CAIE Further Paper 2 2023 November — Question 2 5 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: shifted or transformed variable
DifficultyStandard +0.8 This is a cube root problem requiring conversion to polar form, application of de Moivre's theorem to find three roots, then converting back to a specific non-standard form. While the technique is standard for Further Maths, the algebraic manipulation and the unusual final form (requiring factoring out the -5i) elevate it above routine exercises. It's moderately challenging but well within expected Further Maths scope.
Spec4.02d Exponential form: re^(i*theta)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02k Argand diagrams: geometric interpretation4.02r nth roots: of complex numbers
2 Find the roots of the equation \(( z + 5 i ) ^ { 3 } = 4 + 4 \sqrt { 3 } i\), giving your answers in the form \(r \cos \theta + i ( r \sin \theta - 5 )\), where \(r > 0\) and \(0 < \theta < 2 \pi\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\((z+5i)^3 = 4+4i\sqrt{3} = 8e^{\frac{1}{3}\pi}\)B1 Finds modulus and argument of \(4+4i\sqrt{3}\)
\(z_1 = 2\left(\cos\frac{1}{9}\pi + i\sin\frac{1}{9}\pi\right)-5i = 2\cos\frac{1}{9}\pi + i\left(2\sin\frac{1}{9}\pi-5\right)\)M1 A1 Finds one root.
\(z_2 = 2\cos\frac{7}{9}\pi + i\left(2\sin\frac{7}{9}\pi-5\right)\), \(z_3 = 2\cos\frac{13}{9}\pi + i\left(2\sin\frac{13}{9}\pi-5\right)\)A1 FT Finds other two roots. FT on their modulus.
A1 FT
Total: 5 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(z+5i)^3 = 4+4i\sqrt{3} = 8e^{\frac{1}{3}\pi}$ | B1 | Finds modulus and argument of $4+4i\sqrt{3}$ |
| $z_1 = 2\left(\cos\frac{1}{9}\pi + i\sin\frac{1}{9}\pi\right)-5i = 2\cos\frac{1}{9}\pi + i\left(2\sin\frac{1}{9}\pi-5\right)$ | M1 A1 | Finds one root. |
| $z_2 = 2\cos\frac{7}{9}\pi + i\left(2\sin\frac{7}{9}\pi-5\right)$, $z_3 = 2\cos\frac{13}{9}\pi + i\left(2\sin\frac{13}{9}\pi-5\right)$ | A1 FT | Finds other two roots. FT on their modulus. |
| | A1 FT | |
| **Total: 5** | | |
2 Find the roots of the equation $( z + 5 i ) ^ { 3 } = 4 + 4 \sqrt { 3 } i$, giving your answers in the form $r \cos \theta + i ( r \sin \theta - 5 )$, where $r > 0$ and $0 < \theta < 2 \pi$.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q2 [5]}}
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Q1 4 Q2 5 Q3 6 Q4 10 Q5 10 Q6 14 Q7 11 Q8 15