Given that \(z = \frac{{2 - {\text{i}}}}{{1 + {\text{i}}}} - \frac{{6 + 8{\text{i}}}}{{u + {\text{i}}}}\), find the values of u, u \( \in \mathbb{R}\), such that \(\operatorname{Re} z = \operatorname{Im} z\).

METHOD 1

\(\frac{{2 - {\text{i}}}}{{1 + {\text{i}}}} = \frac{{1 - 3{\text{i}}}}{2}\)     A1

\(\frac{{6 + 8{\text{i}}}}{{u + {\text{i}}}} \times \frac{{u - {\text{i}}}}{{u - {\text{i}}}} = \frac{{6u + 8 + (8u - 6){\text{i}}}}{{{u^2} + 1}}\)     M1A1

\( \Rightarrow \frac{{2 - {\text{i}}}}{{1 + {\text{i}}}} - \frac{{6 + 8u}}{{u + {\text{i}}}} = \frac{1}{2} - \frac{{6u + 8}}{{{u^2} + 1}} - \left( {\frac{3}{2} + \frac{{8u - 6}}{{{u^2} + 1}}} \right){\text{i}}\)

\(\operatorname{Im} z = \operatorname{Re} z\)

\( \Rightarrow \frac{1}{2} - \frac{{6u + 8}}{{{u^2} + 1}} = - \frac{3}{2} - \frac{{8u - 6}}{{{u^2} + 1}}\)     A1

(sketch from gdc, or algebraic method)     (M1)

u = −3; u = 2     A1A1     N2

[7 marks]

METHOD 2

\(\frac{{2 - {\text{i}}}}{{1 + {\text{i}}}} - \frac{{6 + 8{\text{i}}}}{{u + {\text{i}}}} = \frac{{(2 - {\text{i}})(u + {\text{i}}) - (1 + {\text{i}})(6 + 8{\text{i}})}}{{(u - 1) + {\text{i}}(u + 1)}}\)     M1A1

\( = \frac{{(2 - {\text{i}})(u + {\text{i}}) - (1 + {\text{i}})(6 + 8{\text{i}})}}{{(u - 1) + {\text{i}}(u + 1)}} \cdot \frac{{(u - 1) - {\text{i}}(u + 1)}}{{(u - 1) - {\text{i}}(u + 1)}}\)     M1

\( = \frac{{{u^2} - 12u - 15 + {\text{i}}( - 3{u^2} - 16u + 9)}}{{2({u^2} + 1)}}\)     A1

\(\operatorname{Re} z = \operatorname{Im} z \Rightarrow {u^2} - 12u - 15 = - 3{u^2} - 16u + 9\)     M1

u = −3; u = 2     A1A1     N2

[7 marks]

Many candidates failed to access their GDC early enough to avoid huge algebraic manipulations, often carried out with many errors. Some candidates failed to separate and equate the real and imaginary parts of the expression obtained.