Example 7 Given a set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$

Then the number of complex numbers in $D$ whose real and imaginary parts are both integers is $\qquad$ [s]
Solve: Establish a complex plane, let
$$
z=x+y \mathrm{i}(x, y \in \mathbf{R}) \text {. }
$$

From $\left|z_{1}\right|=1$, we know $z_{1}^{2}$ is also on the unit circle $\odot 0$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \\
=\left|z_{1}^{2}-1\right|^{2}
\end{array}
$$

This represents the square of the distance from a point on the unit circle $\odot O$ to the point $(1,0)$ (as shown in Figure 7), with a value range of $[0,4]$.
$$
\begin{array}{l}
\text { Therefore }|z-2005-2006 \mathrm{i}| \leqslant 4 \\
\Rightarrow(x-2005)^{2}+(y-2006)^{2} \leqslant 16,
\end{array}
$$

The number of integer points (both coordinates are integers) is the same as the number of integer points in
$$
x^{2}+y^{2} \leqslant 16
$$

which is 49.