| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Modulus and argument calculations |
| Difficulty | Standard +0.8 This question requires finding an unknown parameter from a given argument condition (involving algebraic manipulation and understanding of complex division), then converting to Cartesian form, and finally to exponential form with exact values. The multi-step nature, need to handle the argument condition algebraically, and requirement for exact exponential form elevate this above standard complex number exercises, though it remains within typical Further Maths scope. |
| Spec | 4.02b Express complex numbers: cartesian and modulus-argument forms4.02f Convert between forms: cartesian and modulus-argument4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Multiply numerator and denominator by \((3-ai)\) | M1 | Must perform complete multiplications but need not simplify \(i^2\). \(\frac{(5a-2i)(3-ai)}{9-a^2}=\frac{13a-i(5a^2+6)}{9-a^2}\). No working so unsure if denominator multiplied by \(3-ai\): M0 |
| Use \(i^2=-1\) at least once and separate real and imaginary parts | M1 | |
| Obtain \(\frac{13a-i(5a^2+6)}{9+a^2}\) or \(\frac{13a-5a^2i-6i}{9+a^2}\) | A1 | OE. If \(15a-2a=13a\) seen later award this A1 |
| Use \(\arg z\) to form equation in \(a\): \(-\frac{5a^2+6}{13a}=\pm\tan(\pm\frac{\pi}{4})\) or \(-\frac{13a}{5a^2+6}=\pm\tan(\pm\frac{\pi}{4})\) or \(\tan^{-1}(-\frac{5a^2+6}{13a})=\pm\frac{\pi}{4}\) or \(\tan^{-1}(-\frac{13a}{5a^2+6})=\pm\frac{\pi}{4}\) | M1 | Allow expression given in answer column or \(5a^2+6=\pm 13a\) or use \(-(x\pm xi)=(13a-i(5a^2+6))/(9+a^2)\) and eliminate \(x\) so \(5a^2+6=\pm 13a\) |
| Obtain \(a=2\) | A1 | Need to reject \(a=\frac{3}{5}\) or ignore it in future work. May not see second root, but if present, must be \(\frac{3}{5}\) |
| Obtain \(z=2-2i\) only | A1 | Allow \(z=-2i+2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\arg z = \arg(5a-2i)-\arg(3+ai)\) | M1 | |
| \(=\tan^{-1}(\frac{-2}{5a})-\tan^{-1}(\frac{a}{3})\) | M1 | Allow one sign error in second M1 |
| \(=\tan^{-1}\left(\frac{-2/5a - a/3}{1+(-2/5a)(a/3)}\right)=\tan^{-1}\left(-\frac{5a^2+6}{13a}\right)\) or \(\tan^{-1}\left(-\frac{13a}{5a^2+6}\right)\) | A1 | |
| \(\pm\frac{\pi}{4}=\tan^{-1}\left(-\frac{5a^2+6}{13a}\right)\) or \(\tan^{-1}\left(-\frac{13a}{5a^2+6}\right)\) | M1 | Equate their \(\tan^{-1}\left(-\frac{5a^2+6}{13a}\right)\) to \(\pm\frac{\pi}{4}\). Then as original scheme for final 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x+iy)(3+ai)=5a-2i\), so \(3x-ay=5a\) and \(ax+3y=-2\) | M1A1 | |
| \(x=\pm y\); find \(x\) or \(y\) in terms of \(a\), e.g. \(x=\frac{2}{3-a}\) or \(x=\frac{5a}{3+a}\) | M1 | |
| Substitute in other equation, e.g. \(3(\frac{2}{3-a})+a(\frac{2}{3-a})=5a\) | M1 | Then as original scheme for final 2 marks |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(\arg(z^3)=-\frac{3}{4}\pi\) or evaluate from \(z=b-bi\) or from \(-2b^3(1+i)\) | B1 | If 2 different values given award B0. Do not ISW |
| Complete method to obtain \(r\) from their \(z\) | M1 | \(\ |
| \(r=16\sqrt{2}\) | A1 | CAO. A1 if \(z=2-2i\) obtained correctly, or \(z=\) used with \(a=2\) found correctly, otherwise A0XP. May see arg and \(r\) given in final answer i.e. \(16\sqrt{2}e^{-\frac{3}{4}\pi i}\). Allow this form for arg and \(r\) to collect full marks, even if \(i\) missing. Ignore answers outside the given interval. If 2 different values given award A0 |
| 3 |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Multiply numerator and denominator by $(3-ai)$ | M1 | Must perform complete multiplications but need not simplify $i^2$. $\frac{(5a-2i)(3-ai)}{9-a^2}=\frac{13a-i(5a^2+6)}{9-a^2}$. No working so unsure if denominator multiplied by $3-ai$: M0 |
| Use $i^2=-1$ at least once and separate real and imaginary parts | M1 | |
| Obtain $\frac{13a-i(5a^2+6)}{9+a^2}$ or $\frac{13a-5a^2i-6i}{9+a^2}$ | A1 | OE. If $15a-2a=13a$ seen later award this A1 |
| Use $\arg z$ to form equation in $a$: $-\frac{5a^2+6}{13a}=\pm\tan(\pm\frac{\pi}{4})$ or $-\frac{13a}{5a^2+6}=\pm\tan(\pm\frac{\pi}{4})$ or $\tan^{-1}(-\frac{5a^2+6}{13a})=\pm\frac{\pi}{4}$ or $\tan^{-1}(-\frac{13a}{5a^2+6})=\pm\frac{\pi}{4}$ | M1 | Allow expression given in answer column or $5a^2+6=\pm 13a$ or use $-(x\pm xi)=(13a-i(5a^2+6))/(9+a^2)$ and eliminate $x$ so $5a^2+6=\pm 13a$ |
| Obtain $a=2$ | A1 | Need to reject $a=\frac{3}{5}$ or ignore it in future work. May not see second root, but if present, must be $\frac{3}{5}$ |
| Obtain $z=2-2i$ only | A1 | Allow $z=-2i+2$ |
### Alternative Method 1 for Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\arg z = \arg(5a-2i)-\arg(3+ai)$ | M1 | |
| $=\tan^{-1}(\frac{-2}{5a})-\tan^{-1}(\frac{a}{3})$ | M1 | Allow one sign error in second M1 |
| $=\tan^{-1}\left(\frac{-2/5a - a/3}{1+(-2/5a)(a/3)}\right)=\tan^{-1}\left(-\frac{5a^2+6}{13a}\right)$ or $\tan^{-1}\left(-\frac{13a}{5a^2+6}\right)$ | A1 | |
| $\pm\frac{\pi}{4}=\tan^{-1}\left(-\frac{5a^2+6}{13a}\right)$ or $\tan^{-1}\left(-\frac{13a}{5a^2+6}\right)$ | M1 | Equate their $\tan^{-1}\left(-\frac{5a^2+6}{13a}\right)$ to $\pm\frac{\pi}{4}$. Then as original scheme for final 2 marks |
### Alternative Method 2 for Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+iy)(3+ai)=5a-2i$, so $3x-ay=5a$ and $ax+3y=-2$ | M1A1 | |
| $x=\pm y$; find $x$ or $y$ in terms of $a$, e.g. $x=\frac{2}{3-a}$ or $x=\frac{5a}{3+a}$ | M1 | |
| Substitute in other equation, e.g. $3(\frac{2}{3-a})+a(\frac{2}{3-a})=5a$ | M1 | Then as original scheme for final 2 marks |
| | **6** | |
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $\arg(z^3)=-\frac{3}{4}\pi$ or evaluate from $z=b-bi$ or from $-2b^3(1+i)$ | B1 | If 2 different values given award B0. Do not ISW |
| Complete method to obtain $r$ from their $z$ | M1 | $\|z^3\|=(\sqrt{x^2+y^2})^3$. If $z$ correct, may see $\|z^3\|=(\sqrt{2^2+(-2)^2})^3$ or $\|z^3\|=\sqrt{(-16)^2+(-16)^2}$ |
| $r=16\sqrt{2}$ | A1 | CAO. A1 if $z=2-2i$ obtained correctly, or $z=$ used with $a=2$ found correctly, otherwise A0XP. May see arg and $r$ given in final answer i.e. $16\sqrt{2}e^{-\frac{3}{4}\pi i}$. Allow this form for arg and $r$ to collect full marks, even if $i$ missing. Ignore answers outside the given interval. If 2 different values given award A0 |
| | **3** | |11 The complex number $z$ is defined by $z = \frac { 5 a - 2 \mathrm { i } } { 3 + a \mathrm { i } }$, where $a$ is an integer. It is given that $\arg z = - \frac { 1 } { 4 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and hence express $z$ in the form $x + \mathrm { i } y$, where $x$ and $y$ are real.
\item Express $z ^ { 3 }$ in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $- \pi < \theta \leqslant \pi$. Give the simplified exact values of $r$ and $\theta$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q11 [9]}}