| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Square roots with follow-up application |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (1) cubing a complex number to find parameters, (2) finding all cube roots using de Moivre's theorem and exponential form, and (3) explaining why conjugate root theorem doesn't apply. While the techniques are standard for Further Maths, the combination of algebraic manipulation, polar form conversion, and conceptual understanding of when theorems apply makes this moderately challenging, though not requiring novel insight. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02g Conjugate pairs: real coefficient polynomials4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (a) | DR |
| Answer | Marks |
|---|---|
| = 2 + 2i [so a = 2 and b = 2] | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | expanding (–1 + i)3 |
| correct expression | or 2i(1 + i) |
| Answer | Marks | Guidance |
|---|---|---|
| –1 + i = √2(cos 3π/4 + i sin 3π/4) | B1 | z = 2, arg z = 3π/4 |
| (–1 + i)3 = 2√2(cos 9π/4 + i sin 9π/4) | M1 | z3 = z3, arg(z3) = 3arg z |
| = 2√2(1/√2 + 1/√2) = 2 + 2i | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (b) | B1 |
| z = √2eiπ/12 , 2e3iπ/4 , √2e7iπ/12 | z = √2eiπ/12 , 2e3iπ/4 , √2e7iπ/12 | B1B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (c) | The statement only applies to polynomial |
| equations with real coefficients. | B1 | 2.3 |
Question 10:
10 | (a) | DR
(–1 + i)3 = (–1)3 + 3(–1)2i + 3(–1)i2 + i3
= 2 + 2i [so a = 2 and b = 2] | M1
A1
A1 | 3.1a
1.1b
2.2a | expanding (–1 + i)3
correct expression | or 2i(1 + i)
Alternative solution
–1 + i = √2(cos 3π/4 + i sin 3π/4) | B1 | z = 2, arg z = 3π/4 | or 2 e3iπ/4
(–1 + i)3 = 2√2(cos 9π/4 + i sin 9π/4) | M1 | z3 = z3, arg(z3) = 3arg z
= 2√2(1/√2 + 1/√2) = 2 + 2i | A1
[3]
10 | (b) | B1 | 2.5 | 3 roots with modulus 2 oe | oe eg 81/6
z = √2eiπ/12 , 2e3iπ/4 , √2e7iπ/12 | z = √2eiπ/12 , 2e3iπ/4 , √2e7iπ/12 | B1B1B1 | 1.1b | 1.1b | oe, e.g. 81/6e17iπ/12, etc | oe, e.g. 81/6e17iπ/12, etc | 0< arg <2π or π < arg < π | 0< arg <2π or π < arg < π
[4]
10 | (c) | The statement only applies to polynomial
equations with real coefficients. | B1 | 2.3
[1]10 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item You are given that $- 1 + \mathrm { i }$ is a root of the equation $z ^ { 3 } = a + b \mathrm { i }$, where $a$ and $b$ are real numbers. Find $a$ and $b$.
\item Find all the roots of the equation in part (a), giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $r$ and $\theta$ are exact.
\item Chris says "the complex roots of a polynomial equation come in complex conjugate pairs". Explain why this does not apply to the polynomial equation in part (a).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q10 [8]}}