Statement on simply connected domains
Theorem. If \(U\) is simply connected and \(f\) is holomorphic on \(U\), then \(\int_\gamma f(z)\,\mathrm{d}z=0\) for every closed piecewise \(C^1\) contour \(\gamma\) in \(U\). Equivalently, holomorphic functions on simply connected domains have primitives.
Triangle version
Theorem. If \(f\) is holomorphic on a domain containing a triangle and its interior, then the integral of \(f\) around the triangle is zero. The proof subdivides the triangle repeatedly and uses differentiability at the limiting point.
Homotopy form
Proposition. If two closed contours are homotopic in \(U\) and \(f\) is holomorphic on \(U\), then their integrals are equal. Cauchy's theorem is therefore a statement about topology as well as analysis.
Goursat proof idea
Proof. Goursat's proof avoids assuming continuity of \(f'\). Repeatedly divide a triangle into four smaller triangles and choose one where the integral is at least one fourth of the previous absolute integral. The nested triangles shrink to a point, where complex differentiability makes the remaining integral negligible.
Holes and obstructions
Example. The function \(1/z\) is holomorphic on \(\mathbb C\setminus\{0\}\), but \(\int_{|z|=1}z^{-1}\,\mathrm{d}z=2\pi i\). The puncture is a topological obstruction to a global primitive.
Why it matters
Remark. Cauchy's theorem is the engine behind Cauchy's integral formula, analyticity of holomorphic functions, residue calculus, and the deformation of contours.
Goursat's triangle theorem
Theorem (Goursat). If \(f\) is holomorphic on an open set containing a closed triangle \(\Delta\), then \[ \int_{\partial\Delta}f(z)\,\mathrm{d}z=0. \] No continuity assumption on \(f'\) is needed.
Proof. Subdivide \(\Delta\) into four similar triangles. Internal edges cancel, so one subtriangle has boundary integral at least one quarter of the original magnitude. Iterate to obtain nested triangles \(\Delta_n\) with diameters \(2^{-n}\operatorname{diam}\Delta\) and \[ \left|\int_{\partial\Delta_n}f\right| \ge4^{-n}\left|\int_{\partial\Delta}f\right|. \] Their intersection is a point \(z_*\). Complex differentiability gives \[ f(z)=f(z_*)+f'(z_*)(z-z_*)+\varepsilon(z)(z-z_*), \qquad\varepsilon(z)\to0. \] The constant and linear terms have zero integral around a triangle. Hence \[ \left|\int_{\partial\Delta_n}f\right| \le \sup_{\Delta_n}|\varepsilon|\, \operatorname{diam}(\Delta_n)L(\partial\Delta_n) =o(4^{-n}). \] Comparison with the lower bound forces the original integral to vanish.
Simply connected domains
Theorem (Cauchy). If \(\Omega\) is simply connected and \(f\) is holomorphic on \(\Omega\), then every closed piecewise \(C^1\) contour \(\gamma\subset\Omega\) satisfies \[ \int_\gamma f(z)\,\mathrm{d}z=0. \] Equivalently, every holomorphic function on \(\Omega\) has a primitive.
One proof triangulates a homotopy filling the contour and applies Goursat; another constructs a primitive by path integration and uses homotopy invariance of contour integrals. The topological hypothesis ensures every closed path can be contracted without crossing the complement.
Homology form
The sharp planar statement is homological: if a cycle \(\Gamma\) in \(\Omega\) has \[ \operatorname{Ind}(\Gamma,a)=0 \quad\text{for every }a\notin\Omega, \] then \(\int_\Gamma f\,\mathrm{d}z=0\) for every holomorphic \(f\) on \(\Omega\). This allows multiply connected domains as long as the cycle winds zero times around every hole.
Example (the puncture obstruction). On \(\mathbb C\setminus\{0\}\), \(f(z)=1/z\) is holomorphic, but \[ \int_{|z|=1}\frac{\mathrm{d}z}{z}=2\pi i. \] The circle is not null-homologous in the punctured plane, and \(1/z\) has no global primitive there. On a slit plane, a branch of \(\log z\) supplies such a primitive.
Deformation invariance
If closed contours \(\gamma_0,\gamma_1\) are homotopic through contours in \(\Omega\) and \(f\) is holomorphic, then \[ \int_{\gamma_0}f\,\mathrm{d}z =\int_{\gamma_1}f\,\mathrm{d}z. \] This makes contour deformation rigorous: a contour may move only through a region of holomorphy, and crossing a singularity changes the integral by its residue contribution.