Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 12 / c / i / Solution

Multiplying by $z^2-1$ gives $z^6-1=0$. The sixth roots of unity $1,-1$ do not solve the original equation, since substituting either gives three. The other four do, because the extra factor is nonzero. HenceTheir real parts are $\pm 1/2$ and their imaginary parts are $\pm \sqrt{3}/2$, giving the four corners in the left Argand diagram below.