| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Linear equations in z and z* |
| Difficulty | Moderate -0.8 This is a straightforward algebraic manipulation question requiring substitution of z = x + iy and z* = x - iy, then equating real and imaginary parts to zero. It involves only basic complex number operations with no conceptual difficulty or problem-solving insight needed—purely mechanical execution of standard techniques taught early in FP1. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | \(i(z+7) + 3(z^* - i) =\) | M1 |
| 3(a) | \(i(x + iy + 7) + 3(x - iy - i)\) | M1 |
| 3(a) | \(= ix - y + 7i + 3x - 3iy - 3i\) | |
| 3(a) | \(= 3x - y + i(x - 3y + 4)\) | A1 |
| 3(b) | \(3x - y = 0, \quad x - 3y + 4 = 0\) | M1 |
| 3(b) | \(x - 9y + 4 = 0\) (or eg \(y - 9y + 12 = 0\)) | A1 |
| 3(b) | Solving to give \(z = \frac{1}{2} + \frac{3}{2}i\) | A1 |
3(a) | $i(z+7) + 3(z^* - i) =$ | M1 | M1 for use of $z^* = x - iy$
3(a) | $i(x + iy + 7) + 3(x - iy - i)$ | M1 | M1 for $i^2 y = -y$
3(a) | $= ix - y + 7i + 3x - 3iy - 3i$ | |
3(a) | $= 3x - y + i(x - 3y + 4)$ | A1 | If the five terms correct but not grouped into Real and Imaginary parts, allow A1 retrospectively provided the correct two expressions used in the M1 line in (b)
3(b) | $3x - y = 0, \quad x - 3y + 4 = 0$ | M1 | Attempting to equate all Real parts to zero and all Imaginary parts to zero
3(b) | $x - 9y + 4 = 0$ (or eg $y - 9y + 12 = 0$) | A1 | A correct equation in either x or y. PI by correct final answer
3(b) | Solving to give $z = \frac{1}{2} + \frac{3}{2}i$ | A1 | Allow $x = \frac{1}{2}, y = \frac{3}{2}$3 It is given that $z = x + \mathrm { i } y$, where $x$ and $y$ are real numbers.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $x$ and $y$, the real and imaginary parts of
$$\mathrm { i } ( z + 7 ) + 3 \left( z ^ { * } - \mathrm { i } \right)$$
\item Hence find the complex number $z$ such that
$$\mathrm { i } ( z + 7 ) + 3 \left( z ^ { * } - \mathrm { i } \right) = 0$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q3 [6]}}