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 $a,b\in \mathbb{Z}$.

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

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 $9a^2(b+2)-(b+2{)}^3=216$

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

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

$\therefore b=-8$     A1

$(a,b)=(0,-8)$

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

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

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

$b+2=3$

$b=1$

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

$\therefore (a,b)=(\pm 3,1)$     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]