| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Roots of unity with derived equations |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on roots of unity requiring multiple standard techniques: finding 5th roots of unity (routine), manipulating the equation (z+1)^5 = z^5 to derive the polynomial, and expressing roots in the given form. While it involves several steps and Further Maths content, each step follows a well-established method with clear guidance from the 'hence' structure. The conceptual demand is moderate—recognizing how to use roots of unity to solve the derived equation—but this is a standard FP3 technique rather than requiring novel insight. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1, e^{\frac{2}{5}\pi i}, e^{\frac{4}{5}\pi i}, e^{\frac{6}{5}\pi i}, e^{\frac{8}{5}\pi i}\) oe polar form | M1, A1 | e.g. gives roots in incorrect form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^5=(z+1)^5=z^5+5z^4+10z^3+10z^2+5z+1\) | M1 | |
| \(\Leftrightarrow 5z^4+10z^3+10z^2+5z+1=0\) | A1 | |
| so \(z+1=ze^{\frac{2k}{5}\pi i}\), \(k=0,1,2,3,4\) | M1 | |
| \(k=0\) no solution | B1 | soi |
| \(z=\frac{1}{e^{\frac{2k}{5}\pi i}-1}\), \(k=1,2,3,4\) | A1 | If B0, then give A1 ft for correct solution plus \(k=0\) |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1, e^{\frac{2}{5}\pi i}, e^{\frac{4}{5}\pi i}, e^{\frac{6}{5}\pi i}, e^{\frac{8}{5}\pi i}$ oe polar form | M1, A1 | e.g. gives roots in incorrect form |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^5=(z+1)^5=z^5+5z^4+10z^3+10z^2+5z+1$ | M1 | |
| $\Leftrightarrow 5z^4+10z^3+10z^2+5z+1=0$ | A1 | |
| so $z+1=ze^{\frac{2k}{5}\pi i}$, $k=0,1,2,3,4$ | M1 | |
| $k=0$ no solution | B1 | soi |
| $z=\frac{1}{e^{\frac{2k}{5}\pi i}-1}$, $k=1,2,3,4$ | A1 | If B0, then give A1 ft for correct solution plus $k=0$ |
---5 (i) Solve the equation $z ^ { 5 } = 1$, giving your answers in polar form.\\
(ii) Hence, by considering the equation $( z + 1 ) ^ { 5 } = z ^ { 5 }$, show that the roots of
$$5 z ^ { 4 } + 10 z ^ { 3 } + 10 z ^ { 2 } + 5 z + 1 = 0$$
can be expressed in the form $\frac { 1 } { \mathrm { e } ^ { \mathrm { i } \theta } - 1 }$, stating the values of $\theta$.
\hfill \mbox{\textit{OCR FP3 2013 Q5 [7]}}