 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/14/a/iv/solution

e^{2 πi /3}

. Squaring doubles its argument, giving

e^{4 πi /3} = - 1/2 - i 3 /2

. Hence

ℜ z = - \frac{1}{2}, ℑ z = - \frac{3}{2} .

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