 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-316/3/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

\overset{e}{˙} = \sum_{j} A_{j} e_{j} sin (ϖ - ϖ_{j}), \overset{ϖ}{˙} = A + \sum_{j} A_{j} \frac{e _{j}}{e} cos (ϖ - ϖ_{j}) .

z = e e^{iϖ}

\overset{z}{˙} = e^{iϖ} (\overset{e}{˙} + i e \overset{ϖ}{˙}) .

Use

sin δ + i cos δ = i e^{- i δ}

. Each planetary term becomes

i A_{j} e_{j} e^{i ϖ_{j}} = i A_{j} z_{j}

, so

\overset{z}{˙} = i A z + i j = 1 \sum N A_{j} z_{j} .

Unlike the polar orbital element equations, this linear equation is nonsingular at

e = 0

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