WJEC Further Unit 4 Specimen — Question 4 9 marks

Exam BoardWJEC
ModuleFurther Unit 4 (Further Unit 4)
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: general complex RHS
DifficultyChallenging +1.2 This is a standard Further Maths complex numbers question requiring conversion to modulus-argument form, application of De Moivre's theorem for cube roots, and conversion back to Cartesian form. While it involves multiple steps and careful angle arithmetic (finding three roots with 120° separation), it's a well-practiced technique with no novel insight required, making it moderately above average difficulty but routine for Further Maths students.
Spec4.02r nth roots: of complex numbers
Find the three cube roots of the complex number \(2 + 3i\), giving your answers in Cartesian form. [9]
AnswerMarks Guidance
\(z = \sqrt{13}\)
\(\arg(z) = \tan^{-1}1.5 = 0.98279...\)B1 AO3
First cube root: \(z = \sqrt{13}(\cos 0.98279... + i\sin 0.98279...)\)
AnswerMarks Guidance
\(= 13^{1/6}(\cos 0.32759... + i\sin 0.32759...)\)M1, m1 AO3
\(= 1.45 + 0.493i\)A1 AO3
Second cube root: \(= 13^{1/6}(\cos(0.32759... + 2\pi/3) + i\sin(0.32759... + 2\pi/3))\)M1 AO3
\(= -1.15 + 1.01i\)A1 AO3
Third cube root: \(= 13^{1/6}(\cos(0.32759... + 4\pi/3) + i\sin(0.32759... + 4\pi/3))\)M1 AO3
\(= -0.298 - 1.50i\)A1 AO3
Total: [9]
$|z| = \sqrt{13}$ | B1 | AO3

$\arg(z) = \tan^{-1}1.5 = 0.98279...$ | B1 | AO3

**First cube root:** $z = \sqrt{13}(\cos 0.98279... + i\sin 0.98279...)$

$= 13^{1/6}(\cos 0.32759... + i\sin 0.32759...)$ | M1, m1 | AO3

$= 1.45 + 0.493i$ | A1 | AO3

**Second cube root:** $= 13^{1/6}(\cos(0.32759... + 2\pi/3) + i\sin(0.32759... + 2\pi/3))$ | M1 | AO3

$= -1.15 + 1.01i$ | A1 | AO3

**Third cube root:** $= 13^{1/6}(\cos(0.32759... + 4\pi/3) + i\sin(0.32759... + 4\pi/3))$ | M1 | AO3

$= -0.298 - 1.50i$ | A1 | AO3

**Total: [9]**

---
Find the three cube roots of the complex number $2 + 3i$, giving your answers in Cartesian form. [9]

\hfill \mbox{\textit{WJEC Further Unit 4  Q4 [9]}}
This paper (12 questions)
View full paper
Q1 7 Q2 6 Q3 5 Q4 9 Q5 8 Q6 7 Q7 10 Q8 10 Q9 14 Q10 11 Q11 17 Q12 16