Giải chi tiết:

Ta có : \(\left| {{z^2} - 2z + 5} \right| = \left| {\left( {z - 1 + 2i} \right)\left( {z + 3i - 1} \right)} \right|\)\( \Leftrightarrow \left| {\left( {z - 1 - 2i} \right)\left( {z - 1 + 2i} \right)} \right| = \left| {\left( {z - 1 + 2i} \right)\left( {z + 3i - 1} \right)} \right|\,\,\left( * \right)\)

TH1 : \(\left| {z - 1 + 2i} \right| = 0 \Leftrightarrow z = 1 - 2i\)\( \Rightarrow w = z - 2 + 2i = 1 - 2i - 2 + 2i =  - 1\)\( \Rightarrow \left| w \right| = 1\).

TH2 : \(\left| {z - 1 + 2i} \right| > 0\) thì \(\left( * \right) \Leftrightarrow \left| {z - 1 - 2i} \right| = \left| {z - 1 + 3i} \right|\,\,\left( {**} \right)\)

Đặt \(z = x + yi\left( {x,y \in \mathbb{R}} \right)\) thì \(\left( {**} \right) \Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2}}  = \sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y + 3} \right)}^2}} \)

\( \Leftrightarrow {\left( {y - 2} \right)^2} = {\left( {y + 3} \right)^2} \Leftrightarrow y =  - \dfrac{1}{2}\)\( \Rightarrow z = x - \dfrac{1}{2}i\)

\( \Rightarrow w = z - 2 + 2i = x - \dfrac{1}{2}i - 2 + 2i = \left( {x - 2} \right) + \dfrac{3}{2}i\)\( \Rightarrow \left| w \right| = \sqrt {{{\left( {x - 2} \right)}^2} + \dfrac{9}{4}}  \ge \sqrt {0 + \dfrac{9}{4}}  = \dfrac{3}{2}\).

Kết hợp cả TH1 và TH2 ta thấy \(\min \left| w \right| = 1\) khi \(z = 1 - 2i\).

Chọn B