Đặt \(z = a + bi\,\,\,(a,b \in \mathbb{R})\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{5(\bar z + i)}}{{z + i}} = 2 - i\\
 \Rightarrow \frac{{5(a - bi + i)}}{{a + bi + 1}} = 2 - i
\end{array}\\
{ \Rightarrow 5(a - bi + i) = \left( {2 - i} \right)\left( {a + bi + 1} \right)}\\
\begin{array}{l}
 \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{5a = 2(a + 1) + b}\\
{ - 5b + 5 = 2b - (a + 1)}
\end{array}} \right.\\
 \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{3a - b = 2}\\
{a - 7b =  - 6}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{a = 1}\\
{b = 1}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy \(z = 1 + i \Rightarrow {z^2} = 2i \)

\(\Rightarrow \omega = 1 + (1 + i) + 2i = 2 + 3i.\)