| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Modulus and argument with equations |
| Difficulty | Moderate -0.3 This is a straightforward Further Pure 1 question testing basic complex number operations: modulus/argument calculation (standard formulas), algebraic manipulation to solve for z (routine division), and using argument condition to find a real parameter (direct application of tan = imaginary/real). All parts are textbook exercises requiring only direct application of standard techniques with no problem-solving insight needed. Slightly easier than average A-level due to its routine nature, though being FP1 places it at the easier end of further maths content. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \( | w | = \left\{ \sqrt{10^2 + (-5)^2} \right\} = \sqrt{125}\) or \(5\sqrt{5}\) or 11.1803... |
| (b) \(\arg w = -\tan^{-1}\left(\frac{5}{10}\right)\) | M1 | Use of \(\tan^{-1}\) or \(\tan\) |
| \(= -0.463647609... = -0.46\) (2 dp) | A1 oe | awrt \(-0.46\) or awrt 5.82 |
| Answer | Marks | Guidance |
|---|---|---|
| \(z + 3i = \frac{10 - 5i}{(2+i)}\) | B1 | Simplifies to give \(* = \frac{\text{complex no.}}{(2+i)}\) |
| \(z + 3i = \frac{(10-5i)}{(2+i)} \times \frac{(2-i)}{(2-i)}\) | M1 | Multiplies by \(\frac{\text{their }(2-i)}{\text{their }(2-i)}\) |
| \(z + 3i = \frac{20 - 10i - 10i - 5}{1+4}\) | M1 | Simplifies realising that a real number is needed on the denominator and applies \(i^2 = -1\) on their numerator expression and denominator expression. |
| \(z + 3i = \frac{15 - 20i}{5}\) | ||
| \(z + 3i = 3 - 4i\) | ||
| \(z = 3 - 7i\) | A1 | \(z = 3 - 7i\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda + 9i + w = \lambda + 9i + 10 - 5i = (\lambda + 10) + 4i\) | M1 | Combines real and imaginary parts and puts "Real part = Imaginary part" i.e. \(\frac{\lambda + 10}{4} = 1\) or \(\frac{4}{\lambda + 10} = 1\) o.e. |
| \(\arg(\lambda + 9i + w) = \frac{\pi}{4} \Rightarrow \lambda + 10 = 4\) | A1 | \(-6\) |
| So, \(\lambda = -6\) | [2] |
**(a)** $|w| = \left\{ \sqrt{10^2 + (-5)^2} \right\} = \sqrt{125}$ or $5\sqrt{5}$ or 11.1803... | B1 | $\sqrt{125}$ or $5\sqrt{5}$ or awrt 11.2 | [1]
**(b)** $\arg w = -\tan^{-1}\left(\frac{5}{10}\right)$ | M1 | Use of $\tan^{-1}$ or $\tan$
$= -0.463647609... = -0.46$ (2 dp) | A1 oe | awrt $-0.46$ or awrt 5.82 | [2]
**(c)** $(2+i)(z+3i) = w$
$z + 3i = \frac{10 - 5i}{(2+i)}$ | B1 | Simplifies to give $* = \frac{\text{complex no.}}{(2+i)}$
$z + 3i = \frac{(10-5i)}{(2+i)} \times \frac{(2-i)}{(2-i)}$ | M1 | Multiplies by $\frac{\text{their }(2-i)}{\text{their }(2-i)}$
$z + 3i = \frac{20 - 10i - 10i - 5}{1+4}$ | M1 | Simplifies realising that a real number is needed on the denominator and applies $i^2 = -1$ on their numerator expression and denominator expression.
$z + 3i = \frac{15 - 20i}{5}$ | |
$z + 3i = 3 - 4i$ | |
$z = 3 - 7i$ | A1 | $z = 3 - 7i$ | [4]
**(d)** $\arg(\lambda + 9i + w) = \frac{\pi}{4}$
$\lambda + 9i + w = \lambda + 9i + 10 - 5i = (\lambda + 10) + 4i$ | M1 | Combines real and imaginary parts and puts "Real part = Imaginary part" i.e. $\frac{\lambda + 10}{4} = 1$ or $\frac{4}{\lambda + 10} = 1$ o.e.
$\arg(\lambda + 9i + w) = \frac{\pi}{4} \Rightarrow \lambda + 10 = 4$ | A1 | $-6$
So, $\lambda = -6$ | | [2]
**Alt 1: Scheme as above:**
$(2 + i)z + 6i + 3i^2 = 10 - 5i \Rightarrow (2+i)z = 13 - 11i$
B1 for $z = \frac{13-11i}{2+i}$; M1 for $z = \frac{(13-11i) \times (2-i)}{(2+i) \times (2-i)}$; M1 for $z = \frac{26-13i-22i-11}{4+1}$; A1 for $z = 3 - 7i$
**Alt 2: Let** $z = a + ib$ gives $(2+i)(a+ib+3i) = 10-5i$ for B1. Equating real and imaginary parts to form two equations both involving $a$ and $b$ for M1. Solves simultaneous equations as far as $a = $ or $b = $ for M1. $a = 3, b = -7$ or $z = 3 - 7i$ for A1.
**Total: [9]**
---The complex number $w$ is given by
$$w = 10 - 5\text{i}$$
\begin{enumerate}[label=(\alph*)]
\item Find $|w|$. [1]
\item Find $\arg w$, giving your answer in radians to 2 decimal places. [2]
\end{enumerate}
The complex numbers $z$ and $w$ satisfy the equation
$$(2 + \text{i})(z + 3\text{i}) = w$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use algebra to find $z$, giving your answer in the form $a + b\text{i}$, where $a$ and $b$ are real numbers. [4]
\end{enumerate}
Given that
$$\arg(\lambda + 9\text{i} + w) = \frac{\pi}{4}$$
where $\lambda$ is a real constant,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item find the value of $\lambda$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q9 [9]}}