19M.2.AHL.TZ2.H_8

pestleMathematicsAAHLPaper 219M· ahl-1-13-polar-and-euler-formsource ↗

Find the roots of the equation ${w^3} = 8{\text{i}}$ , $w \in \mathbb{C}$ . Give your answers in Cartesian form.

[4]

One of the roots ${w_1}$ satisfies the condition ${\text{Re}}\left( {{w_1}} \right) = 0$ .

Given that ${w_1} = \frac{z}{{z - {\text{i}}}}$ , express $z$ in the form $a + b{\text{i}}$ , where $a$ , $b \in \mathbb{Q}$ .

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

${w^3} = 8{\text{i}}$

writing $8{\text{i}} = 8\left( {{\text{cos}}\left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{i}}\,{\text{sin}}\left( {\frac{\pi }{2} + 2\pi k} \right)} \right)$ (M1)

Note: Award M1 for an attempt to find cube roots of $w$ using modulus-argument form.

cube roots $w = 2\left( {{\text{cos}}\left( {\frac{{\frac{\pi }{2} + 2\pi k}}{3}} \right) + {\text{i}}\,{\text{sin}}\left( {\frac{{\frac{\pi }{2} + 2\pi k}}{3}} \right)} \right)$ (M1)

i.e. $w = \sqrt 3 + {\text{i,}}\,\, - \sqrt 3 + {\text{i,}}\,\, - 2{\text{i}}$ A2

Note: Award A2 for all 3 correct, A1 for 2 correct.

Note: Accept $w = 1.73 + {\text{i}}$ and $w = - 1.73 + {\text{i}}$ .

METHOD 2

${w^3} + {\left( {2{\text{i}}} \right)^3} = 0$

$\left( {w + 2{\text{i}}} \right)\left( {{w^2} - 2w{\text{i}} - 4} \right) = 0$ M1

$w = \frac{{2{\text{i}} \pm \sqrt {12} }}{2}$ M1

$w = \sqrt 3 + {\text{i,}}\,\, - \sqrt 3 + {\text{i,}}\,\, - 2{\text{i}}$ A2

Note: Award A2 for all 3 correct, A1 for 2 correct.

Note: Accept $w = 1.73 + {\text{i}}$ and $w = - 1.73 + {\text{i}}$ .

[4 marks]

${w_1} = - 2{\text{i}}$

$\frac{z}{{z - {\text{i}}}} = - 2{\text{i}}$ M1

$z = - 2{\text{i}}\left( {z - {\text{i}}} \right)$

$z\left( {1 + 2{\text{i}}} \right) = - 2$

$z = \frac{{ - 2}}{{1 + 2{\text{i}}}}$ A1

$z = - \frac{2}{5} + \frac{4}{5}{\text{i}}$ A1

Note: Accept $a = - \frac{2}{5}{\text{,}}\,\,b = \frac{4}{5}$ .

[3 marks]

Examiners’ report

[N/A]