Find three distinct roots of the equation $8{z^3} + 27 = 0,{\text{ }}z \in \mathbb{C}$ giving your answers in modulus-argument form.

The roots are represented by the vertices of a triangle in an Argand diagram.

Show that the area of the triangle is $\frac{{27\sqrt 3 }}{{16}}$.

METHOD 1

${z^3} =  - \frac{{27}}{8} = \frac{{27}}{8}(\cos \pi  + {\text{i}}\sin \pi )$     M1(A1)

$= \frac{{27}}{8}\left( {\cos (\pi  + 2n\pi ) + {\text{i}}\sin (\pi  + 2n\pi )} \right)$     (A1)

$z = \frac{3}{2}\left( {\cos \left( {\frac{{\pi  + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\pi  + 2n\pi }}{3}} \right)} \right)$     M1

${z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)$,

${z_2} = \frac{3}{2}(\cos \pi  + {\text{i}}\sin \pi )$,

${z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)$.     A2

 

Note:     Accept $- \frac{\pi }{3}$ as the argument for ${z_3}$.

 

Note:     Award A1 for $2$ correct roots.

 

Note:     Allow solutions expressed in Eulerian $(r{e^{{\text{i}}\theta }})$ form.

 

Note:     Allow use of degrees in mod-arg (r-cis) form only.

 

METHOD 2

$8{z^3} + 27 = 0$

$\Rightarrow z =  - \frac{3}{2}$ so $(2z + 3)$ is a factor

Attempt to use long division or factor theorem:     M1

$\Rightarrow 8{z^3} + 27 = (2z + 3)(4{z^2} - 6z + 9)$

$\Rightarrow 4{z^2} - 6z + 9 = 0$     A1

Attempt to solve quadratic:     M1

$z = \frac{{3 \pm 3\sqrt 3 {\text{i}}}}{4}$     A1

${z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)$,

${z_2} = \frac{3}{2}(\cos \pi  + {\text{i}}\sin \pi )$,

${z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)$.     A2

 

Note:     Accept $- \frac{\pi }{3}$ as the argument for ${z_3}$.

 

Note:     Award A1 for $2$ correct roots.

 

Note:     Allow solutions expressed in Eulerian $(r{e^{{\text{i}}\theta }})$ form.

 

Note:     Allow use of degrees in mod-arg (r-cis) form only.

 

METHOD 3

$8{z^3} + 27 = 0$

Substitute $z = x + {\text{i}}y$     M1

$8({x^3} + 3{\text{i}}{x^2}y - 3x{y^2} - {\text{i}}{y^3}) + 27 = 0$

$\Rightarrow 8{x^3} - 24x{y^2} + 27 = 0$ and $24{x^2}y - 8{y^3} = 0$     A1

Attempt to solve simultaneously:     M1

$8y(3{x^2} - {y^2}) = 0$

$y = 0,{\text{ }}y = x\sqrt 3 ,{\text{ }}y =  - x\sqrt 3$

$\Rightarrow \left( {x =  - \frac{3}{2},{\text{ }}y = 0} \right),{\text{ }}x = \frac{3}{4},{\text{ }}y =  \pm \frac{{3\sqrt 3 }}{4}$     A1

${z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)$,

${z_2} = \frac{3}{2}(\cos \pi  + {\text{i}}\sin \pi )$,

${z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)$.     A2

 

Note:     Accept $- \frac{\pi }{3}$ as the argument for ${z_3}$.

 

Note:     Award A1 for $2$ correct roots.

 

Note:     Allow solutions expressed in Eulerian $(r{e^{{\text{i}}\theta }})$ form.

 

Note:     Allow use of degrees in mod-arg (r-cis) form only.

[6 marks]

EITHER

Valid attempt to use ${\text{area}} = 3\left( {\frac{1}{2}ab\sin C} \right)$     M1

$= 3 \times \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{{\sqrt 3 }}{2}$     A1A1

 

Note:     Award A1 for correct sides, A1 for correct sin $C$.

 

OR

Valid attempt to use ${\text{area}} = \frac{1}{2}{\text{base}} \times {\text{height}}$     M1

${\text{area}} = \frac{1}{2} \times \left( {\frac{3}{4} + \frac{3}{2}} \right) \times \frac{{6\sqrt 3 }}{4}$     A1A1

 

Note:     A1 for correct height, A1 for correct base.

 

THEN

$= \frac{{27\sqrt 3 }}{{16}}$     AG

[3 marks]

Total [9 marks]

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