| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Invariant points under transformations |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to find an invariant point by solving z = w, leading to a quadratic equation in z. While the algebraic manipulation is substantial and involves complex arithmetic, the method is standard for F2 invariant point problems. The presence of surds adds computational complexity but doesn't require novel insight. |
| Spec | 4.02l Geometrical effects: conjugate, addition, subtraction |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(i(1+\sqrt{3}) = \frac{i(1+\sqrt{3})+pi}{i^2(1+\sqrt{3})+3}\) | M1 | Substitute \(i(1+\sqrt{3})\) for \(w\) and \(z\) |
| \(-i(1+\sqrt{3})^2 + 3i(1+\sqrt{3}) = i(1+\sqrt{3}) + pi\) | dM1 | Solve to \(p = \ldots\) |
| \(-1-2\sqrt{3}-3+3+3\sqrt{3} = 1+\sqrt{3}+p\) | dM1 | |
| \(p = -2\) | A1 | Correct value for \(p\) |
## Question 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $i(1+\sqrt{3}) = \frac{i(1+\sqrt{3})+pi}{i^2(1+\sqrt{3})+3}$ | M1 | Substitute $i(1+\sqrt{3})$ for $w$ and $z$ |
| $-i(1+\sqrt{3})^2 + 3i(1+\sqrt{3}) = i(1+\sqrt{3}) + pi$ | dM1 | Solve to $p = \ldots$ |
| $-1-2\sqrt{3}-3+3+3\sqrt{3} = 1+\sqrt{3}+p$ | dM1 | |
| $p = -2$ | A1 | Correct value for $p$ |
---\begin{enumerate}
\item The transformation $T$ from the $z$-plane, where $z = x + \mathrm { i } y$, to the $w$-plane, where $w = u + \mathrm { i } v$, is given by
\end{enumerate}
$$w = \frac { z + p \mathrm { i } } { \mathrm { i } z + 3 } \quad z \neq 3 \mathrm { i } \quad p \in \mathbb { Z }$$
The point representing $\mathrm { i } ( 1 + \sqrt { 3 } )$ is invariant under $T$.\\
Determine the value of $p$.\\
\hfill \mbox{\textit{Edexcel F2 2021 Q1 [3]}}