| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Argument calculations and identities |
| Difficulty | Moderate -0.3 This is a straightforward error-spotting question on argument calculations. Students need to identify that arg(z₁/z₂) = arg(z₁) - arg(z₂) (not addition) and correctly calculate arg(z₁) = -π/4 (not +π/4). While it tests understanding of argument properties, it requires only basic recall and simple calculations with standard complex numbers, making it slightly easier than average. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{arg(z_1) =\} \tan^{-1}\left(\frac{-3}{3}\right)\) or \(\{arg(z_1) =\} \tan^{-1}(-1)\) or \(\{arg(z_1) =\} -\tan^{-1}\left(\frac{3}{3}\right)\) or \(\{arg(z_1) =\} -\frac{\pi}{4}\) or \(\{arg(z_1) =\} 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\) or states should be \(-3\) not \(3\) on top | B1 | AO2.3. See scheme, condone \(-45\). Any incorrect arguments seen is B0. \(arg(z_1) = \tan^{-1}\left(\frac{3}{-3}\right)\) is B0. Note: if they used 3 instead of \(-3\), there are two 3's in line 1 — do they mean both should be \(-3\)? It should be negative is B0. |
| States that \(\left\{arg\left(\frac{z_1}{z_2}\right) =\right\} arg(z_1) - arg(z_2)\) or states that the arguments should be subtracted | B1 | AO2.3. See scheme. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{arg\left(\frac{z_1}{z_2}\right)\right\} = \left(\text{their } -\frac{\pi}{4}\right) - \frac{\pi}{6} = -\frac{5\pi}{12}\) OR \(\left\{arg\left(\frac{z_1}{z_2}\right)\right\} = \left(\text{their } \frac{7\pi}{4}\right) - \frac{\pi}{6} = \frac{19\pi}{12}\) | B1ft | AO2.2a. States a correct value for \(arg\left(\frac{z_1}{z_2}\right)\). Follow through on their answer to part (a)(i). Do not ISW. |
# Question 1:
## Part (a)(i) and (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{arg(z_1) =\} \tan^{-1}\left(\frac{-3}{3}\right)$ or $\{arg(z_1) =\} \tan^{-1}(-1)$ or $\{arg(z_1) =\} -\tan^{-1}\left(\frac{3}{3}\right)$ or $\{arg(z_1) =\} -\frac{\pi}{4}$ or $\{arg(z_1) =\} 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$ or states should be $-3$ not $3$ on top | B1 | AO2.3. See scheme, condone $-45$. Any incorrect arguments seen is B0. $arg(z_1) = \tan^{-1}\left(\frac{3}{-3}\right)$ is B0. Note: if they used 3 instead of $-3$, there are two 3's in line 1 — do they mean both should be $-3$? It should be negative is B0. |
| States that $\left\{arg\left(\frac{z_1}{z_2}\right) =\right\} arg(z_1) - arg(z_2)$ or states that the arguments should be subtracted | B1 | AO2.3. See scheme. |
**Total: (2 marks)**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{arg\left(\frac{z_1}{z_2}\right)\right\} = \left(\text{their } -\frac{\pi}{4}\right) - \frac{\pi}{6} = -\frac{5\pi}{12}$ OR $\left\{arg\left(\frac{z_1}{z_2}\right)\right\} = \left(\text{their } \frac{7\pi}{4}\right) - \frac{\pi}{6} = \frac{19\pi}{12}$ | B1ft | AO2.2a. States a correct value for $arg\left(\frac{z_1}{z_2}\right)$. Follow through on their answer to part (a)(i). Do not ISW. |
**Total: (1 mark)**
**(3 marks total)**\begin{enumerate}
\item A student was asked to answer the following:
\end{enumerate}
For the complex numbers $z _ { 1 } = 3 - 3 \mathrm { i }$ and $z _ { 2 } = \sqrt { 3 } + \mathrm { i }$, find the value of $\arg \left( \frac { z _ { 1 } } { z _ { 2 } } \right)$ The student's attempt is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{33292670-3ad0-4125-a3bb-e4b7b21ed5f4-02_798_1109_534_338}
The student made errors in line 1 and line 3\\
Correct the error that the student made in\\
(a) (i) line 1\\
(ii) line 3\\
(b) Write down the correct value of $\arg \left( \frac { z _ { 1 } } { z _ { 2 } } \right)$
\hfill \mbox{\textit{Edexcel CP2 2022 Q1 [3]}}