= 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. Hence
$$
\boxed{z=e^{ik\pi/3},\qquad k=1,2,4,5.}
$$
Their <real parts> are $\pm1/2$ and their <imaginary parts> are $\pm\sqrt3/2$, giving the four corners in the left <Argand diagram> below.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1-complex-roots.png]
{title=Four roots on the unit circle and the two horizontal rows of complex sine solutions}
Back to article page
