A2 June 2025 Paper 2 Q1
1.
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Given that
\[z = 2 - 2\sqrt{3}\,\mathrm{i} \quad \text{and} \quad w = -1 + \sqrt{3}\,\mathrm{i}\]show that
| Scheme | Marks | AO |
|---|---|---|
| \(\lvert z\rvert = \left(\sqrt{2^2 + \left(-2\sqrt{3}\right)^2} =\right) 4\) and \(\lvert w\rvert = \left(\sqrt{(-1)^2 + \left(\sqrt{3}\right)^2} =\right) 2\) Alt: \(z = -2\left(-1 + \sqrt{3}\mathrm{i}\right) = -2w\) | B1 | 1.1b |
| \(\dfrac{z}{w} = \dfrac{2 - 2\sqrt{3}\mathrm{i}}{-1 + \sqrt{3}\mathrm{i}} = -2 \Rightarrow \left\lvert\dfrac{z}{w}\right\rvert = 2\) | M1 | 2.1 |
| \(\dfrac{\lvert z\rvert}{\lvert w\rvert} = \dfrac{4}{2} = 2\) therefore \(\left\lvert\dfrac{z}{w}\right\rvert = \dfrac{\lvert z\rvert}{\lvert w\rvert}\) | A1cso | 2.4 |
| (3) |
Notes
B1: Correct values for \(\lvert z\rvert\) and \(\lvert w\rvert\). No need for method, may see modulus-argument forms (polar or exponential) used. Alternatively, correctly identifies \(z = -2w\)
M1: Finds \(\dfrac{z}{w}\) and then \(\left\lvert\dfrac{z}{w}\right\rvert\). Look for multiplication of numerator and denominator by the conjugate of the denominator for finding \(\dfrac{z}{w}\). In the look for alternative factoring out the \(-2\) and cancelling \(z\) oe. Allow this mark if modulus-argument forms (polar or exponential) are used to deduce the modulus.
A1: Correctly shows that \(\dfrac{\lvert z\rvert}{\lvert w\rvert} = 2\) and \(\left\lvert\dfrac{z}{w}\right\rvert = 2\) hence state that \(\left\lvert\dfrac{z}{w}\right\rvert = \dfrac{\lvert z\rvert}{\lvert w\rvert}\) cso. If the result is not stated at the end, we require both \(\left\lvert\dfrac{z}{w}\right\rvert\) and \(\dfrac{\lvert z\rvert}{\lvert w\rvert}\) to have explicitly been seen at some stage in their work and a minimal conclusion (e.g. “LHS=RHS”) to be given. Note \(\dfrac{z}{w} = \ldots = -2\) followed by \(\lvert -2\rvert = 2\) without ever seeing \(\left\lvert\dfrac{z}{w}\right\rvert\) will be A0.
If modulus-argument forms (polar or exponential) were used all working must have been correct.
| Scheme | Marks | AO |
|---|---|---|
| \(\arg(z) = -\dfrac{\pi}{3}\) or \(\dfrac{5\pi}{3}\) and \(\arg(w) = \dfrac{2\pi}{3}\) Alt: \(\arg(w) = \dfrac{2\pi}{3}\) and \(\arg\left(w^2\right) = \dfrac{4\pi}{3}\) | B1 | 1.1b |
| \(zw = \left(2 - 2\sqrt{3}\mathrm{i}\right)\left(-1 + \sqrt{3}\mathrm{i}\right) = 4 + 4\sqrt{3}\mathrm{i} \Rightarrow \arg(zw) = \tan^{-1}\left(\dfrac{4\sqrt{3}}{4}\right) = \dfrac{\pi}{3}\) (or \(zw = -2w^2 = -2\left(-2 - 2\sqrt{3}\mathrm{i}\right) \Rightarrow \arg(zw) = \ldots\)) Alt: \(\arg z = \arg(-2w) = \arg w \pm \pi,\ \arg(zw) = \arg\left(-2w^2\right) = \arg w^2 \pm \pi\) | M1 | 2.1 |
| \(\arg(z) + \arg(w) = -\dfrac{\pi}{3} + \dfrac{2\pi}{3} = \dfrac{\pi}{3}\) therefore \(\arg(zw) = \arg(z) + \arg(w)\) | A1cso | 2.4 |
| (3) | ||
| (6 marks) |
Notes
B1: Correct values for \(\arg(z)\) and \(\arg(w)\). Accept degrees equivalents here and throughout.
Alternatively, correct values for \(\arg(w)\) and \(\arg\left(w^2\right)\).
M1: Finds \(zw\) and then \(\arg(zw)\). Do not allow attempts via modulus-argument forms (polar or exponential) that use the sum of arguments to prove the sum of arguments of the product.
Alternatively, finds both \(\arg(z)\) and \(\arg(zw)\) in terms of \(\arg(w)\) using \(\arg(-p) = \arg(p) \pm \pi\).
A1: Shows that \(\arg(z) + \arg(w) = \dfrac{\pi}{3}\) and following \(\arg(zw) = \dfrac{\pi}{3}\) hence that \(\arg(zw) = \arg(z) + \arg(w)\) cso If the result is not stated at the end, we require both sides of the result to have been seen at some stage in their work and a minimal conclusion (e.g. “LHS=RHS”) to be given.
Alternatively, correctly establishes the result using \(\arg(zw) = \arg\left(w^2\right) \pm \pi = 2\arg(w) \pm \pi\) and \(\arg z + \arg w = \arg w \pm \pi + \arg w = 2\arg w \pm \pi\) with \(\arg w^2 = 2\arg w\) verified.