Given that \(\frac{z}{{z + 2}} = 2 - {\text{i}}\) , \(z \in \mathbb{C}\) , find z in the form \(a + {\text{i}}b\) .

METHOD 1
\(z = \left( {2 - {\text{i}}} \right)\left( {z + 2} \right)\)     M1
\( = 2z + 4 - {\text{i}}z - 2{\text{i}}\)
\(z\left( {1 - {\text{i}}} \right) = - 4 + 2{\text{i}}\)
\(z = \frac{{ - 4 + 2{\text{i}}}}{{1 - {\text{i}}}}\)     A1
\(z = \frac{{ - 4 + 2{\text{i}}}}{{1 - {\text{i}}}} \times \frac{{1 + {\text{i}}}}{{1 + {\text{i}}}}\)     M1
\( = - 3 - {\text{i}}\)     A1

METHOD 2

let \(z = a + {\text{i}}b\)
\(\frac{{a + {\text{i}}b}}{{a + {\text{i}}b + 2}} = 2 - {\text{i}}\)     M1
\(a + {\text{i}}b = \left( {2 - i} \right)\left( {\left( {a + 2} \right) + {\text{i}}b} \right)\)
\(a + {\text{i}}b = 2\left( {a + 2} \right) + 2b{\text{i}} - {\text{i}}\left( {a + 2} \right) + b\)
\(a + {\text{i}}b = 2a + b + 4 + \left( {2b - a - 2} \right){\text{i}}\)
attempt to equate real and imaginary parts     M1
\(a = 2a + b + 4\left( { \Rightarrow a + b + 4 = 0} \right)\)
and \(b = 2b - a - 2\left( { \Rightarrow - a + b - 2 = 0} \right)\)     A1

Note: Award Al for two correct equations.

\(b = - 1\); \(a = - 3\)     A1

\(z = - 3 - {\text{i}}\)

[4 marks]

A number of different methods were adopted in this question with some candidates working through their method to a correct answer. However many other candidates either stopped with \(z\) still expressed as a quotient of two complex numbers or made algebraic mistakes.