Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 11

Expand the square using $i^2=-1$:Thus the real part and imaginary part are

Square the result of part (a)(i), keeping the cross term:Hence

Write $z=X+iY$. Its coordinates are $X=t$, $Y=1/t$, so $XY=1$. For the given positive range of $t$, this traces a segment of the first-quadrant hyperbola, moving rightward and downward:The complex conjugate is $z^{\ast }=t-i/t$. It is the reflection in the real axis, tracing $XY=-1$ in the fourth quadrant, rightward and upward:Neither trajectory crosses either axis, because $t$ and $1/t$ remain nonzero. The left panel of the figure marks the endpoints and directions.

Squaring givesThus $\mathrm{Im}u=2$ is constant, while $d(\mathrm{Re}u)/dt=2t+2t^{-3}>0$. The trajectory is the horizontal line segmentIt crosses the imaginary axis at $\boxed{(0,2)}$, when $t=1$, and never crosses the real axis. The middle panel shows the rightward direction.

Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, soThe parameter $s$ increases strictly from $-15/4$ to $15/4$. Therefore the imaginary part $Y=4s$ increases from $-15$ to $15$, while the real part is $X=s^2-4$. The endpoints areThe real axis crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two imaginary axis crossings satisfy $s^2=4$, givingIn order, the trajectory passes from its lower endpoint through $(0,-8)$, then the leftmost point $(-4,0)$, then $(0,8)$, to its upper endpoint. It is a right-opening parabola segment, traversed continuously upward as shown in the right panel.

From $X=s^2-4$ and $Y=4s$, elimination of $s$ givesThis makes the real part a quadratic function of the imaginary part. There is no sign ambiguity because $Y$ determines $s=Y/4$ uniquely. The calculation illustrates power maps of a reciprocal-coordinate hyperbola.