| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Quadratic from one complex root |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths question testing basic complex number manipulation (rationalizing denominator, squaring) and using the sum/product of roots to find a quadratic equation. All steps are routine applications of standard techniques with no problem-solving insight required, making it slightly easier than an average A-level question despite being from FP1. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02f Convert between forms: cartesian and modulus-argument4.02g Conjugate pairs: real coefficient polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(z = \dfrac{4(1-i)}{(1+i)(1-i)}\) | M1 | Multiplies numerator and denominator by \(1-i\) or \(-1+i\) |
| \(z = 2(1-i)\) or \(2 - 2i\) | A1 | cao |
| (b) \(z^2 = (2-2i)(2-2i) = 4 - 8i + 4i^2\) | M1 | Squares their \(z\), or given \(z = \dfrac{4}{1+i}\), to produce at least 3 terms |
| \(= -8i\) | A1 | \(-8i\) or \(0 - 8i\) only |
| (c) If \(z\) is a root so is \(z^*\), so \((x - 2 + 2i)(x - 2 - 2i)\) (or \(x^2 - 2\text{Re}(z)\cdot x + | z | ^2\)) |
| \((x-2+2i)(x-2-2i) = 0\) and so \(p = q =\) | M1 | Multiplies two factors and obtains \(p=\) or \(q=\) |
| Equation is \(x^2 - 4x + 8 (= 0)\), or \(p = -4\) and \(q = 8\) | A1 | Both correct required |
| ALT1(c): Substitute \(z = 2-2i\) and \(z^2 = -8i\) into quadratic, equate real and imaginary parts: \(2p + q = 0\) and \(-2p - 8 = 0\) | M1 | Substitutes and equates real and imaginary parts |
| Solve simultaneously to get \(p = -4\), \(q = 8\) | M1 A1 | |
| ALT2(c): \(p = -\text{sum of roots}\); \(q = \text{product of roots}\) giving \(p=-4\), \(q=8\), equation \(x^2-4x+8(=0)\) | M1, M1, A1 | |
| ALT3(c): \(x - 2 = \pm 2i\); \((x-2)^2 = -4 \Rightarrow x^2 - 4x + 4 = -4\), equation \(x^2 - 4x + 8(=0)\), \(p=-4\), \(q=8\) | M1, M1, A1 |
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $z = \dfrac{4(1-i)}{(1+i)(1-i)}$ | M1 | Multiplies numerator and denominator by $1-i$ or $-1+i$ |
| $z = 2(1-i)$ or $2 - 2i$ | A1 | cao |
| **(b)** $z^2 = (2-2i)(2-2i) = 4 - 8i + 4i^2$ | M1 | Squares their $z$, or given $z = \dfrac{4}{1+i}$, to produce at least 3 terms |
| $= -8i$ | A1 | $-8i$ or $0 - 8i$ only |
| **(c)** If $z$ is a root so is $z^*$, so $(x - 2 + 2i)(x - 2 - 2i)$ (or $x^2 - 2\text{Re}(z)\cdot x + |z|^2$) | M1 | Uses their $z$ and $z^*$ in $(x-z)(x-z^*)$ |
| $(x-2+2i)(x-2-2i) = 0$ and so $p = q =$ | M1 | Multiplies two factors and obtains $p=$ or $q=$ |
| Equation is $x^2 - 4x + 8 (= 0)$, or $p = -4$ and $q = 8$ | A1 | Both correct required |
| **ALT1(c):** Substitute $z = 2-2i$ and $z^2 = -8i$ into quadratic, equate real and imaginary parts: $2p + q = 0$ and $-2p - 8 = 0$ | M1 | Substitutes and equates real and imaginary parts |
| Solve simultaneously to get $p = -4$, $q = 8$ | M1 A1 | |
| **ALT2(c):** $p = -\text{sum of roots}$; $q = \text{product of roots}$ giving $p=-4$, $q=8$, equation $x^2-4x+8(=0)$ | M1, M1, A1 | |
| **ALT3(c):** $x - 2 = \pm 2i$; $(x-2)^2 = -4 \Rightarrow x^2 - 4x + 4 = -4$, equation $x^2 - 4x + 8(=0)$, $p=-4$, $q=8$ | M1, M1, A1 | |4.
$$z = \frac { 4 } { 1 + \mathrm { i } }$$
Find, in the form $a + \mathrm { i } b$ where $a , b \in \mathbb { R }$
\begin{enumerate}[label=(\alph*)]
\item $Z$
\item $z ^ { 2 }$
Given that $z$ is a complex root of the quadratic equation $x ^ { 2 } + p x + q = 0$, where $p$ and $q$ are real integers,
\item find the value of $p$ and the value of $q$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2016 Q4 [7]}}