11. Given the complex number $z$ satisfies $z \cdot \bar{z}-z-\bar{z}=3$, and $\arg (z-1)$ $=\frac{\pi}{3}$. Then $z=$ $\qquad$ .
$$
11.2+\sqrt{3} i \text {. }
$$

Let $w=z-1$, and $|w|=r$, then $z=w+1$.
$$
\text { Hence }(w+1)(\bar{w}+1)-(w+1)-(\bar{w}+1)=3 \text {. }
$$

From this, we get $r^{2}=4, r=2$. Therefore,
$$
\begin{array}{l}
w=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=1+\sqrt{3} i \\
\therefore z=2+\sqrt{3} i
\end{array}
$$