Edexcel FP1 2011 June — Question 2 8 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeStandard quadratic with real coefficients
DifficultyModerate -0.8 This is a routine FP1 question testing basic complex number skills: modulus/argument calculation using standard formulas, solving a quadratic with the quadratic formula to get complex roots, and plotting on an Argand diagram. All parts are straightforward applications of standard techniques with no problem-solving or insight required, making it easier than average even for Further Maths.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02i Quadratic equations: with complex roots4.02k Argand diagrams: geometric interpretation

2. $$z _ { 1 } = - 2 + \mathrm { i }$$
  1. Find the modulus of \(z _ { 1 }\).
  2. Find, in radians, the argument of \(z _ { 1 }\), giving your answer to 2 decimal places. The solutions to the quadratic equation $$z ^ { 2 } - 10 z + 28 = 0$$ are \(z _ { 2 }\) and \(z _ { 3 }\).
  3. Find \(z _ { 2 }\) and \(z _ { 3 }\), giving your answers in the form \(p \pm i \sqrt { } q\), where \(p\) and \(q\) are integers.
  4. Show, on an Argand diagram, the points representing your complex numbers \(z _ { 1 } , z _ { 2 }\) and \(z _ { 3 }\).

Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z_1 = \sqrt{(-2)^2 + 1^2} = \sqrt{5} = 2.236...\)
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\arg z = \pi - \tan^{-1}\left(\frac{1}{2}\right)\)M1 \(\tan^{-1}\!\left(\frac{1}{2}\right)\) or \(\tan^{-1}\!\left(\frac{2}{1}\right)\) or \(\cos^{-1}\!\left(\frac{2}{\sqrt{5}}\right)\) or \(\sin^{-1}\!\left(\frac{2}{\sqrt{5}}\right)\) or \(\sin^{-1}\!\left(\frac{1}{\sqrt{5}}\right)\) or \(\cos^{-1}\!\left(\frac{1}{\sqrt{5}}\right)\)
\(= 2.677945045... = 2.68\) (2 dp)A1 oe awrt 2.68
> Note: \(\arg z = \tan^{-1}\!\left(\frac{1}{2}\right) = -0.46\) on its own is M0; but \(\pi + \tan^{-1}\!\left(\frac{1}{2}\right) = 2.68\) scores M1A1; \(\pi - \tan^{-1}\!\left(\frac{1}{2}\right)\) is M0 as is \(\pi - \tan\!\left(\frac{1}{2}\right)\) (2.60)
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z^2 - 10z + 28 = 0\), \(z = \dfrac{10 \pm \sqrt{100 - 4(1)(28)}}{2(1)}\)M1 An attempt to use the quadratic formula (usual rules)
\(= \dfrac{10 \pm \sqrt{-12}}{2}\) then attempt to simplify \(\sqrt{-12}\) in terms of \(i\)M1 E.g. \(i\sqrt{12}\) or \(i\sqrt{3 \times 4}\). If \(b^2 - 4ac > 0\) then only first M1 available
\(z = 5 \pm \sqrt{3}\,i\), \(\{p=5,\, q=3\}\)A1 oe \(5 \pm \sqrt{3}\,i\)
> Note: Correct answers with no working scores full marks.
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Point \((-2, 1)\) plotted correctly on Argand diagramB1 With or without label. Points are \((-2,1)\), \((5, \sqrt{3})\) and \((5, -\sqrt{3})\)
Distinct points \(z_2\) and \(z_3\) plotted correctly and symmetrically about the \(x\)-axisB1\(\checkmark\) With or without label
> Note: The second B mark in (d) depends on having obtained complex numbers in (c).
# Question 2:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|z_1| = \sqrt{(-2)^2 + 1^2} = \sqrt{5} = 2.236...$ | B1 | $\sqrt{5}$ or awrt 2.24 |

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\arg z = \pi - \tan^{-1}\left(\frac{1}{2}\right)$ | M1 | $\tan^{-1}\!\left(\frac{1}{2}\right)$ or $\tan^{-1}\!\left(\frac{2}{1}\right)$ or $\cos^{-1}\!\left(\frac{2}{\sqrt{5}}\right)$ or $\sin^{-1}\!\left(\frac{2}{\sqrt{5}}\right)$ or $\sin^{-1}\!\left(\frac{1}{\sqrt{5}}\right)$ or $\cos^{-1}\!\left(\frac{1}{\sqrt{5}}\right)$ |
| $= 2.677945045... = 2.68$ (2 dp) | A1 oe | awrt 2.68 |

> **Note:** $\arg z = \tan^{-1}\!\left(\frac{1}{2}\right) = -0.46$ on its own is M0; but $\pi + \tan^{-1}\!\left(\frac{1}{2}\right) = 2.68$ scores M1A1; $\pi - \tan^{-1}\!\left(\frac{1}{2}\right)$ is M0 as is $\pi - \tan\!\left(\frac{1}{2}\right)$ (2.60)

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z^2 - 10z + 28 = 0$, $z = \dfrac{10 \pm \sqrt{100 - 4(1)(28)}}{2(1)}$ | M1 | An attempt to use the quadratic formula (usual rules) |
| $= \dfrac{10 \pm \sqrt{-12}}{2}$ then attempt to simplify $\sqrt{-12}$ in terms of $i$ | M1 | E.g. $i\sqrt{12}$ or $i\sqrt{3 \times 4}$. If $b^2 - 4ac > 0$ then only first M1 available |
| $z = 5 \pm \sqrt{3}\,i$, $\{p=5,\, q=3\}$ | A1 oe | $5 \pm \sqrt{3}\,i$ |

> **Note:** Correct answers with no working scores full marks.

## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Point $(-2, 1)$ plotted correctly on Argand diagram | B1 | With or without label. Points are $(-2,1)$, $(5, \sqrt{3})$ and $(5, -\sqrt{3})$ |
| Distinct points $z_2$ and $z_3$ plotted correctly and symmetrically about the $x$-axis | B1$\checkmark$ | With or without label |

> **Note:** The second B mark in (d) depends on having obtained complex numbers in (c).

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2.

$$z _ { 1 } = - 2 + \mathrm { i }$$
\begin{enumerate}[label=(\alph*)]
\item Find the modulus of $z _ { 1 }$.
\item Find, in radians, the argument of $z _ { 1 }$, giving your answer to 2 decimal places.

The solutions to the quadratic equation

$$z ^ { 2 } - 10 z + 28 = 0$$

are $z _ { 2 }$ and $z _ { 3 }$.
\item Find $z _ { 2 }$ and $z _ { 3 }$, giving your answers in the form $p \pm i \sqrt { } q$, where $p$ and $q$ are integers.
\item Show, on an Argand diagram, the points representing your complex numbers $z _ { 1 } , z _ { 2 }$ and $z _ { 3 }$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP1 2011 Q2 [8]}}