Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 17

The point has positive real part and negative imaginary part. Its complex modulus is $\sqrt{1+3}=2$, and its principal complex argument is $-\pi /3$. Hence

Use Euler's formula to evaluate the two exponentials: $e^{i\pi /2}=i$ and $\sqrt{2}{e}^{i\pi /4}=1+i$. Thus $z=1+2i$ lies in the first quadrant, givingIts principal argument is approximately $1.10715$ radians.

Write $1+i=\sqrt{2}{e}^{i\pi /4}$ and multiply the phases:ThereforeThe displayed representative already lies in the principal argument interval.

Set $w=z^3$. Then $w^2+w+1=0$ gives $w=e^{2\pi i/3}$ or $e^{-2\pi i/3}$. Taking all three cube roots of each yields six distinct solutions.

Equivalently, multiply the original polynomial by $z^3-1$ to obtain $z^9-1$. The original roots are exactly the ninth roots of unity that are not cube roots of unity: a cube root would give $1+1+1=3$, not zero. ThusAll six have modulus one, and their distinct arguments show that no roots have been repeated or omitted from the degree-six polynomial.