= Solution
Write $27i=27e^{i(\pi/2+2\pi k)}$. Every <cube root> has <complex modulus> $3$ and an angle one third of one of these <complex arguments>:
$$
\boxed{z_k=3e^{i(\pi/6+2\pi k/3)},\quad k=0,1,2.}
$$
Equivalently the three <complex roots> are
$$
\boxed{\frac{3\sqrt3}{2}+\frac32i,\qquad
-\frac{3\sqrt3}{2}+\frac32i,\qquad -3i.}
$$
They lie on the circle of radius $3$ centred at the origin in the <Argand plane>, spaced by $2\pi/3$. Further integer values of $k$ repeat these three points. Cubing any of them multiplies its angle by three and cubes its <complex modulus> from $3$ to $27$, giving $27i$.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013-resit-complex-roots-and-locus.png]
{title=Original <Argand diagrams> for the three <cube roots> and the argument-constrained <hyperbola>}
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