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Question

HLPaper 1

Let the roots of the equation ${z^3} = - 3 + \sqrt 3 \text{i}$ be $a$ , $b$ and $c$ .

On an Argand diagram, $a$ , $b$ and $c$ are represented by the points X, Y and Z respectively.

Find the area of triangle $XYZ$ .

[4]

Verified

Solution

METHOD 1 attempting to find the total area of (congruent) triangles XOY, YOZ and XOZM1 Area $= 3\left( {\frac{1}{2}} \right)\left( {12^{\frac{1}{6}}} \right)\left( {12^{\frac{1}{6}}} \right)\sin\frac{2\pi}{3}$ A1A1 Note: Award A1 for $\left( {12^{\frac{1}{6}}} \right)\left( {12^{\frac{1}{6}}} \right)$ and A1 for $\sin\frac{2\pi}{3}$ = $\frac{3\sqrt 3 }{4}\left( {12^{\frac{1}{3}}} \right)$ (or equivalent) A1

METHOD2 XY^2 $= {\left( {12^{\frac{1}{6}}} \right)^2} + {\left( {12^{\frac{1}{6}}} \right)^2} - 2\left( {12^{\frac{1}{6}}} \right)\left( {12^{\frac{1}{6}}} \right)\cos\frac{2\pi }{3}$ (or equivalent) A1 XY $= \sqrt 3 \left( {12^{\frac{1}{6}}} \right)$ (or equivalent) A1 attempting to find the area of XYZ using Area = $\frac{1}{2}$ × XY × YZ × $\sin\alpha$ for exampleM1 Area $= \frac{1}{2}\left( {\sqrt 3 \times {12^{\frac{1}{6}}}} \right)\left( {\sqrt 3 \times {12^{\frac{1}{6}}}} \right)\sin\frac{\pi }{3}$ = $\frac{3\sqrt 3 }{4}\left( {12^{\frac{1}{3}}} \right)$ (or equivalent) A1

[4 marks]

Express $- 3 + \sqrt{3} i$ in the form $r e^{i\theta}$ , where $r > 0$ and $-\pi < \theta \leqslant \pi$ .

[5]

Verified

Solution

attempt to find modulus (M1) $r = 2\sqrt{3} \left( { = \sqrt{12} } \right)$ A1 attempt to find argument in the correct quadrant (M1) $\theta = \pi + \text{arctan}\left( { - \frac{\sqrt{3}}{3}} \right)$ A1 $= \frac{5\pi}{6}$ A1 $- 3 + \sqrt{3}i = \sqrt{12} \, \text{e}^{\frac{5\pi i}{6}}\left( { = 2\sqrt{3} \, \text{e}^{\frac{5\pi i}{6}}} \right)$ [5 marks]

Find $a$ , $b$ and $c$ expressing your answers in the form $r\left(e^{i\theta}\right)$ , where $r > 0$ and $-\pi < \theta \leq \pi$ .

[5]

Verified

Solution

attempt to find a root using de Moivre’s theorem M1 $12^{\frac{1}{6}} \mathrm{e}^{\frac{5\pi \mathrm{i}}{18}}$ A1 attempt to find further two roots by adding and subtracting $\frac{2\pi}{3}$ to the argument *** M1*** $12^{\frac{1}{6}} \mathrm{e}^{ - \frac{7\pi \mathrm{i}}{18}}$ A1 $12^{\frac{1}{6}} \mathrm{e}^{\frac{17\pi \mathrm{i}}{18}}$ A1 Note: Ignore labels for $a$ , $b$ and $c$ at this stage.

[5 marks]

By considering the sum of the roots $a$ , $b$ and $c$ , show that

${\cos}\frac{{5\pi }}{{18}} + {\cos}\frac{{7\pi }}{{18}} + {\cos}\frac{{17\pi }}{{18}} = 0$ .

[4]

Verified

Solution

$a + b + c = 0$ R1 $12^{\frac{1}{6}}\left( \cos\left( - \frac{7\pi }{18} \right) + \text{i}\,\sin\left( - \frac{7\pi }{18} \right) + \cos\frac{5\pi }{18} + \text{i}\,\sin\frac{5\pi }{18} + \cos\frac{17\pi }{18} + \text{i}\,\sin\frac{17\pi }{18} \right) = 0$ A1 consideration of real parts M1 $12^{\frac{1}{6}}\left( \cos\left( - \frac{7\pi }{18} \right) + \cos\frac{5\pi }{18} + \cos\frac{17\pi }{18} \right) = 0$ $\cos\left( - \frac{7\pi }{18} \right) = \cos\frac{17\pi }{18}$ explicitly stated A1 $\cos\frac{5\pi }{18} + \cos\frac{7\pi }{18} + \cos\frac{17\pi }{18} = 0$ AG [4 marks]

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