| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | nth roots via factorization |
| Difficulty | Standard +0.8 This is a Further Maths question requiring factorization of a complex polynomial, then finding 8th roots using De Moivre's theorem. Part (a) is straightforward pattern matching, but part (b) requires systematic application of nth roots in exponential form across two separate equations. The multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 4.02i Quadratic equations: with complex roots4.02j Cubic/quartic equations: conjugate pairs and factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 1, \quad b = \text{i}\) | B1 | |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = e^{i\frac{2k}{5}\pi}\) | M1 | Finds one fifth root of unity. |
| \(z = e^{i\frac{2k}{5}\pi}, \quad k = 0, 1, 2, 3, 4\) | A1 | Gives all fifth roots of unity. Accept 1 not in exponential form. A0 if \(r = 1\) not seen or implied. |
| Argument of \(\text{i}\) is \(\frac{1}{2}\pi\) | B1 | |
| \(z = e^{i\frac{1}{6}\pi}\) | M1 A1 | Finds one root of \(z^3 = \text{i}\). A0 if first root not given in exponential form. |
| \(z = e^{i\frac{5}{6}\pi}, \quad e^{i\frac{3}{2}\pi}\) | A1 | Finds other two roots of \(z^3 = \text{i}\). A0 if \(r = 1\) not seen or implied. Withhold final A1 if \(a\) and \(b\) reversed in part (b) but all other work correct. |
| 6 | SC If \(z = e^{i(\frac{1}{10}\pi + \frac{2k}{5}\pi)}, \quad k = 0,1,2,3,4\) award M1 A1 A1 only. SC If \(z = e^{i\left(\frac{2k}{3}\pi\right)}, \quad k = 0,1,2\) award M1 A1 only. |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 1, \quad b = \text{i}$ | **B1** | |
| | **1** | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = e^{i\frac{2k}{5}\pi}$ | **M1** | Finds one fifth root of unity. |
| $z = e^{i\frac{2k}{5}\pi}, \quad k = 0, 1, 2, 3, 4$ | **A1** | Gives all fifth roots of unity. Accept 1 not in exponential form. A0 if $r = 1$ not seen or implied. |
| Argument of $\text{i}$ is $\frac{1}{2}\pi$ | **B1** | |
| $z = e^{i\frac{1}{6}\pi}$ | **M1 A1** | Finds one root of $z^3 = \text{i}$. A0 if first root not given in exponential form. |
| $z = e^{i\frac{5}{6}\pi}, \quad e^{i\frac{3}{2}\pi}$ | **A1** | Finds other two roots of $z^3 = \text{i}$. A0 if $r = 1$ not seen or implied. Withhold final A1 if $a$ and $b$ reversed in part **(b)** but all other work correct. |
| | **6** | **SC** If $z = e^{i(\frac{1}{10}\pi + \frac{2k}{5}\pi)}, \quad k = 0,1,2,3,4$ award M1 A1 A1 only. **SC** If $z = e^{i\left(\frac{2k}{3}\pi\right)}, \quad k = 0,1,2$ award M1 A1 only. |1
\begin{enumerate}[label=(\alph*)]
\item Find $a$ and $b$ such that
$$z ^ { 8 } - i z ^ { 5 } - z ^ { 3 } + i = \left( z ^ { 5 } - a \right) \left( z ^ { 3 } - b \right) .$$
\item Hence find the roots of
$$z ^ { 8 } - i z ^ { 5 } - z ^ { 3 } + i = 0$$
giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r > 0$ and $0 \leqslant \theta < 2 \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q1 [7]}}