WJEC Further Unit 4 2023 June — Question 7 7 marks

Exam BoardWJEC
ModuleFurther Unit 4 (Further Unit 4)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: general complex RHS
DifficultyChallenging +1.2 Finding cube roots of a complex number requires converting to modulus-argument form, applying De Moivre's theorem with three roots, then converting back to Cartesian form—a multi-step process involving several techniques. While this is a standard Further Maths topic with a well-defined method, it requires careful calculation with non-trivial arithmetic (finding |z| and arg(z) for 11-2i, then cube roots and trigonometric evaluations), making it moderately harder than average A-level questions but still routine for Further Maths students.
Spec4.02r nth roots: of complex numbers
Find the cube roots of the complex number \(z = 11 - 2i\), giving your answers in the form \(x + iy\), where \(x\) and \(y\) are real and correct to three decimal places. [7]
AnswerMarks Guidance
\(\z\ = \sqrt{11^2 + (-2)^2} = 5\sqrt{5}\)
\(\arg(z) = \tan^{-1}\left(-\frac{2}{11}\right) = -0.179(85\ldots)\)B1 si
Therefore cube roots:
1st root: \(\sqrt[3]{5\sqrt{5}}(\cos -0.05995 + i\sin -0.05995) = 2.232 - 0.134i\)M1, A1 Use of de Moivre. FT \(\
2nd root: \(\sqrt[3]{5\sqrt{5}}(\cos 2.0344 + i\sin 2.0344) = -1.000 + 2.000i\)M1, A1
3rd root: \(\sqrt[3]{5\sqrt{5}}(\cos 4.1288 + i\sin 4.1288) = -1.232 - 1.866i\)A1
[7]Note: \(\sqrt[3]{5\sqrt{5}} = \sqrt{5}\) 
| $\|z\| = \sqrt{11^2 + (-2)^2} = 5\sqrt{5}$ | B1 | si |
| $\arg(z) = \tan^{-1}\left(-\frac{2}{11}\right) = -0.179(85\ldots)$ | B1 | si |
| | | |
| Therefore cube roots: | | |
| 1st root: $\sqrt[3]{5\sqrt{5}}(\cos -0.05995 + i\sin -0.05995) = 2.232 - 0.134i$ | M1, A1 | Use of de Moivre. FT $\|z\|$ and $\arg(z)$; $\frac{2\pi}{3}$ |
| 2nd root: $\sqrt[3]{5\sqrt{5}}(\cos 2.0344 + i\sin 2.0344) = -1.000 + 2.000i$ | M1, A1 | |
| 3rd root: $\sqrt[3]{5\sqrt{5}}(\cos 4.1288 + i\sin 4.1288) = -1.232 - 1.866i$ | A1 | |
| | [7] | Note: $\sqrt[3]{5\sqrt{5}} = \sqrt{5}$ |
Find the cube roots of the complex number $z = 11 - 2i$, giving your answers in the form $x + iy$, where $x$ and $y$ are real and correct to three decimal places. [7]

\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q7 [7]}}
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