6. Let $a, \mathbf{b}$ be real numbers, and the complex numbers $z_{1}=\sqrt{3} a-1+(\sqrt{3}-b) i$ and $_{2}^{z_{2}}=2-\sqrt{3} a+b i$ have equal magnitudes, and ${ }^{z_{1} \overline{z_{2}}}$ is a pure imaginary number, then $a+\mathbf{b}=$ . $\qquad$
【Answer】 $\sqrt{3} \pm 1$
【Solution】【Detailed】 $\left\{\begin{array}{c}\sqrt{3} a-1=-b, \\ \sqrt{3}-b=\sqrt{3} a-2 \text {. Solving, we get }\end{array} a=b=\frac{\sqrt{3}-1}{2}\right.$ or $a=b=\frac{\sqrt{3}+1}{2}$, hence $a+b=\sqrt{3} \pm 1$. Therefore, the answer is: $\sqrt{3} \pm 1$