt{3}+i$,

$\therefore\left\{\begin{array}{l}2(\cos \theta+\cos \alpha)=\sqrt{3} \\ 2(\sin \theta+\sin \alpha)=1 \text {, 两式平方作和得: } 4(2+2 \cos \theta \cos \alpha+2 \sin \theta \sin \alpha)=4 \text {, }\end{array}\right.$

化简得: $\cos \theta \cos \alpha+\sin \theta \sin \alpha=-\frac{1}{2}$

$\therefore\left|z_{1}-z_{2}\right|=|2(\cos \theta-\cos \alpha)+2(\sin \theta-\sin \alpha) \cdot i|$

$=\sqrt{4(\cos \theta-\cos \alpha)^{2}+4(\sin \theta-\sin \alpha)^{2}}=\sqrt{8-8(\cos \theta \cos \alpha+\sin \theta \sin \alpha)}=\sqrt{8+4}=2 \sqrt{3}$.

故答案为: $2 \sqrt{3}$.

【分析】令 $z_{1}=2 \cos \theta+2 \sin \theta \cdot i, \quad z_{2}=2 \cos \alpha+2 \sin \alpha \cdot i$, 根据复数的相等可求得 $\cos \theta \cos \alpha+\sin \theta \sin \alpha=-\frac{1}{2}$, 代入复数模长的公式中即可得到结果.