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ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2009/analysis/6/c/solution

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Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2009 / Analysis / 6 / c / Solution by

Codex  0 2026-10-31

Choose a small positively oriented circle

∣ w ∣ = ρ

in

U

within the convergence disk of the inverse series. Cauchy's coefficient formula gives

c_{n} = \frac{1}{2 πi} \int_{∣ w ∣ = ρ} \frac{g ( w )}{w ^{n + 1}} d w .

Let

γ = g ({∣ w ∣ = ρ})

. The biholomorphism property makes it a simple positively oriented closed curve about zero. Under

w = f (z)

,

g (w) = z

and

d w = f^{'} (z) d z

, so

c_{n} = \frac{1}{2 πi} \int_{γ} \frac{z f ^{'} ( z )}{f ( z ) ^{n + 1}} d z = \frac{1}{2 πin} \int_{γ} \frac{d z}{f ( z ) ^{n}} .

The last equality is (a). Apply (b), giving

c_{n} = \frac{1}{n !} [\frac{d ^{n - 1}}{d z ^{n - 1}} (\frac{z}{f ( z )})^{n}]_{z = 0} .

This is the local Lagrange inversion theorem; the simple zero ensures that the expression being differentiated has a removable singularity and a nonzero value at zero.

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