 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-1/1c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

z \neq = 0

z^{w} = exp (w Log z)

Log z = lo g ∣ z ∣ + i Arg z

with

- π < Arg z \leq π

(cos θ + i sin θ)^{n} = cos (n θ) + i sin (n θ)

for every integer

n

Since

3 + i = 2 e^{iπ /6}

, the six roots are

z = 2^{1/6} exp (i (\frac{π}{36} + \frac{kπ}{3})), k = 0, 1, \dots, 5.

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