| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Argument calculations and identities |
| Difficulty | Standard +0.3 This is a multi-part question covering standard Core Pure AS topics. Part (a) requires recall of the complex conjugate root theorem. Part (b) is a routine argument calculation using the geometric interpretation. Part (c) follows directly from (b) by computing individual arguments. Part (d) is a standard perpendicular bisector locus. All parts are textbook exercises requiring no novel insight, making this slightly easier than average. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.02k Argand diagrams: geometric interpretation4.02m Geometrical effects: multiplication and division4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Complex roots of a real polynomial occur in conjugate pairs | M1 | 1.2 |
| So a polynomial with \(z_1\), \(z_2\) and \(z_3\) as roots also needs \(z_2^*\) and \(z_3^*\) as roots, so 5 roots in total, but a quartic has at most 4 roots, so no quartic can have \(z_1\), \(z_2\) and \(z_3\) as roots. | A1 | 2.4 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{z_2-z_1}{z_3-z_1} = \frac{-1+2i-(-2)}{1+i-(-2)} = \frac{1+2i}{3+i} \times \frac{3-i}{3-i} = \ldots\) | M1 | 1.1b |
| \(= \frac{3-i+6i+2}{9+1} = \frac{5+5i}{10} = \frac{1}{2}+\frac{1}{2}i\) oe | A1 | 1.1b |
| As \(\frac{1}{2}+\frac{1}{2}i\) is in the first quadrant, hence \(\arg\left(\frac{z_2-z_1}{z_3-z_1}\right) = \arctan\left(\frac{1/2}{1/2}\right) = \arctan(1) = \frac{\pi}{4}\) | A1* | 2.1 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\arg\left(\frac{z_2-z_1}{z_3-z_1}\right) = \arg(z_2-z_1)-\arg(z_3-z_1) = \arg(1+2i)-\arg(3+i)\) | M1 | 1.1b |
| Hence \(\arctan(2)-\arctan\left(\frac{1}{3}\right) = \frac{\pi}{4}\) | A1* | 2.1 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Line passing through \(z_2\) and the negative imaginary axis drawn. | B1 | 1.1b |
| Area below and left of their line shaded, where the line must have negative gradient passing through negative imaginary axis but need not pass through \(z_2\) | B1 | 1.1b |
| (2) | (9 marks) |
## Question 5:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Complex roots of a real polynomial occur in conjugate pairs | M1 | 1.2 |
| So a polynomial with $z_1$, $z_2$ and $z_3$ as roots also needs $z_2^*$ and $z_3^*$ as roots, so 5 roots in total, but a quartic has at most 4 roots, so no quartic can have $z_1$, $z_2$ and $z_3$ as roots. | A1 | 2.4 |
| **(2)** | | |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{z_2-z_1}{z_3-z_1} = \frac{-1+2i-(-2)}{1+i-(-2)} = \frac{1+2i}{3+i} \times \frac{3-i}{3-i} = \ldots$ | M1 | 1.1b |
| $= \frac{3-i+6i+2}{9+1} = \frac{5+5i}{10} = \frac{1}{2}+\frac{1}{2}i$ oe | A1 | 1.1b |
| As $\frac{1}{2}+\frac{1}{2}i$ is in the first quadrant, hence $\arg\left(\frac{z_2-z_1}{z_3-z_1}\right) = \arctan\left(\frac{1/2}{1/2}\right) = \arctan(1) = \frac{\pi}{4}$ | A1* | 2.1 |
| **(3)** | | |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\arg\left(\frac{z_2-z_1}{z_3-z_1}\right) = \arg(z_2-z_1)-\arg(z_3-z_1) = \arg(1+2i)-\arg(3+i)$ | M1 | 1.1b |
| Hence $\arctan(2)-\arctan\left(\frac{1}{3}\right) = \frac{\pi}{4}$ | A1* | 2.1 |
| **(2)** | | |
### Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| Line passing through $z_2$ and the negative imaginary axis drawn. | B1 | 1.1b |
| Area below and left of their line shaded, where the line must have negative gradient passing through negative imaginary axis but need not pass through $z_2$ | B1 | 1.1b |
| **(2)** | | **(9 marks)** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9312b91c-bca7-4427-a1f7-cb03065ee5e5-10_483_528_260_772}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The complex numbers $z _ { 1 } = - 2 , z _ { 2 } = - 1 + 2 \mathrm { i }$ and $z _ { 3 } = 1 + \mathrm { i }$ are plotted in Figure 1, on an Argand diagram for the complex plane with $z = x + \mathrm { i } y$
\begin{enumerate}[label=(\alph*)]
\item Explain why $z _ { 1 } , z _ { 2 }$ and $z _ { 3 }$ cannot all be roots of a quartic polynomial equation with real coefficients.
\item Show that $\arg \left( \frac { z _ { 2 } - z _ { 1 } } { z _ { 3 } - z _ { 1 } } \right) = \frac { \pi } { 4 }$
\item Hence show that $\arctan ( 2 ) - \arctan \left( \frac { 1 } { 3 } \right) = \frac { \pi } { 4 }$
A copy of Figure 1, labelled Diagram 1, is given on page 12.
\item Shade, on Diagram 1, the set of points of the complex plane that satisfy the inequality
$$| z + 2 | \leqslant | z - 1 - \mathrm { i } |$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9312b91c-bca7-4427-a1f7-cb03065ee5e5-12_479_524_296_776}
\end{center}
\section*{Diagram 1}
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP AS 2019 Q5 [9]}}