| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 7 |
| Topic | Complex Numbers Argand & Loci |
| Type | Modulus-argument form conversion |
| Difficulty | Moderate -0.8 This is a straightforward conversion from modulus-argument form to Cartesian form using standard formulas (z = r(cos θ + i sin θ)). Part (i) requires direct application, part (ii) uses De Moivre's theorem or simple multiplication, and part (iii) involves basic complex conjugate manipulation. All values involve standard angles (π/6) with exact trigonometric values. This is routine practice requiring recall and careful arithmetic rather than problem-solving or insight. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02f Convert between forms: cartesian and modulus-argument |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(z = 2\cos\frac{\pi}{6} + 2i\sin\frac{\pi}{6} = \sqrt{3} + i\) | M1, A1 [2] | |
| (ii) \(z^2 = (\sqrt{3}+i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i\) | M1, A1ft, A1ft [2] | For \((their z)^2\); FT their \(z\) |
| Or \(z = (2, \frac{\pi}{6}) \Rightarrow z^2 = (4, \frac{\pi}{3}) = 4(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = 2 + 2\sqrt{3}i\) | M1, A1ft [2] | FT their \(z\) |
| (iii) \(z^6 = \sqrt{3} - i\) | B1ft, M1, A1 [3] | FT their \(z\); Or \(zz^* = |
| Or \(z = (2, \frac{\pi}{6}) \Rightarrow z^* = (2, -\frac{\pi}{6})\); \(\frac{z}{z^*} = (1, \frac{\pi}{3}) = (\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})\) | B1ft, M1 | FT their \(z\); \(\frac{1}{2} + \frac{\sqrt{3}}{2}i\) |
**(i)** $z = 2\cos\frac{\pi}{6} + 2i\sin\frac{\pi}{6} = \sqrt{3} + i$ | M1, A1 [2] |
**(ii)** $z^2 = (\sqrt{3}+i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i$ | M1, A1ft, A1ft [2] | For $(their z)^2$; FT their $z$
**Or** $z = (2, \frac{\pi}{6}) \Rightarrow z^2 = (4, \frac{\pi}{3}) = 4(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = 2 + 2\sqrt{3}i$ | M1, A1ft [2] | FT their $z$
**(iii)** $z^6 = \sqrt{3} - i$ | B1ft, M1, A1 [3] | FT their $z$; Or $zz^* = |z|^2 = 4$ (B1) and $\frac{z^2}{|z|^2} = \frac{2+2\sqrt{3}i}{4} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$ | A1 for showing $it = \frac{1}{2} + \frac{1}{2}\sqrt{3}i$
**Or** $z = (2, \frac{\pi}{6}) \Rightarrow z^* = (2, -\frac{\pi}{6})$; $\frac{z}{z^*} = (1, \frac{\pi}{3}) = (\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$ | B1ft, M1 | FT their $z$; $\frac{1}{2} + \frac{\sqrt{3}}{2}i$ | A1
---1 In this question you must show detailed reasoning.\\
For the complex number $z$ it is given that $| z | = 2$ and $\arg z = \frac { 1 } { 6 } \pi$.\\
Find the following in the form $a + \mathrm { i } b$, where $a$ and $b$ are exact numbers.\\
(i) $z$\\
(ii) $z ^ { 2 }$\\
(iii) $\frac { z } { z ^ { * } }$
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q1 [7]}}