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.