 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/11/b/ii/solution

Squaring gives

u = z^{2} = t^{2} - t^{- 2} + 2 i .

Thus

Im u = 2

is constant, while

d (Re u) / d t = 2 t + 2 t^{- 3} > 0

. The trajectory is the horizontal line segment

u : (- 15/4, 2) ⟶ (15/4, 2), Im u = 2.

It crosses the imaginary axis at

(0, 2)

, when

t = 1

, and never crosses the real axis. The middle panel shows the rightward direction.

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