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]