Question 119, Let $a, b \in R$, if the equation $\left(z^{2}+a z+b\right)\left(z^{2}+a z+2 b\right)=0$ about $z$ has 4 distinct complex roots $z_{1}, z_{2}, z_{3}, z_{4}$, and the points corresponding to them on the complex plane are exactly the four vertices of a square with side length 1, find the value of $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|$.
Question 119, Solution: Let's assume $z_{1}, z_{2}$ are the roots of $z^{2} + a z + b = 0$, and $z_{3}, z_{4}$ are the roots of $z^{2} + a z + 2 b = 0$.

According to Vieta's formulas, we have:
\[
\left\{
\begin{array}{c}
z_{1} + z_{2} = z_{3} + z_{4} = -a \\
z_{1} z_{2} = b \\
z_{3} z_{4} = 2 b
\end{array}
\right.
\]
Thus, the points corresponding to $z_{1}, z_{2}$ are the two vertices of one diagonal of a square, and the points corresponding to $z_{3}, z_{4}$ are the two vertices of the other diagonal of the square, with the center of the square corresponding to the complex number $-\frac{a}{2}$.
Clearly, $z_{1}, z_{2}, z_{3}, z_{4}$ are not all real numbers.
If $z_{1}, z_{2}, z_{3}, z_{4}$ are all imaginary, then $z_{1}, z_{2}$ form a pair of conjugate complex numbers, and $z_{3}, z_{4}$ form another pair of conjugate complex numbers, thus the real parts of $z_{1}, z_{2}, z_{3}, z_{4}$ are all $-\frac{a}{2}$, and their corresponding points cannot form the four vertices of a square. Therefore, among $z_{1}, z_{2}, z_{3}, z_{4}$, there must be a pair of conjugate complex numbers, and the other two are real numbers.

If $b \leq 0$, then $a^{2} - 4 b \geq 0, a^{2} - 8 b \geq 0$, thus $z_{1}, z_{2}, z_{3}, z_{4}$ are all real numbers, leading to a contradiction. Hence, $b > 0$. Therefore, $a^{2} - 4 b > a^{2} - 8 b$, which implies that $a^{2} - 4 b > 0 > a^{2} - 8 b$. This indicates that $z_{1}, z_{2}$ are real numbers, while $z_{3}, z_{4}$ are a pair of conjugate complex numbers. Thus, we can set:
\[
z_{1} = \frac{-a + \sqrt{a^{2} - 4 b}}{2}, \quad z_{2} = \frac{-a - \sqrt{a^{2} - 4 b}}{2}, \quad z_{3} = \frac{-a + \sqrt{8 b - a^{2} i}}{2}, \quad z_{4} = \frac{-a - \sqrt{8 b - a^{2}}}{2}.
\]

According to the given conditions, we have:
\[
\left\{
\begin{array}{l}
|z_{1} - z_{2}| = \sqrt{a^{2} - 4 b} = \sqrt{2} \\
|z_{3} - z_{4}| = \sqrt{8 b - a^{2}} = \sqrt{2}
\end{array}
\right.
\]
Thus, we have:
\[
\left\{
\begin{array}{l}
a^{2} = 6 \\
b = 1
\end{array}
\right.
\]
Noting that $z_{1} z_{2} = b = 1$, we know that $z_{1}, z_{2}$ have the same sign. Therefore:
\[
|z_{1}| + |z_{2}| + |z_{3}| + |z_{4}| = |z_{1} + z_{2}| + 2|z_{3}| = |-a| + \sqrt{a^{2} + (8 b - a^{2})} = \sqrt{6} + 2 \sqrt{2}
\]

In conclusion, $|z_{1}| + |z_{2}| + |z_{3}| + |z_{4}| = \sqrt{6} + 2 \sqrt{2}$.