| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Parameter from modulus condition |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring division of complex numbers by multiplying by the conjugate, then applying the modulus formula to solve for p. While it's FP1 material (inherently slightly harder), it follows a standard algorithmic procedure with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{z_1}{z_2} = \frac{p+2i}{1-2i} \cdot \frac{1+2i}{1+2i}\) | M1 | Multiplying top and bottom by conjugate |
| \(= \frac{p+2pi+2i-4}{5}\) | M1 | At least 3 correct terms in numerator, evidence that \(i^2=-1\) and denominator real |
| \(= \frac{p-4}{5} + \frac{2p+2}{5}i\) | A1, A1 | Real + imaginary with \(i\) factored out. Accept single denominator with numerator in correct form. Accept '\(a=\)' and '\(b=\)' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left | \frac{z_1}{z_2}\right | ^2 = \left(\frac{p-4}{5}\right)^2 + \left(\frac{2p+2}{5}\right)^2\) |
| \(\left(\frac{p-4}{5}\right)^2 + \left(\frac{2p+2}{5}\right)^2 = 13^2\) | dM1 | \(\left |
| \(\frac{p^2-8p+16}{25} + \frac{4p^2+8p+4}{25} = 169\) | ||
| \(5p^2 + 20 = 4225\) | ||
| \(p^2 = 841 \Rightarrow p = \pm 29\) | dM1A1 | dM1: Attempt to solve quadratic in \(p\), dependent on both previous Ms. A1: both 29 and -29 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{ | z_1 | }{ |
| \(\frac{\sqrt{p^2+4}}{\sqrt{5}} = 13\) | dM1 | Equating to 13 |
| \(\frac{p^2+4}{5} = 169\) or \(13^2\) | ||
| \(p^2 = 841 \Rightarrow p = \pm 29\) | dM1A1 | dM1: Attempt to solve quadratic in \(p\), dependent on both previous Ms. A1: both 29 and -29 |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{z_1}{z_2} = \frac{p+2i}{1-2i} \cdot \frac{1+2i}{1+2i}$ | M1 | Multiplying top and bottom by conjugate |
| $= \frac{p+2pi+2i-4}{5}$ | M1 | At least 3 correct terms in numerator, evidence that $i^2=-1$ and denominator real |
| $= \frac{p-4}{5} + \frac{2p+2}{5}i$ | A1, A1 | Real + imaginary with $i$ factored out. Accept single denominator with numerator in correct form. Accept '$a=$' and '$b=$' |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left|\frac{z_1}{z_2}\right|^2 = \left(\frac{p-4}{5}\right)^2 + \left(\frac{2p+2}{5}\right)^2$ | M1 | Accept their answers to part (a). Any erroneous $i$ or $i^2$ award M0 |
| $\left(\frac{p-4}{5}\right)^2 + \left(\frac{2p+2}{5}\right)^2 = 13^2$ | dM1 | $\left|\frac{z_1}{z_2}\right|^2 = 13^2$ or $\left|\frac{z_1}{z_2}\right| = 13$ |
| $\frac{p^2-8p+16}{25} + \frac{4p^2+8p+4}{25} = 169$ | | |
| $5p^2 + 20 = 4225$ | | |
| $p^2 = 841 \Rightarrow p = \pm 29$ | dM1A1 | dM1: Attempt to solve quadratic in $p$, dependent on both previous Ms. A1: **both** 29 **and** -29 |
**OR alternative method:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{|z_1|}{|z_2|} = \frac{\sqrt{p^2+4}}{\sqrt{5}}$ | M1 | Finding moduli. Any erroneous $i$ or $i^2$ award M0 |
| $\frac{\sqrt{p^2+4}}{\sqrt{5}} = 13$ | dM1 | Equating to 13 |
| $\frac{p^2+4}{5} = 169$ or $13^2$ | | |
| $p^2 = 841 \Rightarrow p = \pm 29$ | dM1A1 | dM1: Attempt to solve quadratic in $p$, dependent on both previous Ms. A1: **both** 29 **and** -29 |
---\begin{enumerate}
\item The complex numbers $z _ { 1 }$ and $z _ { 2 }$ are given by
\end{enumerate}
$$z _ { 1 } = p + 2 i \text { and } z _ { 2 } = 1 - 2 i$$
where $p$ is an integer.\\
(a) Find $\frac { z _ { 1 } } { z _ { 2 } }$ in the form $a + b$ i where $a$ and $b$ are real. Give your answer in its simplest form in terms of $p$.
Given that $\left| \frac { z _ { 1 } } { z _ { 2 } } \right| = 13$,\\
(b) find the possible values of $p$.\\
\hfill \mbox{\textit{Edexcel FP1 2014 Q1 [8]}}