11. Given the complex number $z$ satisfies $z \cdot \bar{z}-z-\bar{z}=3$, and $\arg (z-1)=\frac{\pi}{3}$, then $z=$
11. $2+\sqrt{3} \mathrm{i}$. Hint: Let $\omega=z-1$, and $|\omega|=r$, then $z=\omega+1$, so $(\omega+1)(\bar{\omega}+1)-(\omega+1)-(\bar{\omega}+1)=3$. From this, we get $r^{2}=4, r=2$. Therefore, $\omega=2\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)=1+\sqrt{3} \mathrm{i}, \therefore z=2+\sqrt{3} \mathrm{i}$.