A2 June 2022 Paper 2 Q1
1. A student was asked to answer the following:
For the complex numbers \(z_1 = 3 - 3\mathrm{i}\) and \(z_2 = \sqrt{3} + \mathrm{i}\), find the value of \(\arg\left(\dfrac{z_1}{z_2}\right)\)
The student’s attempt is shown below.
| Line 1 → | \(\arg(z_1) = \tan^{-1}\left(\dfrac{3}{3}\right) = \dfrac{\pi}{4}\) |
| Line 2 → | \(\arg(z_2) = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}\) |
| Line 3 → | \(\arg\left(\dfrac{z_1}{z_2}\right) = \dfrac{\arg(z_1)}{\arg(z_2)}\) |
| Line 4 → | \(= \left(\dfrac{\pi}{4}\right)\Big/\left(\dfrac{\pi}{6}\right) = \dfrac{3}{2}\) |
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 (2)
(b) Write down the correct value of \(\arg\left(\dfrac{z_1}{z_2}\right)\) (1)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\{\arg(z_1) =\}\tan^{-1}\left(\dfrac{-3}{3}\right)\) or \(\{\arg(z_1) =\}\tan^{-1}(-1)\) or \(\{\arg(z_1) =\} -\tan^{-1}\left(\dfrac{3}{3}\right)\) or \(\{\arg(z_1) =\} -\dfrac{\pi}{4}\) or \(\{\arg(z_1) =\}\, 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}\) or states should be \(-3\) not 3 on top | B1 | 2.3 |
| (ii) States that \(\left\{\arg\left(\dfrac{z_1}{z_2}\right) =\right\}\arg(z_1) - \arg(z_2)\) Or states that the arguments should be subtracted | B1 | 2.3 |
| (2) |
Notes
(i) B1: See scheme, Condone \(-45\)
Any incorrect arguments seen is B0.
\(\arg(z_1) = \tan^{-1}\left(\dfrac{3}{-3}\right)\) is B0
Note: They used 3 instead of \(-3\) is B0, there are two 3’s in line 1 do they mean both should \(-3\)
It should be negative is B0
(ii) B1: See scheme
| Scheme | Marks | AO |
|---|---|---|
| \(\left\{\arg\left(\dfrac{z_1}{z_2}\right) = \left(\text{their } {-\dfrac{\pi}{4}}\right) - \dfrac{\pi}{6} =\right\} -\dfrac{5\pi}{12}\) Or \(\left\{\arg\left(\dfrac{z_1}{z_2}\right) = \left(\text{their } \dfrac{7\pi}{4}\right) - \dfrac{\pi}{6}\right\} = \dfrac{19\pi}{12}\) | B1ft | 2.2a |
| (1) | ||
| (3 marks) |
Notes
B1ft: States a correct value for \(\arg\left(\dfrac{z_1}{z_2}\right)\) Follow through on their answer to part (a) (i), do not ISW