AS June 2023 Paper 1 Q4
4.
(Solutions relying on calculator technology are not acceptable.) (3)
Given that
- \(n\) is a positive integer
- \(\left(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}}\right)^n\) is a real number
Given that
- \(\left|z^{10}\right| = 59\,049\)
- \(\arg\left(z^{10}\right) = -\dfrac{5\pi}{3}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}} \times \dfrac{5 - \mathrm{i}}{5 - \mathrm{i}}\) or \(2 + 3\mathrm{i} = k(1 + \mathrm{i})(5 + \mathrm{i}) = \ldots\) | M1 | 1.1a |
| \(\dfrac{10 - 2\mathrm{i} + 15\mathrm{i} + 3}{25 + 1}\) or \(\dfrac{13 + 13\mathrm{i}}{26}\) or \(2 + 3\mathrm{i} = k(5 + \mathrm{i} + 5\mathrm{i} - 1) = \ldots\) | dM1 | 1.1b |
| \(\dfrac{1}{2}(1 + \mathrm{i})\) cso or \(2 + 3\mathrm{i} = k(4 + 6\mathrm{i})\) therefore \(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}} = k(1 + \mathrm{i})\) where \(k = \dfrac{1}{2}\) cso | A1 | 2.1 |
| (3) |
Notes
M1: Selects the process \(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}} \times \dfrac{5 - \mathrm{i}}{5 - \mathrm{i}}\)
dM1: Evidence of multiplying out brackets
A1: Achieves \(\dfrac{1}{2}(1 + \mathrm{i})\) or \(\dfrac{13}{26}(1 + \mathrm{i})\) with no errors cso, isw.
Note: Correct answer from no working score no marks
Note: Going from \(\dfrac{13 + 13\mathrm{i}}{26}\) and then stating \(k = \dfrac{1}{2}\) is A0, they have not shown the form asked for
Alternative
M1: Multiplies across by \((5 + \mathrm{i})\) and expands the brackets
dM1: Collects terms
A1: Achieves \(2 + 3\mathrm{i} = k(4 + 6\mathrm{i})\) and draws the conclusion that therefore \(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}} = k(1 + \mathrm{i})\) where \(k = \dfrac{1}{2}\)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 4\) | B1 | 2.2a |
| (1) |
Notes
B1: Deduces \(n = 4\) only
| Scheme | Marks | AO |
|---|---|---|
| \(|z| = 3\) | B1 | 1.2 |
| \(\arg\left(z^{10}\right) = 10\arg(z) = -\dfrac{5\pi}{3} \Rightarrow \arg(z) = \ldots\left\{-\dfrac{\pi}{6}\right\}\) \(\arg\left(z^{10}\right) = 10\arg(z) = \dfrac{\pi}{3} \Rightarrow \arg(z) = \ldots\left\{\dfrac{\pi}{30}\right\}\) | M1 | 1.1b |
| \(z = 3\left(\cos\left(-\dfrac{\pi}{6}\right) + \mathrm{i}\sin\left(-\dfrac{\pi}{6}\right)\right) = \ldots\) | M1 | 2.1 |
| \(z = \dfrac{3\sqrt{3}}{2} - \dfrac{3}{2}\mathrm{i}\) or \(a = \dfrac{3\sqrt{3}}{2}\) and \(b = -\dfrac{3}{2}\) | A1 | 1.1b |
| (4) | ||
| (8 marks) |
Notes
Note: Send to review any attempts where they are finding additional solutions such as arguments of \(\boldsymbol{z}\) is \(\dfrac{(6k - 5)\pi}{30}\) For example correctly uses \(\arg(z) = \dfrac{\pi}{30}\)
B1 (M1 on ePen): \(|z| = 3\) can be implied by \(a^2 + b^2 = 9\) isw
M1: Uses \(\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\) to find \(\arg(z) = -\dfrac{5\pi}{3} \div 10\) or \(\arg(z) = \dfrac{\pi}{3} \div 10\)
M1: Uses \(z = \text{their } |z|\left(\cos(\text{their arg}) + \mathrm{i}\sin(\text{their arg})\right)\) to find the complex number \(z\) or values for \(a\) or \(b\). As long as the modulus has changed.
A1: Correct complex number or values for \(a\) and \(b\).
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(a^2 + b^2 = 9\) | B1 | 1.2 |
| \(10\arg z = -\dfrac{5\pi}{3} \Rightarrow \arg z = -\dfrac{5\pi}{3} \div 10\) Or e.g \(10\arg(z) = \dfrac{\pi}{3} \Rightarrow \arg(z) = \ldots\left\{\dfrac{\pi}{30}\right\}\) | M1 | 1.1b |
| Forming and solving simultaneous equations to find a value for \(a\) or \(b\) \(\dfrac{b}{a} = \tan\left(-\dfrac{\pi}{6}\right) \Rightarrow \dfrac{b}{a} = -\dfrac{\sqrt{3}}{3} \Rightarrow b = -a\dfrac{\sqrt{3}}{3}\) or \(\dfrac{b}{a} = \tan\dfrac{\pi}{30} \Rightarrow b = 0.105\ldots a\) | M1 | 2.1 |
| \(z = \dfrac{3\sqrt{3}}{2} - \dfrac{3}{2}\mathrm{i}\) or \(a = \dfrac{3\sqrt{3}}{2}\) and \(b = -\dfrac{3}{2}\) | A1 | 1.1b |
| (4) |
(Corrected from the printed mark scheme: the printed scheme has \(\dfrac{b}{a} = \arctan\left(-\dfrac{\pi}{6}\right)\) and \(\dfrac{b}{a} = \arctan\dfrac{\pi}{30} \Rightarrow b = 0.104\ldots a\); it should be \(\tan\), giving \(b = 0.105\ldots a\).)
B1: \(a^2 + b^2 = 9\) isw
M1: Uses \(\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\) to find \(\arg(z) = -\dfrac{5\pi}{3} \div 10\)
M1: Uses the argument of \(z\) find an equation in \(a\) and \(b\). Then solve simultaneously to find a value for \(a\) or \(b\).
As long as \(\boldsymbol{\sqrt{a^2 + b^2} \neq 59049}\)
A1: Correct complex number or values for \(a\) and \(b\).
Note there are other correct answers
| \(z_1 = \dfrac{3\sqrt{3}}{2} - \dfrac{3}{2}\mathrm{i}\) \(z_2 =\) awrt \(2.98 +\) awrt \(0.314\mathrm{i}\) \(z_3 =\) awrt \(2.23 +\) awrt \(2.01\mathrm{i}\) \(z_4 =\) awrt \(0.624 +\) awrt \(2.93\mathrm{i}\) \(z_5 =\) awrt \(-1.22 +\) awrt \(2.74\mathrm{i}\) | \(z_6 = -\dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}\mathrm{i}\) \(z_7 =\) awrt \(-2.98 +\) awrt \(-0.314\mathrm{i}\) \(z_8 =\) awrt \(-2.23 +\) awrt \(-2.01\mathrm{i}\) \(z_9 =\) awrt \(-0.624 +\) awrt \(-2.93\mathrm{i}\) \(z_{10} =\) awrt \(1.22 +\) awrt \(-2.74\mathrm{i}\) |