Past exam of the mathematical methods I course of the University of Oxford / 2018 / 7

For a purely imaginary candidate $z=it$, $t\in \mathbb{R}$, the imaginary part of the polynomial is $2t(t^2-4)$. The candidate $t=0$ is not a root, while $t=\pm 2$ both make the real part zero. The resulting factorization into quadratic polynomials givesThe roots of a polynomial sum to $0+2=2$ and have product $4\cdot 4=16$, as required by Vieta's formulas for a monic quartic with cubic coefficient $-2$ and constant coefficient $16$.

Expand $(\cos \theta +i\sin \theta {)}^4$ using De Moivre's theorem, then divide its imaginary and real parts by ${\cos }^4\theta$. With $u=\tan \theta$, this givesThe quotient holds where its denominator and $\cos \theta$ are nonzero. At $\theta =\pi /8$, use the undivided real-part identity: $\cos 4\theta =0$ implies $u^4-6u^2+1=0$. Thus $u^2=3\pm 2\sqrt{2}$. Since $0<\tan (\pi /8)<1$, the smaller positive value applies:This limiting step avoids treating an undefined value of $\tan 4\theta$ as an ordinary finite number.

A plane through $r_0$ with nonzero normal vector $n$ consists of the position vectors $r$ satisfying $(r-r_0)\cdot n=0$. Thus $r\cdot n=d$, with $d=r_0\cdot n$. When $n$ is a unit vector, $d$ is the signed distance from the origin in the direction $n$; for an unnormalized normal vector, that distance is $d/\Vert n\Vert$. The displacement within the plane is perpendicular to $n$.

The cross product of the two normal vectors is $(2,6,-3)\times (5,2,-1)=(0,-13,-26)$, so their intersection of two planes has direction $(0,1,2)$. Eliminating $2y-z$ gives $x=2$, and substitution then gives $z=2y-2$. Thus the Cartesian constraints and parametric equation of a straight line areThere is no valid division by a zero $x$ component in a symmetric ratio form; retaining $x=2$ explicitly handles that component.