Determine the roots of the equation ${(z + 2{\text{i}})^3} = 216{\text{i}}$, $z \in \mathbb{C}$, giving the answers in the form $z = a\sqrt 3  + b{\text{i}}$ where $a,{\text{ }}b \in \mathbb{Z}$.

METHOD 1

$216{\text{i}} = 216\left( {\cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}} \right)$     A1

$z + 2{\text{i}} = \sqrt[3]{{216}}{\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) = {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)^{\frac{1}{3}}}$     (M1)

$z + 2{\text{i}} = 6\left( {\cos \left( {\frac{\pi }{6} + \frac{{2\pi k}}{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{{2\pi k}}{3}} \right)} \right)$     A1

${z_1} + 2{\text{i}} = 6\left( {\cos \frac{\pi }{6} + {\text{i}}\sin \frac{\pi }{6}} \right) = 6\left( {\frac{{\sqrt 3 }}{2} + \frac{{\text{i}}}{2}} \right) = 3\sqrt 3  + 3{\text{i}}$

${z_2} + 2{\text{i}} = 6\left( {\cos \frac{{5\pi }}{6} + {\text{i}}\sin \frac{{5\pi }}{6}} \right) = 6\left( {\frac{{ - \sqrt 3 }}{2} + \frac{{\text{i}}}{2}} \right) =  - 3\sqrt 3  + 3{\text{i}}$

${z_3} + 2{\text{i}} = 6\left( {\cos \frac{{3\pi }}{2} + {\text{i}}\sin \frac{{3\pi }}{2}} \right) =  - 6{\text{i}}$     A2

 

Note:     Award A1A0 for one correct root.

 

so roots are ${z_1} = 3\sqrt 3  + {\text{i, }}{z_2} =  - 3\sqrt 3  + {\text{i}}$ and ${z_3} =  - 8{\text{i}}$     M1A1

 

Note:     Award M1 for subtracting 2i from their three roots.

 

METHOD 2

${\left( {a\sqrt 3  + (b + 2){\text{i}}} \right)^3} = 216{\text{i}}$

${\left( {a\sqrt 3 } \right)^3} + 3{\left( {a\sqrt 3 } \right)^2}(b + 2){\text{i}} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} - {\text{i}}{(b + 2)^3} = 216{\text{i}}$     M1A1

${\left( {a\sqrt 3 } \right)^3} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} + {\text{i}}\left( {3{{\left( {a\sqrt 3 } \right)}^2}(b + 2) - {{(b + 2)}^3}} \right) = 216{\text{i}}$

${\left( {a\sqrt 3 } \right)^3} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} = 0$ and $3{\left( {a\sqrt 3 } \right)^2}(b + 2) - {(b + 2)^3} = 216$     M1A1

$a\left( {{a^2} - {{(b + 2)}^2}} \right) = 0$ and $9{a^2}(b + 2) - {(b + 2)^3} = 216$

$a = 0$ or ${a^2} = {(b + 2)^2}$

if $a = 0,{\text{ }} - {(b + 2)^3} = 216 \Rightarrow b + 2 =  - 6$

$\therefore b =  - 8$     A1

$(a,{\text{ }}b) = (0,{\text{ }} - 8)$

if ${a^2} = {(b + 2)^2},{\text{ }}9{(b + 2)^2}(b + 2) - {(b + 2)^3} = 216$

$8{(b + 2)^3} = 216$

${(b + 2)^3} = 27$

$b + 2 = 3$

$b = 1$

$\therefore {a^2} = 9 \Rightarrow a =  \pm 3$

$\therefore (a,{\text{ }}b) = ( \pm 3,{\text{ }}1)$     A1A1

so roots are ${z_1} = 3\sqrt 3  + {\text{i, }}{z_2} =  - 3\sqrt 3  + {\text{i}}$ and ${z_3} =  - 8{\text{i}}$

 

METHOD 3

${(z + 2{\text{i}})^3} - {( - 6{\text{i}})^3} = 0$

attempt to factorise:     M1

$\left( {(z + 2{\text{i}}) - ( - 6{\text{i}})} \right)\left( {{{(z + 2{\text{i}})}^2} + (z + 2{\text{i}})( - 6{\text{i}}) + {{( - 6{\text{i}})}^2}} \right) = 0$     A1

$(z + 8{\text{i}})({z^2} - 2{\text{i}}z - 28) = 0$     A1

$z + 8{\text{i}} = 0 \Rightarrow z =  - 8{\text{i}}$     A1

${z^2} - 2{\text{i}}z - 28 = 0 \Rightarrow z = \frac{{2{\text{i}} \pm \sqrt { - 4 - (4 \times 1 \times  - 28)} }}{2}$     M1

$z = \frac{{2{\text{i}} \pm \sqrt {108} }}{2}$

$z = \frac{{2{\text{i}} \pm 6\sqrt 3 }}{2}$

$z = {\text{i}} \pm 3\sqrt 3$     A1A1

 

Special Case:

Note:     If a candidate recognises that $\sqrt[3]{{216{\text{i}}}} =  - 6{\text{i}}$ (anywhere seen), and makes no valid progress in finding three roots, award A1 only.

 

[7 marks]

[N/A]