Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 12

The original PDF asks for cube roots of $8$. Their complex modulus is $2$, and their arguments are $2k\pi /3$, $k=0,1,2$. Converting to Cartesian form givesCubing any one gives the positive real number $8$.

The quadratic polynomial factors asbecause the two roots sum to $3+i$ and multiply to $2+i$. Therefore

Real coefficients imply that the conjugate $-i$ is also a polynomial root. Write the remaining root as $r$. ThenThe coefficient of $z$ in this product is $-2(2+r)$. That coefficient is zero in the specified polynomial, so $r=-2$. Hence the two other roots areAs a check, the full polynomial is $2(z^2-4)(z^2+1)=2z^4-6z^2-8$, corresponding to $a=0$, $b=-6$ and $c=-8$.

Rewrite the complex position as $z(t)=2+3i/(t-i)$. Differentiating gives the complex velocity and accelerationFor real $t$, $|t-i|=\sqrt{t^2+1}$. Their requested magnitudes are thereforeThere is no pole for real time. The magnitude is the Euclidean length of the two Cartesian components, not the modulus of either component separately.