 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-302/2/v/solution

Since

z + d z = φ (z, θ)

d z^{r} = \frac{\partial φ ^{r} ( z , θ )}{\partial θ ^{a}}_{θ = 0} θ^{a} + O (θ^{2}) .

Thus

d z^{r} = μ_{a}^{r} (z) θ^{a} .

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