10. Let $a, b \in \mathbf{R}$, if the equation $\left(z^{2}+a z+b\right)\left(z^{2}+a z+2 b\right)=0$ has 4 distinct complex roots $z_{1}, z_{2}, z_{3}, z_{4}$, and these roots correspond to the four vertices of a square with side length 1 in the complex plane, find the value of $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|$.
Given that $a, b \in \mathbf{R}$, according to the problem, one of the equations $z^{2}+a z+b=0$ and $z^{2}+a z+2 b=0$ must have two imaginary roots, and the other must have two real roots. Therefore, $b>0 \Rightarrow a^{2}-4 b>0, a^{2}-8 b<0$.
Notice that $\left(-\frac{a}{2}, 0\right)$ is the center of the square, so $\sqrt{a^{2}-4 b}=\sqrt{8 b-a^{2}}=\sqrt{2}$ $\Rightarrow a^{2}=6, b=1$. Let's take $a=\sqrt{6}$, then the roots of $z^{2}+\sqrt{6} z+1=0$ are $z_{1,2}=\frac{-\sqrt{6} \pm \sqrt{2}}{2}$, and the roots of $z^{2}+\sqrt{6} z+2=0$ are $z_{3,4}=\frac{-\sqrt{6} \pm \sqrt{2} \mathrm{i}}{2}$. Therefore, $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|=\sqrt{6}+2 \sqrt{2}$.