Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 2

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- Solution 2

Solution

Write

27 i = 27 e^{i (π /2 + 2 πk)}

. Every cube root has complex modulus

3

and an angle one third of one of these complex arguments:

z_{k} = 3 e^{i (π /6 + 2 πk /3)}, k = 0, 1, 2.

Equivalently the three complex roots are

\frac{3 3}{2} + \frac{3}{2} i, - \frac{3 3}{2} + \frac{3}{2} i, - 3 i .

They lie on the circle of radius

3

centred at the origin in the Argand plane, spaced by

2 π /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

27 i

.

Figure 1.
Original Argand diagrams for the three cube roots and the argument-constrained hyperbola
. Details.

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