| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Standard +0.3 This is a straightforward complex numbers question requiring standard techniques: converting to exponential form, applying conditions about real products and moduli. Part (a) is routine conversion, part (b) involves setting up two simple equations from the given conditions. No novel insight required, just methodical application of A-level complex number properties. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02d Exponential form: re^(i*theta)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(r = 4\) | B1 | \( |
| Correct method for the argument | M1 | \(\theta = 2\tan^{-1}\left(\frac{-\sqrt{3}}{1}\right)\) or \(2 \times \frac{5\pi}{6}\) |
| Obtain \(\theta = -\frac{\pi}{3}\) | A1 | Arg with no working B1 instead of M1 A1. A0 if decimals. Allow separate mod and arg to gain full marks |
| Alternative: \(z^2 = 2 - 2\sqrt{3}\)i so \(r = \sqrt{2^2 + (-2\sqrt{3})^2} = 4\) | B1 | |
| Correct method for the argument | M1 | \(\arg z^2 = \tan^{-1}\frac{-2\sqrt{3}}{2}\) |
| Obtain \(\theta = -\frac{\pi}{3}\) | A1 | Arg with no working B1 instead of M1 A1. A0 if decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of \(\alpha + \text{their}\,\theta = 0\) or \(\alpha + \text{their}\,\theta = -\pi\) or \(\alpha + \text{their}\,\theta = \pi\) | M1 | Seen or implied. Using their \(\theta\) or new value calculated in (b) |
| Use of \(R = \frac{\text{their}\,r}{12}\) | M1 | Seen or implied |
| Obtain \(\frac{1}{3}e^{-i\frac{2\pi}{3}}\) and \(\frac{1}{3}e^{i\frac{\pi}{3}}\) | A1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $r = 4$ | B1 | $|z| = \sqrt{\left((-\sqrt{3})^2 + 1^2\right)}$ so $r = |z|^2 = (-\sqrt{3})^2 + 1^2$ |
| Correct method for the argument | M1 | $\theta = 2\tan^{-1}\left(\frac{-\sqrt{3}}{1}\right)$ or $2 \times \frac{5\pi}{6}$ |
| Obtain $\theta = -\frac{\pi}{3}$ | A1 | Arg with no working B1 instead of M1 A1. A0 if decimals. Allow separate mod and arg to gain full marks |
| **Alternative:** $z^2 = 2 - 2\sqrt{3}$i so $r = \sqrt{2^2 + (-2\sqrt{3})^2} = 4$ | B1 | |
| Correct method for the argument | M1 | $\arg z^2 = \tan^{-1}\frac{-2\sqrt{3}}{2}$ |
| Obtain $\theta = -\frac{\pi}{3}$ | A1 | Arg with no working B1 instead of M1 A1. A0 if decimals |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of $\alpha + \text{their}\,\theta = 0$ or $\alpha + \text{their}\,\theta = -\pi$ or $\alpha + \text{their}\,\theta = \pi$ | M1 | Seen or implied. Using their $\theta$ or new value calculated in (b) |
| Use of $R = \frac{\text{their}\,r}{12}$ | M1 | Seen or implied |
| Obtain $\frac{1}{3}e^{-i\frac{2\pi}{3}}$ and $\frac{1}{3}e^{i\frac{\pi}{3}}$ | A1 | |
---3 It is given that $z = - \sqrt { 3 } + \mathrm { i }$.
\begin{enumerate}[label=(\alph*)]
\item Express $z ^ { 2 }$ in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$.
\item The complex number $\omega$ is such that $z ^ { 2 } \omega$ is real and $\left| \frac { z ^ { 2 } } { \omega } \right| = 12$.
Find the two possible values of $\omega$, giving your answers in the form $R \mathrm { e } ^ { \mathrm { i } \alpha }$, where $R > 0$ and $- \pi < \alpha \leqslant \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q3 [6]}}