Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 11

Take the imaginary part of De Moivre's theorem at exponent five. Writing $s=\sin \theta$ and $c=\cos \theta$ givesFor real $x\in [-1,1]$, put $x=\sin \theta$. The equation becomes $\sin 5\theta =1$, so $\theta =\pi /10+2\pi k/5$. The distinct sine values are $1$, $(\sqrt{5}-1)/4$, and $-(1+\sqrt{5})/4$. To establish completeness over the complex numbers as well, factor the polynomial:Thus all polynomial roots, including their multiplicities, areThe repeated sine values account for the double roots; there are no additional nonreal roots hidden by the trigonometric parametrization.

Set $w=(z+1)/(z-1)$, with $z\ne 1$. For a nonzero integer $n$, all solutions of $w^n=i$ areThis covers negative integers too: $w$ is then nonzero and $w^{|n|}=-i$. No $w_k$ equals one. Inverting the Möbius transformation and using half-angle identities gives the cotangent roots of a powered Möbius transformation:These $|n|$ solutions are distinct and purely imaginary. For $n=0$ there is no solution: every defined zeroth power is one, not $i$. For negative $n$, $z=-1$ must additionally be excluded from the equation's domain; none of the displayed roots equals it.

For the cubic, direct expansion showsSince $z=1$ is not a root, the cubic is precisely the $n=3$ case. Taking $k=0,1,2$ yieldsFor an algebraic check, the cubic factors as $(z-i)(z^2+4iz-1)$, which gives the same three distinct roots.