A2 June 2023 Paper 2 Q2
2 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \(|-24 + 7\mathrm{i}| = \sqrt{(-24)^2 + 7^2}\) or \(\arg(-24 + 7\mathrm{i}) = \tan^{-1}\dfrac{7}{-24}\) oe | M1 | 1.1 |
| \(|-24 + 7\mathrm{i}| = 25\) and awrt \(-0.284\) or \(2.86\) | A1 | 1.1 |
| \(-24 + 7\mathrm{i} = 25(\cos 2.86 + \mathrm{i}\sin 2.86)\) | A1 | 1.1 |
| [3] |
Notes
M1: Attempt to find modulus or argument using a correct formula (values must be real)
If \(\frac{7}{24}\) allow only if supported by explanation, further working or clear diagram. May use alternative trig function
A1: (1st) Condone use of degrees for this mark (\(-16.3^\circ\) or \(163.7^\circ\)). Accept \(\arctan\left(-\frac{7}{24}\right)\)
A1: (2nd) Final answer. Accept equivalent notation, e.g. cis, \((r, \theta)\) or exponential form, not \((-C + \mathrm{i}S)\) nor \(rC + r\mathrm{i}S\).
| Scheme | Marks | AO |
|---|---|---|
| DR \(6\mathrm{i}z + 18w = -42\mathrm{i}\) \(-6\mathrm{i}z - 5w = 3\mathrm{i} - 13\) | *M1 | 1.1 |
| \(13w = -13 - 39\mathrm{i}\) so \(w = -1 - 3\mathrm{i}\) | A1 | 1.1 |
| \(\mathrm{i}z + 3(-1 - 3\mathrm{i}) = -7\mathrm{i}\) or \(-6z + 5\mathrm{i}(-1 - 3\mathrm{i}) = 3 + 13\mathrm{i}\) \(\mathrm{i}z = 3 + 2\mathrm{i}\) or \(-6z = -12 + 18\mathrm{i}\) | dep*M1 | 1.1 |
| \(z = 2 - 3\mathrm{i}\) | A1 | 1.1 |
| [4] |
Notes
*M1: Scaling both equations (using \(\mathrm{i}^2 = -1\)) so that the coefficient of \(z\) or \(w\) is the same in magnitude.
Or \(-5z + 15\mathrm{i}w = 35\) and \(-18z + 15\mathrm{i}w = 9 + 39\mathrm{i}\)
A1: (1st) or \(13z = 26 - 39\mathrm{i}\) so \(z = 2 - 3\mathrm{i}\)
dep*M1: Substituting back into one equation and attempt to solve by collecting real and imaginary parts.
Or \(\mathrm{i}(2 - 3\mathrm{i}) + 3w = -7\mathrm{i}\) or \(-6(2 - 3\mathrm{i}) + 5\mathrm{i}w = 3 + 13\mathrm{i}\), i.e. reaches \(kz = a + b\mathrm{i}\) for real \(a, b\)
Alternative method
| Scheme | Marks |
|---|---|
| \(w = \dfrac{-7\mathrm{i} - \mathrm{i}z}{3}\) or \(w = \dfrac{3 + 13\mathrm{i} + 6z}{5\mathrm{i}}\) | M1 |
| \(-6z + 5\mathrm{i}\left(\dfrac{-7\mathrm{i} - \mathrm{i}z}{3}\right) = 3 + 13\mathrm{i}\) \(\therefore -18z + 35 + 5z = 9 + 39\mathrm{i}\) | M1 |
| \(\therefore (-13z = -26 + 39\mathrm{i}\) so\() \ z = 2 - 3\mathrm{i}\) | A1 |
| \(w = \dfrac{-7\mathrm{i} - \mathrm{i}(2 - 3\mathrm{i})}{3} = -1 - 3\mathrm{i}\) | A1 |
M1: (1st) Using one equation to express one unknown in terms of the other. Or \(z = \frac{-7\mathrm{i} - 3w}{\mathrm{i}}\) or \(z = \frac{3 + 13\mathrm{i} - 5\mathrm{i}w}{-6}\)
M1: (2nd) Substituting into the other equation and using \(\mathrm{i}^2 = -1\) at least once
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\mathrm{i}(a + b\mathrm{i}) + 3(c + d\mathrm{i}) = -7\mathrm{i}\) \(-6(a + b\mathrm{i}) + 5\mathrm{i}(c + d\mathrm{i}) = 3 + 13\mathrm{i}\) | M1 |
| Re: \(-b + 3c = 0,\ -6a - 5d = 3\) Im: \(a + 3d = -7,\ -6b + 5c = 13\) | M1 |
| \(z = 2 - 3\mathrm{i}\) | A1 |
| \(w = -1 - 3\mathrm{i}\) | A1 |
M1: (1st) replaces \(z\) and \(w\) with two Cartesian forms in both equations
M1: (2nd) Takes real and imaginary parts from both complex equations. Condone i’s left in
A1: (1st) Could be BC. \(a = 2\), \(b = -3\) sufficient
A1: (2nd) \(c = -1\), \(d = -3\) sufficient
Alternative method 3
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} \mathrm{i} & 3 \\ -6 & 5\mathrm{i} \end{pmatrix}\begin{pmatrix} z \\ w \end{pmatrix} = \begin{pmatrix} -7\mathrm{i} \\ 3 + 13\mathrm{i} \end{pmatrix}\) \(\begin{pmatrix} z \\ w \end{pmatrix} = \begin{pmatrix} \mathrm{i} & 3 \\ -6 & 5\mathrm{i} \end{pmatrix}^{-1}\begin{pmatrix} -7\mathrm{i} \\ 3 + 13\mathrm{i} \end{pmatrix}\) | M1 |
| where \(\begin{pmatrix} \mathrm{i} & 3 \\ -6 & 5\mathrm{i} \end{pmatrix}^{-1} = \dfrac{1}{13}\begin{pmatrix} 5\mathrm{i} & -3 \\ 6 & \mathrm{i} \end{pmatrix}\) | A1 |
| \(\dfrac{1}{13}\begin{pmatrix} 5\mathrm{i} & -3 \\ 6 & \mathrm{i} \end{pmatrix}\begin{pmatrix} -7\mathrm{i} \\ 3 + 13\mathrm{i} \end{pmatrix} = \dfrac{1}{13}\begin{pmatrix} 35 - 9 - 39\mathrm{i} \\ -42\mathrm{i} + 3\mathrm{i} - 13 \end{pmatrix}\) | M1 |
| \(\Rightarrow z = 2 - 3\mathrm{i},\ w = -1 - 3\mathrm{i}\) | A1 |
M1: (1st) Writes in matrix form and derives an equation for \(\begin{pmatrix} z \\ w \end{pmatrix}\). Must left-multiply by their inverse matrix
A1: (1st) …with correct inverse matrix
M1: (2nd) Expands…
A1: (2nd) … to correct simplified solution; can be in vector form