= Solution
The components are in <phase quadrature> when the axes of the <ellipse> coincide with the coordinate axes. Since the $x$ <wave amplitude> is one and the $y$ <wave amplitude> is two, the <Jones vector> must have $A=\pm2i$.
A handedness convention is essential. Adopt the optical convention in which right-handed <elliptical polarization> rotates clockwise when viewed toward the source, from the positive-$z$ side. With the paper's $e^{+i\omega t}$ <phase convention for complex amplitudes>, this gives
$$
\boxed{A=+2i,\qquad \boldsymbol\Psi(t)=(\cos\omega t,-2\sin\omega t).}
$$
The tip follows $\Psi_x^2+\Psi_y^2/4=1$. In particular,
$$
\boxed{\boldsymbol\Psi(0)=(1,0),\qquad \boldsymbol\Psi(\pi/(2\omega))=(0,-2).}
$$
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2-a2-polarization.png]
{title=An original <elliptical polarization> trajectory and the two requested <vectors>, using $A=+2i$}
The figure states the viewing direction and shows increasing time. Under the alternative convention that calls rotation from $+x$ toward $+y$ right-handed when looking along propagation, the equally consistent answer is $A=-2i$ and the second <vector> is $(0,+2)$. The original paper does not state a handedness convention, so its wording alone cannot select between these labels; the real components and viewing direction remove the ambiguity.
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