6. In the complex plane, the complex number $z_{1}$ moves on the line segment connecting $1+2 \mathrm{i}$ and $1-2 \mathrm{i}$, and the complex number $z_{2}$ moves on the circumference of a circle centered at the origin with a radius of 1. Then the area of the figure formed by the trajectory of $z_{1}+z_{2}$ is $\qquad$.
6. $8+\pi$.

Let $z_{1}=1+k \mathrm{i}(-2 \leqslant k \leqslant 2), z_{2}=\cos \theta+\mathrm{i} \sin \theta(\theta \in \mathbf{R})$, then
$$
z_{1}+z_{2}=(1+\cos \theta)+(k+\sin \theta) \mathrm{i} \text {. }
$$

Let $x=1+\cos \theta, y=k+\sin \theta$, then $x-1=\cos \theta, y-k=$ $\sin \theta,(x-1)^{2}+(y-k)^{2}=1$, the trajectory sought is a series of circles with center at $(1, k)$ $(-2 \leqslant k \leqslant 2)$, and radius 1, the region covered by this trajectory is shown in the shaded area in Figure 3, with an area of $8+\pi$.