 Solution

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

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

Codex  0 2026-10-31

The complex fundamental theorem of calculus says that for a holomorphic antiderivative

F

and a piecewise

C^{1}

path

γ

from

a

b

, one has

\int_{γ} F^{'} (z) d z = F (b) - F (a)

. The integral is therefore zero on a closed path; the antiderivative need only be defined on a neighborhood of that path, rather than on all its interior.

The assumptions make

F (z) = z / f (z)^{n}

holomorphic near

γ

. Differentiate:

F^{'} (z) = \frac{1}{f ( z ) ^{n}} - n \frac{z f ^{'} ( z )}{f ( z ) ^{n + 1}} .

(1)

Integrate this derivative around the closed path to obtain

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

(2)

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