| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Roots of unity applications |
| Difficulty | Challenging +1.2 This is a structured multi-part question on roots of unity with clear scaffolding. Part (a) is straightforward midpoint calculation, (b) is direct verification, and (c) requires applying the standard 6th roots of unity then transforming back via the given substitution. While it involves several steps and combines geometric interpretation with algebraic manipulation, the question guides students through each stage with no novel insight required—it's above average difficulty due to length and the transformation process, but remains a standard textbook-style application of roots of unity to geometric problems. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02k Argand diagrams: geometric interpretation4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = \frac{z_1+z_2}{2} = \frac{35-25i-29+39i}{2} = \ldots\) or equivalent methods | M1 | — |
| \(= 3 + 7i\) | A1* | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\beta(z_1 - \alpha) = \frac{1+i}{64}(32-32i) = \frac{1}{64}(32-32i+32i-32i^2) = \frac{1}{64}(64)\) | M1 | — |
| \(= 1\) | A1* | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{e^0\}, e^{i\frac{\pi}{3}}, e^{i\frac{2\pi}{3}}, e^{i\pi}, e^{i\frac{4\pi}{3}}, e^{i\frac{5\pi}{3}}\) or \(e^{i\frac{k\pi}{3}}\), \(k=0,1,2,3,4,5\) | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \beta(z-\alpha) = e^{i\frac{k\pi}{3}} \Rightarrow z = \frac{e^{i\frac{k\pi}{3}}}{\beta} + \alpha\) | M1 | — |
| \(z = \frac{64(\cos\frac{k\pi}{3}+i\sin\frac{k\pi}{3})(1-i)}{(1+i)(1-i)} + 3+7i\) | M1 | — |
| Two of: \((19+16\sqrt{3})+(-9+16\sqrt{3})i\), \((-13+16\sqrt{3})+(23+16\sqrt{3})i\), \((-13-16\sqrt{3})+(23-16\sqrt{3})i\), \((19-16\sqrt{3})-(9+16\sqrt{3})i\) | A1 | Or four correct decimal answers: \(46.7+18.7i\), \(-40.7-4.7i\), \(14.7+50.7i\), \(-8.7-36.7i\) |
| All four correct values | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(32\sqrt{2}e^{\frac{\pi}{4}i}\) or \(32\sqrt{2}e^{-\frac{\pi}{4}i}\) | ||
| \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix}32\\-32\end{pmatrix}+\begin{pmatrix}3\\7\end{pmatrix}=\ldots\) or \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix}-32\\32\end{pmatrix}+\begin{pmatrix}3\\7\end{pmatrix}=\ldots\) | M1 | 1.1b |
| Or: \(32\sqrt{2}e^{\frac{\pi}{4}i} \times e^{\frac{\pi}{3}i} = \ldots\) then applies \(r(\cos\theta - i\sin\theta)+3+7i=\ldots\) | ||
| \(32\sqrt{2}e^{-\frac{\pi}{4}i} \times e^{\frac{\pi}{3}i} = \ldots\) then applies \(r(\cos\theta - i\sin\theta)+3+7i=\ldots\) | ||
| Two of: \((19+16\sqrt{3})+(-9+16\sqrt{3})i\), \((-13-16\sqrt{3})+(23-16\sqrt{3})i\), \((-13+16\sqrt{3})+(23+16\sqrt{3})i\), \((19-16\sqrt{3})-(9+16\sqrt{3})i\) | A1 | 2.5 |
| Or four correct decimal answers: \(46.7+18.7i\), \(-40.7-4.7i\), \(14.7+50.7i\), \(-8.7-36.7i\) | ||
| All four of: \((19+16\sqrt{3})+(-9+16\sqrt{3})i\), \((-13+16\sqrt{3})+(23+16\sqrt{3})i\), \((-13-16\sqrt{3})+(23-16\sqrt{3})i\), \((19-16\sqrt{3})-(9+16\sqrt{3})i\) | A1 | 2.2a |
# Question 5:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = \frac{z_1+z_2}{2} = \frac{35-25i-29+39i}{2} = \ldots$ or equivalent methods | M1 | — |
| $= 3 + 7i$ | A1* | — |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\beta(z_1 - \alpha) = \frac{1+i}{64}(32-32i) = \frac{1}{64}(32-32i+32i-32i^2) = \frac{1}{64}(64)$ | M1 | — |
| $= 1$ | A1* | — |
**(c)(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{e^0\}, e^{i\frac{\pi}{3}}, e^{i\frac{2\pi}{3}}, e^{i\pi}, e^{i\frac{4\pi}{3}}, e^{i\frac{5\pi}{3}}$ or $e^{i\frac{k\pi}{3}}$, $k=0,1,2,3,4,5$ | B1 | — |
**(c)(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \beta(z-\alpha) = e^{i\frac{k\pi}{3}} \Rightarrow z = \frac{e^{i\frac{k\pi}{3}}}{\beta} + \alpha$ | M1 | — |
| $z = \frac{64(\cos\frac{k\pi}{3}+i\sin\frac{k\pi}{3})(1-i)}{(1+i)(1-i)} + 3+7i$ | M1 | — |
| Two of: $(19+16\sqrt{3})+(-9+16\sqrt{3})i$, $(-13+16\sqrt{3})+(23+16\sqrt{3})i$, $(-13-16\sqrt{3})+(23-16\sqrt{3})i$, $(19-16\sqrt{3})-(9+16\sqrt{3})i$ | A1 | Or four correct decimal answers: $46.7+18.7i$, $-40.7-4.7i$, $14.7+50.7i$, $-8.7-36.7i$ |
| All four correct values | A1 | — |
# Question (c)(ii) [Complex Numbers - Finding Roots]:
| Answer | Mark | Guidance |
|--------|------|----------|
| $32\sqrt{2}e^{\frac{\pi}{4}i}$ or $32\sqrt{2}e^{-\frac{\pi}{4}i}$ | | |
| $\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix}32\\-32\end{pmatrix}+\begin{pmatrix}3\\7\end{pmatrix}=\ldots$ or $\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix}-32\\32\end{pmatrix}+\begin{pmatrix}3\\7\end{pmatrix}=\ldots$ | M1 | 1.1b |
| Or: $32\sqrt{2}e^{\frac{\pi}{4}i} \times e^{\frac{\pi}{3}i} = \ldots$ then applies $r(\cos\theta - i\sin\theta)+3+7i=\ldots$ | | |
| $32\sqrt{2}e^{-\frac{\pi}{4}i} \times e^{\frac{\pi}{3}i} = \ldots$ then applies $r(\cos\theta - i\sin\theta)+3+7i=\ldots$ | | |
| Two of: $(19+16\sqrt{3})+(-9+16\sqrt{3})i$, $(-13-16\sqrt{3})+(23-16\sqrt{3})i$, $(-13+16\sqrt{3})+(23+16\sqrt{3})i$, $(19-16\sqrt{3})-(9+16\sqrt{3})i$ | A1 | 2.5 |
| Or four correct decimal answers: $46.7+18.7i$, $-40.7-4.7i$, $14.7+50.7i$, $-8.7-36.7i$ | | |
| All four of: $(19+16\sqrt{3})+(-9+16\sqrt{3})i$, $(-13+16\sqrt{3})+(23+16\sqrt{3})i$, $(-13-16\sqrt{3})+(23-16\sqrt{3})i$, $(19-16\sqrt{3})-(9+16\sqrt{3})i$ | A1 | 2.2a |
**Notes (c)(ii):**
- **M1:** Finds rotation matrix and subtracts centre from $z_1$ or $z_2$. Or finds exponential for $z_1-\alpha$ or $z_2-\alpha$
- **M1:** Finds Cartesian form by multiplying by rotation matrix and adding centre. Or multiplies by $e^{\frac{\pi}{3}i}$, writes in Cartesian form and adds centre
- **A1:** At least two correct other roots than $z_1$ and $z_2$ in Cartesian form
- **A1:** All four correct in Cartesian form, no extra solutions
- Note: all four correct decimal answers or as coordinates score **A1A0**
- Note: Correct answers implies method marks
---\begin{enumerate}
\item The points representing the complex numbers $z _ { 1 } = 35 - 25 i$ and $z _ { 2 } = - 29 + 39 i$ are opposite vertices of a regular hexagon, $H$, in the complex plane.
\end{enumerate}
The centre of $H$ represents the complex number $\alpha$\\
(a) Show that $\alpha = 3 + 7 \mathrm { i }$
Given that $\beta = \frac { 1 + \mathrm { i } } { 64 }$\\
(b) show that
$$\beta \left( z _ { 1 } - \alpha \right) = 1$$
The vertices of $H$ are given by the roots of the equation
$$( \beta ( z - \alpha ) ) ^ { 6 } = 1$$
(c) (i) Write down the roots of the equation $w ^ { 6 } = 1$ in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$\\
(ii) Hence, or otherwise, determine the position of the other four vertices of $H$, giving your answers as complex numbers in Cartesian form.
\hfill \mbox{\textit{Edexcel CP2 2023 Q5 [9]}}