| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Optimization of modulus on loci |
| Difficulty | Standard +0.8 Part (i) is a standard verification requiring conversion between exponential and Cartesian forms, then back to exponential form—routine but multi-step. Part (ii) requires geometric interpretation of an arg locus as a half-line, then finding the minimum distance from origin to this line using perpendicular distance, which demands spatial reasoning beyond standard textbook exercises. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z_1 = 6\left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right] = \{3 + 3\sqrt{3}i\}\) and \(z_2 = 6\sqrt{3}\left[\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right] = \{-9 + 3\sqrt{3}i\}\), then adds to get \(\{-6 + 6\sqrt{3}i\}\) | M1 | Complete method to find both \(z_1\) and \(z_2\) in \(a+bi\) form and add |
| Shows \(\ | z_1+z_2\ | = \sqrt{6^2+(6\sqrt{3})^2} = 12\) and \(\arg(z_1+z_2) = \pi - \tan^{-1}\left(\frac{6\sqrt{3}}{6}\right) = \frac{2\pi}{3}\) |
| \(z_1 + z_2 = 12e^{\frac{2\pi}{3}i}\) | A1* | Achieves correct answer with no errors or omissions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Draws diagram recognising shortest distance forms a right-angled triangle with \(\ | z_1+z_2\ | =12\) and argument \(\frac{2\pi}{3}\), point at distance 5 on real axis |
| \(\sin\left(\frac{\pi}{3}\right) = \frac{\ | z\ | }{5} \Rightarrow \ |
| \(\ | z\ | = \frac{5\sqrt{3}}{2}\) |
## Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z_1 = 6\left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right] = \{3 + 3\sqrt{3}i\}$ and $z_2 = 6\sqrt{3}\left[\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right] = \{-9 + 3\sqrt{3}i\}$, then adds to get $\{-6 + 6\sqrt{3}i\}$ | M1 | Complete method to find both $z_1$ and $z_2$ in $a+bi$ form and add |
| Shows $\|z_1+z_2\| = \sqrt{6^2+(6\sqrt{3})^2} = 12$ and $\arg(z_1+z_2) = \pi - \tan^{-1}\left(\frac{6\sqrt{3}}{6}\right) = \frac{2\pi}{3}$ | dM1 | Dependent on M1; must show method for both modulus and argument |
| $z_1 + z_2 = 12e^{\frac{2\pi}{3}i}$ | A1* | Achieves correct answer with no errors or omissions |
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## Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Draws diagram recognising shortest distance forms a right-angled triangle with $\|z_1+z_2\|=12$ and argument $\frac{2\pi}{3}$, point at distance 5 on real axis | M1 | Draws diagram and recognises shortest distance forms right-angled triangle |
| $\sin\left(\frac{\pi}{3}\right) = \frac{\|z\|}{5} \Rightarrow \|z\| = \ldots$ | M1 | Uses trigonometry to find shortest length |
| $\|z\| = \frac{5\sqrt{3}}{2}$ | A1 | Correct exact value |
---\begin{enumerate}
\item (i) Given that
\end{enumerate}
$$z _ { 1 } = 6 \mathrm { e } ^ { \frac { \pi } { 3 } \mathrm { i } } \text { and } z _ { 2 } = 6 \sqrt { 3 } \mathrm { e } ^ { \frac { 5 \pi } { 6 } \mathrm { i } }$$
show that
$$z _ { 1 } + z _ { 2 } = 12 \mathrm { e } ^ { \frac { 2 \pi } { 3 } \mathrm { i } }$$
(ii) Given that
$$\arg ( z - 5 ) = \frac { 2 \pi } { 3 }$$
determine the least value of $| z |$ as $z$ varies.
\hfill \mbox{\textit{Edexcel CP2 2022 Q4 [6]}}