| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Direct nth roots: roots with geometric or algebraic follow-up |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring conversion to polar form, finding cube roots using De Moivre's theorem, and recognizing that z₁³ᵏ + z₂³ᵏ + z₃³ᵏ = 3(-1-i)ᵏ through properties of roots. While the techniques are standard for Further Maths, the algebraic manipulation and insight needed for part (b) elevates it above routine exercises. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^3 = -1 - i = 2^{\frac{1}{2}}e^{i\frac{5}{4}\pi}\) | B1 | |
| \(z_1 = 2^{\frac{1}{6}}e^{i\frac{5}{12}\pi}\) | M1 A1 | |
| \(z_2 = 2^{\frac{1}{6}}e^{i\frac{13}{12}\pi}, \quad z_3 = 2^{\frac{1}{6}}e^{i\frac{21}{12}\pi}\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z_1^{3k} + z_2^{3k} + z_3^{3k} = 3\!\left(2^{\frac{3}{2}k}e^{i\frac{5}{4}k\pi}\right)\) | M1 | |
| \(R = \ | w\ | = 3\!\left(2^{\frac{1}{2}k}\right)\) |
| \(\alpha = \frac{5}{4}k\pi\) | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^3 = -1 - i = 2^{\frac{1}{2}}e^{i\frac{5}{4}\pi}$ | B1 | |
| $z_1 = 2^{\frac{1}{6}}e^{i\frac{5}{12}\pi}$ | M1 A1 | |
| $z_2 = 2^{\frac{1}{6}}e^{i\frac{13}{12}\pi}, \quad z_3 = 2^{\frac{1}{6}}e^{i\frac{21}{12}\pi}$ | A1 A1 | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z_1^{3k} + z_2^{3k} + z_3^{3k} = 3\!\left(2^{\frac{3}{2}k}e^{i\frac{5}{4}k\pi}\right)$ | M1 | |
| $R = \|w\| = 3\!\left(2^{\frac{1}{2}k}\right)$ | A1 | |
| $\alpha = \frac{5}{4}k\pi$ | A1 | |
---3
\begin{enumerate}[label=(\alph*)]
\item Find the roots of the equation $z ^ { 3 } = - 1 - \mathrm { i }$, giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $0 \leqslant \theta < 2 \pi$.\\
Let $\mathbf { w } = \mathbf { z } _ { 1 } ^ { 3 \mathrm { k } } + \mathbf { z } _ { 2 } ^ { 3 \mathrm { k } } + \mathbf { z } _ { 3 } ^ { 3 \mathrm { k } }$, where $k$ is a positive integer and $\mathrm { z } _ { 1 } , \mathrm { z } _ { 2 } , \mathrm { z } _ { 3 }$ are the roots of $\mathrm { z } ^ { 3 } = - 1 - \mathrm { i }$.
\item Express $w$ in the form $R \mathrm { e } ^ { \mathrm { i } \alpha }$, where $R > 0$, giving $R$ and $\alpha$ in terms of $k$.\\
\includegraphics[max width=\textwidth, alt={}, center]{1de67949-6262-4ade-b986-02b6563ae404-06_889_824_267_616}
The diagram shows the curve with equation $\mathrm { y } = \mathrm { x } ^ { 2 }$ for $0 \leqslant x \leqslant 1$, together with a set of $n$ rectangles of width $\frac { 1 } { n }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q3 [8]}}