Lời giải chi tiết:

$\begin{array}{l}\frac{{\left( {\left| z \right| - 1} \right)\left( {1 + iz} \right)}}{{z - \frac{1}{{\bar z}}}} = i \Leftrightarrow \frac{{\left( {\left| z \right| - 1} \right)\left( {1 + iz} \right)\overline z }}{{{{\left| z \right|}^2} - 1}} = i\,\,\left( {\left| z \right| \ne 1} \right)\\ \Leftrightarrow \left( {\left| z \right| - 1} \right)\left( {1 + iz} \right)\overline z  = i\left( {\left| z \right| - 1} \right)\left( {\left| z \right| + 1} \right)\\ \Leftrightarrow \left( {\left| z \right| - 1} \right)\left[ {\left( {1 + iz} \right)\overline z  - i\left( {\left| z \right| + 1} \right)} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l}\left| z \right| = 1\,\,\left( {ktm} \right)\\\overline z  + i{\left| z \right|^2} - i\left| z \right| - i = 0\end{array} \right. \Leftrightarrow \overline z  + i\left( {{{\left| z \right|}^2} - \left| z \right| - 1} \right) = 0\\ \Leftrightarrow \overline z  =  - i\left( {{{\left| z \right|}^2} - \left| z \right| - 1} \right) \Leftrightarrow \left| {\overline z } \right| = \left| {{{\left| z \right|}^2} - \left| z \right| - 1} \right|\\ \Leftrightarrow {\left| z \right|^2} = {\left| z \right|^4} + {\left| z \right|^2} + 1 - 2{\left| z \right|^3} - 2{\left| z \right|^2} + 2\left| z \right|\\ \Leftrightarrow {\left| z \right|^4} - 2{\left| z \right|^3} - 2{\left| z \right|^2} + 2\left| z \right| + 1 = 0\\ \Leftrightarrow \left( {\left| z \right| - 1} \right)\left( {\left| z \right| + 1} \right)\left( {{{\left| z \right|}^2} - 2\left| z \right| - 1} \right) = 0\\ \Leftrightarrow {\left| z \right|^2} - 2\left| z \right| - 1 = 0 \Leftrightarrow \left| z \right| = 1 + \sqrt 2 \\ \Rightarrow {a^2} + {b^2} = {\left| z \right|^2} = 3 + 2\sqrt 2 \end{array}$

Chọn C.