11. Given the complex number $z$ satisfies $|z|=1$, then the minimum value of $\left|z^{2}-2 z+3\right|$ is $\qquad$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
\text { Let } z=\cos \theta+\mathrm{i} \sin \theta \Rightarrow z^{2}-2 z+3=\cos 2 \theta+\mathrm{i} \sin 2 \theta-2(\cos \theta+\mathrm{i} \sin \theta)+3 \\
=\cos 2 \theta-2 \cos \theta+3+\mathrm{i}(\sin 2 \theta-2 \sin \theta),
\end{array}
$$

Then $\left|z^{2}-2 z+3\right|^{2}=(\cos 2 \theta-2 \cos \theta+3)^{2}+(\sin 2 \theta-2 \sin \theta)^{2}$
$$
\begin{array}{l}
=(\cos 2 \theta-2 \cos \theta)^{2}+(\sin 2 \theta-2 \sin \theta)^{2}+6(\cos 2 \theta-2 \cos \theta)+9 \\
=6 \cos 2 \theta-16 \cos \theta+14=12 \cos ^{2} \theta-16 \cos \theta+8=12\left(\cos \theta-\frac{2}{3}\right)^{2}+\frac{8}{3},
\end{array}
$$

Therefore, the minimum value of $\left|z^{2}-2 z+3\right|$ is $\frac{2 \sqrt{6}}{3}$.