OCR FP3 2006 June — Question 2 7 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2006
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeModulus and argument calculations
DifficultyModerate -0.8 This is a straightforward application of exponential and polar form rules for complex numbers. Part (a) requires multiplying/dividing moduli and adding/subtracting arguments—direct formula application. Part (b) uses De Moivre's theorem with negative exponent. Both parts are routine FP3 exercises with no problem-solving or conceptual challenges, making this easier than average even for Further Maths.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02e Arithmetic of complex numbers: add, subtract, multiply, divide

  1. Given that \(z_1 = 2e^{\frac{5\pi i}{6}}\) and \(z_2 = 3e^{\frac{2\pi i}{3}}\), express \(z_1z_2\) and \(\frac{z_1}{z_2}\) in the form \(re^{i\theta}\), where \(r > 0\) and \(0 \leq \theta < 2\pi\). [4]
  2. Given that \(w = 2(\cos \frac{1}{3}\pi + i \sin \frac{1}{3}\pi)\), express \(w^{-5}\) in the form \(r(\cos \theta + i \sin \theta)\), where \(r > 0\) and \(0 \leq \theta < 2\pi\). [3]

AnswerMarks Guidance
(a) \((z_1z_2) = 6e^{i\frac{23}{12}\pi}\); \(\frac{z_1}{z_2} = \frac{2}{3}e^{-i\frac{1}{12}\pi} = \frac{2}{3}e^{i\frac{23}{12}\pi}\)B1; B1; M1 A1 4 For modulus = 6; For argument = \(\frac{23}{12}\pi\); For subtracting arguments; For correct answer
(b) \((w^{-5}) = 2^{-5}\cis(-\frac{5}{8}\pi)\) = \(\frac{1}{32}(\cos\frac{11}{8}\pi + i\sin\frac{11}{8}\pi)\)M1; A1; A1 3 [7] For use of de Moivre; For \(-\frac{5}{8}\pi\) seen or implied; For correct answer (allow \(2^{-5}\) and \(\cis\frac{11}{8}\pi\))
**(a)** $(z_1z_2) = 6e^{i\frac{23}{12}\pi}$; $\frac{z_1}{z_2} = \frac{2}{3}e^{-i\frac{1}{12}\pi} = \frac{2}{3}e^{i\frac{23}{12}\pi}$ | B1; B1; M1 A1 4 | For modulus = 6; For argument = $\frac{23}{12}\pi$; For subtracting arguments; For correct answer

**(b)** $(w^{-5}) = 2^{-5}\cis(-\frac{5}{8}\pi)$ = $\frac{1}{32}(\cos\frac{11}{8}\pi + i\sin\frac{11}{8}\pi)$ | M1; A1; A1 3 [7] | For use of de Moivre; For $-\frac{5}{8}\pi$ seen or implied; For correct answer (allow $2^{-5}$ and $\cis\frac{11}{8}\pi$)

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\begin{enumerate}[label=(\alph*)]
\item Given that $z_1 = 2e^{\frac{5\pi i}{6}}$ and $z_2 = 3e^{\frac{2\pi i}{3}}$, express $z_1z_2$ and $\frac{z_1}{z_2}$ in the form $re^{i\theta}$, where $r > 0$ and $0 \leq \theta < 2\pi$. [4]

\item Given that $w = 2(\cos \frac{1}{3}\pi + i \sin \frac{1}{3}\pi)$, express $w^{-5}$ in the form $r(\cos \theta + i \sin \theta)$, where $r > 0$ and $0 \leq \theta < 2\pi$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR FP3 2006 Q2 [7]}}