4. Given complex numbers $z, z_{1}, z_{2}\left(z_{1} \neq z_{2}\right)$ satisfy:
$$
z_{1}^{2}=z_{2}^{2}=-2-2 \sqrt{3} \mathrm{i},\left|z-z_{1}\right|=\left|z-z_{2}\right|=4 \text {. }
$$

Then $|z|=$ $\qquad$
4. $2 \sqrt{3}$.

First, find the square root of the complex number $-2-2 \sqrt{3} \mathrm{i}$.
Let $(x+y \mathrm{i})^{2}=-2-2 \sqrt{3} \mathrm{i}(x, y \in \mathbf{R})$. Then $\left(x^{2}-y^{2}\right)+2 x y \mathrm{i}=-2-2 \sqrt{3} \mathrm{i}$
$\Rightarrow\left\{\begin{array}{l}x^{2}-y^{2}=-2 \\ 2 x y=-2 \sqrt{3}\end{array}\right.$
$\Rightarrow\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=-\sqrt{3} ;\end{array}\left\{\begin{array}{l}x_{2}=-1, \\ y_{2}=\sqrt{3} .\end{array}\right.\right.$
Since $z_{1}^{2}=z_{2}^{2}=-2-2 \sqrt{3} \mathrm{i}, z_{1} \neq z_{2}$, we know that $z_{1}$ and $z_{2}$ are the two square roots of the complex number $-2-2 \sqrt{3} \mathrm{i}$.
By symmetry, we can assume
$z_{1}=1-\sqrt{3} \mathrm{i}, z_{2}=-1+\sqrt{3} \mathrm{i}$.
Thus, $\left|z_{1}-z_{2}\right|=\left|z-z_{1}\right|=\left|z-z_{2}\right|=4$.
Therefore, the points $Z, Z_{1}, Z_{2}$ corresponding to the complex numbers $z, z_{1}, z_{2}$ form an equilateral triangle with side length 4.

Since the points $Z_{1}, Z_{2}$ corresponding to the complex numbers $z_{1}, z_{2}$ are symmetric about the origin $O$, $O Z$ is the height of $\triangle Z Z_{1} Z_{2}$.
Hence, $|z|=|O Z|=2 \sqrt{3}$.