| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Parameter from argument condition |
| Difficulty | Standard +0.3 This is a standard FP1 complex number question requiring multiplication by conjugate to reach Cartesian form, then using the argument condition to solve for p. The techniques are routine for Further Maths students, though part (b) requires recognizing tan θ = 1 means the real and imaginary parts are equal, leading to a simple equation in p. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = \frac{p+2i}{3+pi} \cdot \frac{3-pi}{3-pi}\) | M1 | Multiplying top and bottom by conjugate |
| \(= \frac{3p - p^2i + 6i + 2p}{9 + p^2}\) | M1 | At least 3 correct terms in numerator, evidence that \(i^2 = -1\) and denominator real |
| \(= \frac{5p}{p^2+9} + \frac{6-p^2}{p^2+9}i\) | A1, A1 | Real + imaginary with \(i\) factored out. Accept single denominator with numerator in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\arg(z) = \arctan\left(\frac{\frac{6-p^2}{p^2+9}}{\frac{5p}{p^2+9}}\right)\) | M1 | Correct method for the argument, can be implied by correct equation for \(p\) |
| \(\frac{6-p^2}{5p} = 1\) | M1 | Their \(\arg(z)\) in terms of \(p = 1\) |
| \(p^2 + 5p - 6 = 0\) | A1 | Correct 3TQ |
| \((p+6)(p-1) = 0\) | M1 | Attempt to solve their quadratic in \(p\) |
| \(p = 1,\ p = -6\) | A1 | both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a + bi = \frac{p+2i}{3+pi}\) | M1 | Equate to \(a+bi\) then rearrange and equate real and imaginary parts |
| \(3a - pb = p,\ ap + 3b = 2\) | dM1 | Two equations for \(a\) and \(b\) in terms of \(p\), attempt to solve for \(a\) and \(b\) in terms of \(p\) |
| \(= \frac{5p}{p^2+9} + \frac{6-p^2}{p^2+9}i\) | A1, A1 | Real + imaginary with \(i\) factored out |
## Question 4:
### Part (a) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = \frac{p+2i}{3+pi} \cdot \frac{3-pi}{3-pi}$ | M1 | Multiplying top and bottom by conjugate |
| $= \frac{3p - p^2i + 6i + 2p}{9 + p^2}$ | M1 | At least 3 correct terms in numerator, evidence that $i^2 = -1$ and denominator real |
| $= \frac{5p}{p^2+9} + \frac{6-p^2}{p^2+9}i$ | A1, A1 | Real + imaginary with $i$ factored out. Accept single denominator with numerator in correct form |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\arg(z) = \arctan\left(\frac{\frac{6-p^2}{p^2+9}}{\frac{5p}{p^2+9}}\right)$ | M1 | Correct method for the argument, can be implied by correct equation for $p$ |
| $\frac{6-p^2}{5p} = 1$ | M1 | Their $\arg(z)$ in terms of $p = 1$ |
| $p^2 + 5p - 6 = 0$ | A1 | Correct 3TQ |
| $(p+6)(p-1) = 0$ | M1 | Attempt to solve their quadratic in $p$ |
| $p = 1,\ p = -6$ | A1 | **both** |
### Part (a) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a + bi = \frac{p+2i}{3+pi}$ | M1 | Equate to $a+bi$ then rearrange and equate real and imaginary parts |
| $3a - pb = p,\ ap + 3b = 2$ | dM1 | Two equations for $a$ and $b$ in terms of $p$, attempt to solve for $a$ and $b$ in terms of $p$ |
| $= \frac{5p}{p^2+9} + \frac{6-p^2}{p^2+9}i$ | A1, A1 | Real + imaginary with $i$ factored out |
---4. The complex number $z$ is given by
$$z = \frac { p + 2 \mathrm { i } } { 3 + p \mathrm { i } }$$
where $p$ is an integer.
\begin{enumerate}[label=(\alph*)]
\item Express $z$ in the form $a + b \mathrm { i }$ where $a$ and $b$ are real. Give your answer in its simplest form in terms of $p$.
\item Given that $\arg ( z ) = \theta$, where $\tan \theta = 1$ find the possible values of $p$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q4 [9]}}