 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/12/b/solution

u + i v

, with

u, v

real. Equating the real part and imaginary part of its square gives

u^{2} - v^{2} = - 1

and

2 uv = 1

. Taking complex modulus also gives

u^{2} + v^{2} = 2

. Therefore

u^{2} = \frac{2 - 1}{2}, v^{2} = \frac{2 + 1}{2} .

Since

uv > 0

, the signs agree. The two roots are

\pm \frac{2 - 1}{2} + i \frac{2 + 1}{2} .

The

\pm

changes both signs together, rather than independently.

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