CAIE P3 2004 June — Question 8 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2004
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeStandard quadratic with real coefficients
DifficultyModerate -0.3 This is a straightforward multi-part question testing standard complex number techniques: solving a quadratic using the formula, converting to modulus-argument form, and verifying a simple algebraic property. All parts are routine applications of basic complex number theory with no novel insight required, making it slightly easier than average.
Spec4.02b Express complex numbers: cartesian and modulus-argument forms4.02i Quadratic equations: with complex roots4.02q De Moivre's theorem: multiple angle formulae

8
  1. Find the roots of the equation \(z ^ { 2 } - z + 1 = 0\), giving your answers in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
  2. Obtain the modulus and argument of each root.
  3. Show that each root also satisfies the equation \(z ^ { 3 } = - 1\).

Question 8:
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i) EITHER: Solve the quadratic and use \(\sqrt{-1} = i\)M1
Obtain roots \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\), or equivalentA1
OR: Substitute \(x + iy\) and solve for \(x\) or \(y\)M1
Obtain correct rootsA1 Total: 2
(ii) State that the modulus of each root is equal to 1B1\(\sqrt{}\)
State that the arguments are \(\frac{1}{3}\pi\) and \(-\frac{1}{3}\pi\) respectivelyB1\(\sqrt{}\) + B1\(\sqrt{}\) Total: 3
[Accept degrees and \(\frac{5}{3}\pi\) instead of \(-\frac{1}{3}\pi\). Accept modulus in form \(\sqrt{\frac{p}{q}}\) or \(\sqrt{n}\), where \(p, q, n\) are integers.]
(iii) EITHER: Verify \(z^3 = -1\) for each rootB1 + B1
OR: State \(z^3 + 1 = (z+1)(z^2 - z + 1)\)B1
Justify the given statementB1
OR: Obtain \(z^3 = z^2 - z\)B1
Justify the given statementB1 Total: 2
## Question 8:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i) EITHER:** Solve the quadratic and use $\sqrt{-1} = i$ | M1 | |
| Obtain roots $\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\frac{1}{2} - i\frac{\sqrt{3}}{2}$, or equivalent | A1 | |
| **OR:** Substitute $x + iy$ and solve for $x$ or $y$ | M1 | |
| Obtain correct roots | A1 | **Total: 2** |
| **(ii)** State that the modulus of each root is equal to 1 | B1$\sqrt{}$ | |
| State that the arguments are $\frac{1}{3}\pi$ and $-\frac{1}{3}\pi$ respectively | B1$\sqrt{}$ + B1$\sqrt{}$ | **Total: 3** |
| [Accept degrees and $\frac{5}{3}\pi$ instead of $-\frac{1}{3}\pi$. Accept modulus in form $\sqrt{\frac{p}{q}}$ or $\sqrt{n}$, where $p, q, n$ are integers.] | | |
| **(iii) EITHER:** Verify $z^3 = -1$ for each root | B1 + B1 | |
| **OR:** State $z^3 + 1 = (z+1)(z^2 - z + 1)$ | B1 | |
| Justify the given statement | B1 | |
| **OR:** Obtain $z^3 = z^2 - z$ | B1 | |
| Justify the given statement | B1 | **Total: 2** |

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8 (i) Find the roots of the equation $z ^ { 2 } - z + 1 = 0$, giving your answers in the form $x + \mathrm { i } y$, where $x$ and $y$ are real.\\
(ii) Obtain the modulus and argument of each root.\\
(iii) Show that each root also satisfies the equation $z ^ { 3 } = - 1$.

\hfill \mbox{\textit{CAIE P3 2004 Q8 [7]}}