| Date | May 2012 | Marks available | 4 | Reference code | 12M.1.hl.TZ2.6 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Given that \((4 - 5{\text{i}})m + 4n = 16 + 15{\text{i}}\) , where \({{\text{i}}^2} = - 1\), find m and n if
m and n are real numbers;
m and n are conjugate complex numbers.
Markscheme
attempt to equate real and imaginary parts M1
equate real parts: \(4m + 4n = 16\); equate imaginary parts: \( -5m = 15\) A1
\( \Rightarrow m = -3,{\text{ }}n = 7\) A1
[3 marks]
let \(m = x + {\text{i}}y,{\text{ }}n = x - {\text{i}}y\) M1
\( \Rightarrow (4 - 5{\text{i}})(x + {\text{i}}y) + 4(x - {\text{i}}y) = 16 + 15{\text{i}}\)
\( \Rightarrow 4x - 5{\text{i}}x + 4{\text{i}}y + 5y + 4x - 4{\text{i}}y = 16 + 15{\text{i}}\)
attempt to equate real and imaginary parts M1
\(8x + 5y = 16,{\text{ }} -5x = 15\) A1
\( \Rightarrow x = -3,{\text{ }}y = 8\) A1
\(( \Rightarrow m = -3 + 8{\text{i}},{\text{ }}n = -3 - 8{\text{i}})\)
[4 marks]
Examiners report
Part (a) was generally well answered. In (b), however, some candidates put \(m = a + {\text{i}}b\) and \(n = c + {\text{i}}d\) which gave four equations for two unknowns so that no further progress could be made.
Part (a) was generally well answered. In (b), however, some candidates put \(m = a + {\text{i}}b\) and \(n = c + {\text{i}}d\) which gave four equations for two unknowns so that no further progress could be made.
