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Cauchy's integral formula for a null-homologous cycle
Statement
Let be open, let be holomorphic, and let be a complex chain which is a cycle, with trace in and null-homologous in . Then for every
No connectedness of is assumed.
Facts & Assumptions
Given: An open , a holomorphic , and a cycle with which is null-homologous in .
With the filled difference quotient of , the function equal to on and to on is a well-defined entire function; it is bounded, and for every there is with whenever (Dixon's glued function is entire and vanishes at infinity).
Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).
If is continuous on the trace of a complex chain, then ; and for one has (Integration over a complex chain and the index of a chain). A chain is a finite list of integer-weighted contours with trace the union of the having (Complex chains, their traces, and cycles).
A cycle with trace in is null-homologous in when for every (Null-homologous cycles and homologous cycles in an open set).
Complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand); finite sums in the additive commutative monoid of are additive, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, is a field, every element is uniquely , and every nonzero element has inverse ).
The filled difference quotient of a holomorphic on equals off the diagonal and on it (The filled difference quotient of a holomorphic function is jointly continuous).
A function holomorphic on all of is entire (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).
A holomorphic function is continuous (Complex differentiability at a point implies continuity there).
Proof
By [L1] the glued function is entire and bounded, so [L2] makes it a constant .
By [L1], for every there is with for ; such exist, so for every and therefore .
Let . Since is holomorphic on , [L8] makes it continuous on , hence on the trace of . Then for every , so [L6] gives on the trace, and [L3] with [L5] splits the defining integral into .
Steps 1.1, 1.2 and 1.3 give , which is the stated formula; nothing in the argument used connectedness of , and the hypothesis that is null-homologous entered only through [L1] and [L4].
Depends on
- Dixon's glued function is entire and vanishes at infinity
- Liouville's theorem: every bounded entire function is constant
- Integration over a complex chain and the index of a chain
- Null-homologous cycles and homologous cycles in an open set
- Complex chains, their traces, and cycles
- Complex line integrals are linear in the integrand
- The filled difference quotient of a holomorphic function is jointly continuous
- Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions
- A finite sum in a commutative monoid indexed by an arbitrary finite set
- $\mathbb C=\mathbb R[x]/(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)/(a^2+b^2)$
- Complex differentiability at a point implies continuity there
Used by
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Sources
- J. Lebl, Complex Analysis, Ch. 4 §4.2 (standard reference, not scraped)
- M. Weber, Complex Analysis (Indiana University), Ch. 4 §4.1 (standard reference, not scraped)