$$\left( z+1 \right)+\text{i}\left( 7-z \right)=\left( 1-\text{i} \right)z+\left( 1+7\text{i} \right)=\left( 1-\text{i} \right)\left( z-3+4\text{i} \right)$$，  即$$\left\textbar{} \left( z+1 \right)+\text{i}\left( 7-z \right) \right\textbar=\left\textbar{} 1-\text{i} \right\textbar\cdot \left\textbar{} z-3+4\text{i} \right\textbar=\sqrt{2}\cdot \left\textbar{} z-\left( 3-4\text{i} \right) \right\textbar$$，  即可以表示点$$z$$到$$3-4\text{i}$$的距离的$$\sqrt{2}$$倍，  ∵$$z$$在单位圆上，  ∴$$\left\textbar{} \left( z+1 \right)+\text{i}\left( 7-z \right) \right\textbar$$的取值范围是$$\left[ 4\sqrt{2},6\sqrt{2} \right]$$．  故选：$$\text{D}$$． 
设复数$$z$$满足$$\left\textbar{} z \right\textbar=1$$，则$$\left\textbar{} \left( z+1 \right)+\text{i}\left( 7-z \right) \right\textbar$$的值不可能为．
Options:
* $$4\sqrt{2}$$ 
* $$4\sqrt{3}$$ 
* $$5\sqrt{2}$$ 
* $$5\sqrt{3}$$ 