= Solution
Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, so
$$
v=u^2=(s^2-4)+4si.
$$
The 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 are
$$
\boxed{v:\ (161/16,-15)\longrightarrow(161/16,15).}
$$
The <real axis> crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two <imaginary axis> crossings satisfy $s^2=4$, giving
$$
\boxed{(0,-8)\text{ at }t=\sqrt{\sqrt2-1},\qquad(0,8)\text{ at }t=\sqrt{\sqrt2+1}.}
$$
In 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.
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