Ta có \({\left( { - 2 + i} \right)^2} + a\left( { - 2 + i} \right) + b = 0\)
\(\Leftrightarrow {i^2} - 4i + 4 - 2a + ai + b = 0 \Leftrightarrow 3 - 2a + b + ai - 4i = 0\)
\(\Leftrightarrow \left( {3 - 2a + b} \right) + i\left( {a - 4} \right) = 0 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {3 - 2a + b = 0}\\ {a - 4 = 0} \end{array}} \right.\)
\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {a = 4}\\ {b = 5} \end{array}} \right. \Rightarrow a - b = 1\)
