A2 June 2023 Paper 2 Q5
5. The points representing the complex numbers \(z_1 = 35 - 25\mathrm{i}\) and \(z_2 = -29 + 39\mathrm{i}\) are opposite vertices of a regular hexagon, \(H\), in the complex plane.
The centre of \(H\) represents the complex number \(\alpha\)
Given that \(\beta = \dfrac{1 + \mathrm{i}}{64}\)
The vertices of \(H\) are given by the roots of the equation
\[\left(\beta(z - \alpha)\right)^6 = 1\]| Scheme | Marks | AO |
|---|---|---|
| \(\alpha = \dfrac{z_1 + z_2}{2} = \dfrac{35 - 25\mathrm{i} - 29 + 39\mathrm{i}}{2} = \ldots\) \(\alpha = z_1 + \dfrac{1}{2}\overrightarrow{z_1z_2} = 35 - 25\mathrm{i} + \dfrac{1}{2}(-64 + 64\mathrm{i}) = \ldots\) \(\alpha = z_2 + \dfrac{1}{2}\overrightarrow{z_2z_1} = -29 + 39\mathrm{i} + \dfrac{1}{2}(64 - 64\mathrm{i}) = \ldots\) (corrected from the printed mark scheme: \(-29 + 39\mathrm{i}\) is printed as \(-29 + 39\)) | M1 | 1.1b |
| \(= 3 + 7\mathrm{i}\,*\) | A1* | 1.1b |
| (2) |
Notes
M1: Attempts the midpoint of \(z_1\) and \(z_2\)
A1*: Correct point.
| Scheme | Marks | AO |
|---|---|---|
| \(\beta(z_1 - \alpha) = \left(\dfrac{1 + \mathrm{i}}{64}\right)\left(35 - 25\mathrm{i} - (3 + 7\mathrm{i})\right) = \left(\dfrac{1 + \mathrm{i}}{64}\right)(32 - 32\mathrm{i}) =\) \(= \dfrac{1}{64}\left(32 - 32\mathrm{i} + 32\mathrm{i} - 32\mathrm{i}^2\right) = \dfrac{1}{64}(32 - 32\mathrm{i} + 32\mathrm{i} + 32)\) | M1 | 1.1b |
| \(= \dfrac{1}{64}(64) = 1\,*\) | A1* | 1.1b |
| (2) |
Notes
M1: Substitutes into the equation with \(z_1\) and \(\alpha\) and \(\beta\), simplifies and expands and applies \(\mathrm{i}^2 = -1\), this may be implied by their working.
A1*: Completes the proof to find the correct answer with no errors seen, all necessary brackets as required
| Scheme | Marks | AO |
|---|---|---|
| (i) Roots are \(\left\{\mathrm{e}^0\left(\text{or } 1 \text{ or } \mathrm{e}^{\mathrm{i}2\pi}\right)\right\}, \mathrm{e}^{\mathrm{i}\frac{\pi}{3}}, \mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}, \mathrm{e}^{\mathrm{i}\pi}, \mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}, \mathrm{e}^{\mathrm{i}\frac{5\pi}{3}}\) or \(\mathrm{e}^{\mathrm{i}\frac{k\pi}{3}},\ k = 0, 1, 2, 3, 4, 5\) \(\left\{\mathrm{e}^0\left(\text{or } 1 \text{ or } \mathrm{e}^{\mathrm{i}2\pi}\right)\right\}, \mathrm{e}^{\mathrm{i}\frac{\pi}{3}}, \mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}, \mathrm{e}^{\mathrm{i}\pi}, \mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}, \mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}\) or \(\mathrm{e}^{\mathrm{i}\frac{k\pi}{3}},\ k = -2, -1, 0, 1, 2, 3,\) | B1 | 1.1b |
| (1) | ||
| (ii) \(w = \beta(z - \alpha) = \mathrm{e}^{\mathrm{i}\frac{k\pi}{3}} \Rightarrow z = \dfrac{\mathrm{e}^{\mathrm{i}\frac{k\pi}{3}}}{\beta} + \alpha\) | M1 | 3.1a |
| \(\Rightarrow z = \dfrac{64\left(\cos\dfrac{k\pi}{3} + \mathrm{i}\sin\dfrac{k\pi}{3}\right)(1 - \mathrm{i})}{(1 + \mathrm{i})(1 - \mathrm{i})} + 3 + 7\mathrm{i} = \ldots\) | M1 | 1.1b |
| Two of \(\left(19 + 16\sqrt{3}\right) + \left(-9 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 - 16\sqrt{3}\right) + \left(23 - 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 + 16\sqrt{3}\right) + \left(23 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(19 - 16\sqrt{3}\right) - \left(9 + 16\sqrt{3}\right)\mathrm{i}\) Or four correct decimal answers 46.7 + 18.7i – 40.7 – 4.7i 14.7 + 50.7i – 8.7 – 36.7i | A1 | 2.5 |
| All four of \(\left(19 + 16\sqrt{3}\right) + \left(-9 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 + 16\sqrt{3}\right) + \left(23 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 - 16\sqrt{3}\right) + \left(23 - 16\sqrt{3}\right)\mathrm{i}\) \(\left(19 - 16\sqrt{3}\right) - \left(9 + 16\sqrt{3}\right)\mathrm{i}\) | A1 | 2.2a |
| (4) | ||
| (11 marks) |
Notes
Alternative 1
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\beta(z - \alpha)\right)^6 = 1^6 \Rightarrow (z - \alpha)^6 = \dfrac{1}{\beta^6} = 8589934592\mathrm{i}\) \(r = \sqrt[6]{8589934592} = 32\sqrt{2}\) or 45.25... and \(\theta = \dfrac{\pi}{12} + \dfrac{k\pi}{3}\) or \(\theta = -\dfrac{\pi}{4} + \dfrac{k\pi}{3}\) (corrected from the printed mark scheme: \(\dfrac{1}{\beta^6} = 2^{33}\mathrm{i}\) is printed as 8589934459i, and the number under the root as 858993459) | M1 | 3.1a |
| \(z = r(\cos\theta + \mathrm{i}\sin\theta) + 3 + 7\mathrm{i} = \ldots\) (corrected from the printed mark scheme: printed as \(r(\cos\theta - \mathrm{i}\sin\theta)\), which with these values of \(\theta\) does not give the vertices) | M1 | 1.1b |
| Two of \(\left(19 + 16\sqrt{3}\right) + \left(-9 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 - 16\sqrt{3}\right) + \left(23 - 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 + 16\sqrt{3}\right) + \left(23 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(19 - 16\sqrt{3}\right) - \left(9 + 16\sqrt{3}\right)\mathrm{i}\) Or four correct decimal answers 46.7 + 18.7i – 40.7 – 4.7i 14.7 + 50.7i – 8.7 – 36.7i | A1 | 2.5 |
| All four of \(\left(19 + 16\sqrt{3}\right) + \left(-9 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 + 16\sqrt{3}\right) + \left(23 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 - 16\sqrt{3}\right) + \left(23 - 16\sqrt{3}\right)\mathrm{i}\) \(\left(19 - 16\sqrt{3}\right) - \left(9 + 16\sqrt{3}\right)\mathrm{i}\) | A1 | 2.2a |
| (4) |
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| Rotation matrix \(\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\) and \(\begin{pmatrix}35\\ -25\end{pmatrix} - \begin{pmatrix}3\\ 7\end{pmatrix}\) or \(\begin{pmatrix}-29\\ 39\end{pmatrix} - \begin{pmatrix}3\\ 7\end{pmatrix}\) Or find the exponential form for \(\begin{pmatrix}35\\ -25\end{pmatrix} - \begin{pmatrix}3\\ 7\end{pmatrix}\) or \(\begin{pmatrix}-29\\ 39\end{pmatrix} - \begin{pmatrix}3\\ 7\end{pmatrix}\) \(32\sqrt{2}\mathrm{e}^{\frac{\pi}{4}}\) or \(32\sqrt{2}\mathrm{e}^{-\frac{\pi}{4}}\) | M1 | 3.1a |
| \(\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}32\\ -32\end{pmatrix} + \begin{pmatrix}3\\ 7\end{pmatrix} = \ldots\) or \(\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}-32\\ 32\end{pmatrix} + \begin{pmatrix}3\\ 7\end{pmatrix} = \ldots\) Or \(32\sqrt{2}\mathrm{e}^{\frac{\pi}{4}\mathrm{i}} \times \mathrm{e}^{\frac{\pi}{3}\mathrm{i}} = \ldots\) then applies \(r(\cos\theta - \mathrm{i}\sin\theta) + 3 + 7\mathrm{i} = \ldots\) \(32\sqrt{2}\mathrm{e}^{-\frac{\pi}{4}\mathrm{i}} \times \mathrm{e}^{\frac{\pi}{3}\mathrm{i}} = \ldots\) then applies \(r(\cos\theta - \mathrm{i}\sin\theta) + 3 + 7\mathrm{i} = \ldots\) | M1 | 1.1b |
| Two of \(\left(19 + 16\sqrt{3}\right) + \left(-9 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 - 16\sqrt{3}\right) + \left(23 - 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 + 16\sqrt{3}\right) + \left(23 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(19 - 16\sqrt{3}\right) - \left(9 + 16\sqrt{3}\right)\mathrm{i}\) Or four correct decimal answers 46.7 + 18.7i – 40.7 – 4.7i 14.7 + 50.7i – 8.7 – 36.7i Or as coordinates | A1 | 2.5 |
| All four of \(\left(19 + 16\sqrt{3}\right) + \left(-9 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 + 16\sqrt{3}\right) + \left(23 + 16\sqrt{3}\right)\mathrm{i}\) \(\left(-13 - 16\sqrt{3}\right) + \left(23 - 16\sqrt{3}\right)\mathrm{i}\) \(\left(19 - 16\sqrt{3}\right) - \left(9 + 16\sqrt{3}\right)\mathrm{i}\) | A1 | 2.2a |
| (4) |
Notes
Mark (c) as one
(c) (i)
B1: Correct roots, accept all 6 listed or given in general form as in the scheme. Need not show the 1.
(c)(ii)
M1: Realises the need to set the roots of unity equal to \(\beta(z - \alpha)\) and solve for \(z\). Must be attempted at least once with any of their roots.
M1: Finds the Cartesian form for their equation for at least one of the roots other than \(z_1\) and \(z_2\)
A1: At least two correct other roots than \(z_1\) and \(z_2\) in Cartesian form.
A1: Deduces all four correct in Cartesian form and no extra solutions
Alternative 1
M1: Finds the modulus and argument of \((z - \alpha)^6\)
M1: Finds the Cartesian form for one of their modulus and arguments
A1A1: same as above
Alternative 2
M1: Finds the rotation matrix and subtracts the centre from \(z_1\) or \(z_2\). Or finds the exponential from for \(z_1 - \alpha\) or \(z_2 - \alpha\)
M1: Finds the Cartesian form by multiplying by the rotation matrix and adding the centre. Or multiplies by \(\mathrm{e}^{\frac{\pi}{3}\mathrm{i}}\) write in Cartesian form and adds on the centre
A1A1: same as above
Note all four correct decimal answers or written as coordinates score A1A0
46.7 + 18.7i – 40.7 – 4.7i 14.7 + 50.7i – 8.7 – 36.7i
Note: Correct answers implies the method marks