复数 $z_{1}=1+i \cos \theta, z_{2}=\sin \theta-i, i$ 为虚数单位, 则 $\left|z_{1}-z_{2}\right|$ 的最大值为 ( )

$\begin{array}{lll}\text {
$ecause z_{1}=1+\mathrm{i} \cos 	heta, z_{2}=\sin 	heta-i$ ，$\left|z_{1}-z_{2}
ight|=|(1-\sin 	heta)+(\cos 	heta+1) \mathrm{i}|=\sqrt{(1-\sin 	heta)^{2}+(1+\cos 	heta)^{2}}$$=\sqrt{3+2(\cos 	heta-\sin 	heta)}=\sqrt{3+2 \sqrt{2} \cos \left(	heta+rac{\pi}{4}
ight)}$,$ecause\left|\cos \left(	heta+rac{\pi}{4}
ight)
ight|_{\max }=1, \quad 	herefore\left|z_{1}-z_{2}
ight|_{\max }=\sqrt{3+2 \sqrt{2}}=\sqrt{2}+1$, 故选 D.