Phương pháp giải:

Cho \({z_1} = {a_1} + {b_1}i;\,\,{z_2} = {a_2} + {b_2}i\,\,\,\left( {{a_1},\,\,{a_2},\,\,{b_1},\,\,{b_2} \in \mathbb{R}} \right).\) Ta có: \({z_1} + {z_2} = {a_1} + {a_2} + \left( {{b_1} + {b_2}} \right)i.\)

\( \Rightarrow \left| {{z_1} + {z_2}} \right| = \sqrt {{{\left( {{a_1} + {a_2}} \right)}^2} + {{\left( {{b_1} + {b_2}} \right)}^2}} .\)

Lời giải chi tiết:

Ta có: \(\left\{ \begin{array}{l}{z_1} = 2 + 3i\\{z_2} = 1 - i\end{array} \right. \Rightarrow {z_1} + {z_2} = 3 + 2i\) \( \Rightarrow \left| {{z_1} + {z_2}} \right| = \sqrt {{3^2} + {2^2}}  = \sqrt {13} .\)

Chọn D.