FP1 June 2013 (R) Q9
9. The complex number \(w\) is given by \[w = 10 - 5\mathrm{i}\]
The complex numbers \(z\) and \(w\) satisfy the equation \[(2 + \mathrm{i})(z + 3\mathrm{i}) = w\]
where \(a\) and \(b\) are real numbers. (4)
Given that \[\arg(\lambda + 9\mathrm{i} + w) = \frac{\pi}{4}\] where \(\lambda\) is a real constant,
| Scheme | Marks |
|---|---|
| \(w = 10 - 5\mathrm{i}\) | |
| \(|w| = \left\{\sqrt{10^2 + (-5)^2}\right\} = \sqrt{125}\) or \(5\sqrt{5}\) or \(11.1803\ldots\) \(\underline{\sqrt{125}}\) or \(\underline{5\sqrt{5}}\) or awrt 11.2 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\arg w = -\tan^{-1}\left(\tfrac{5}{10}\right)\) Use of \(\tan^{-1}\) or \(\tan\) | M1 |
| \(= -0.463647609\ldots = -0.46\ (2\text{ dp})\) awrt \(-0.46\) or awrt 5.82 | A1 oe |
| (2) |
| Scheme | Marks |
|---|---|
| \((2 + \mathrm{i})(z + 3\mathrm{i}) = w\) | |
| \(z + 3\mathrm{i} = \dfrac{10 - 5\mathrm{i}}{(2 + \mathrm{i})}\) Simplifies to give \(* = \dfrac{\text{complex no.}}{(2 + \mathrm{i})}\) | B1 |
| \(z + 3\mathrm{i} = \dfrac{(10 - 5\mathrm{i})}{(2 + \mathrm{i})} \times \dfrac{(2 - \mathrm{i})}{(2 - \mathrm{i})}\) Multiplies by \(\dfrac{\text{their } (2 - \mathrm{i})}{\text{their } (2 - \mathrm{i})}\) | M1 |
| \(z + 3\mathrm{i} = \dfrac{20 - 10\mathrm{i} - 10\mathrm{i} - 5}{1 + 4}\) Simplifies realising that a real number is needed on the denominator and applies \(\mathrm{i}^2 = -1\) on their numerator expression and denominator expression. | M1 |
| \(z + 3\mathrm{i} = \dfrac{15 - 20\mathrm{i}}{5}\) | |
| \(z + 3\mathrm{i} = 3 - 4\mathrm{i}\) \(z = 3 - 7\mathrm{i}\) (Note: \(a = 3, b = -7\).) \(z = 3 - 7\mathrm{i}\) | A1 |
| (4) |
Notes
Alt 1: Scheme as above:
\((2 + \mathrm{i})z + 6\mathrm{i} + 3\mathrm{i}^2 = 10 - 5\mathrm{i} \Rightarrow (2 + \mathrm{i})z = 13 - 11\mathrm{i}\)
B1 for \(z = \dfrac{13 - 11\mathrm{i}}{2 + \mathrm{i}}\); M1 for \(z = \dfrac{(13 - 11\mathrm{i})}{(2 + \mathrm{i})} \times \dfrac{(2 - \mathrm{i})}{(2 - \mathrm{i})}\); M1 for \(z = \dfrac{26 - 13\mathrm{i} - 22\mathrm{i} - 11}{4 + 1}\);
A1 for \(z = 3 - 7\mathrm{i}\)
Alt 2: Let \(z = a + \mathrm{i}b\) gives \((2 + \mathrm{i})(a + \mathrm{i}b + 3\mathrm{i}) = 10 - 5\mathrm{i}\) for B1
Equating real and imaginary parts to form two equations both involving \(a\) and \(b\) for M1
Solves simultaneous equations as far as \(a =\) or \(b =\) for M1
\(a = 3, b = -7\) or \(z = 3 - 7\mathrm{i}\) for A1
| Scheme | Marks |
|---|---|
| \(\arg(\lambda + 9\mathrm{i} + w) = \dfrac{\pi}{4}\) | |
| \(\lambda + 9\mathrm{i} + w = \lambda + 9\mathrm{i} + 10 - 5\mathrm{i} = (\lambda + 10) + 4\mathrm{i}\) | |
| \(\arg(\lambda + 9\mathrm{i} + w) = \dfrac{\pi}{4} \Rightarrow \lambda + 10 = 4\) Combines real and imaginary parts and puts “Real part = Imaginary part” i.e. \(\dfrac{\lambda + 10}{4} = 1\) or \(\dfrac{4}{\lambda + 10} = 1\) o.e. | M1 |
| So, \(\lambda = -6\) \(-6\) | A1 |
| (2) | |
| (9 marks) |