User:Anghel/Sandbox

Proof
This proof assumes that the derivative $f' : D \to \C$ is continuous.

This assumption is always true, as can be seen from Holomorphic Function is Continuously Differentiable.

However, the proof of that theorem depends on Holomorphic Function is Analytic, which depends on Cauchy's Integral Formula, which again depends on the Cauchy-Goursat Theorem.

It follows that this proof is circular, unless $f$ is known to be continuously differentiable.

This proof also references the Cauchy-Riemann Equations for results about the partial derivatives of $f$.

The part of the proof for the Cauchy-Riemann Equations that show these partial derivatives are continuous still depends on the Holomorphic Function is Continuously Differentiable theorem.

We rewrite $f$ as a sum of its real and imaginary parts.

Let $u, v: \R^2 \to \R$ be defined by:


 * $\map f {x + i y} = \map u {x, y} + i \map v {x, y}$

By the Cauchy-Riemann Equations, the partial derivatives of $u$ and $v$ fulfil these equations:


 * $\map {\dfrac {\partial u} {\partial x} } {x, y} = \map {\dfrac {\partial v} {\partial y} } {x, y} = \map \Re {\map {f'} z}$
 * $\map {\dfrac {\partial u} {\partial y} } {x, y} = -\map {\dfrac {\partial v} {\partial x} } {x, y} = -\map \Im {\map {f'} z}$

From Composite of Continuous Mappings is Continuous and Real and Imaginary Part Projections are Continuous, it follows that $\map \Re f$ and $\map \Im f$ are continuous.

Thus, all the partial derivatives of $u$ and $v$ are continuous.

Suppose that $C$ is a simple closed staircase contour.

From Contour Integral as Line Integrals, there exists a piecewiese continuously differentiable Jordan curve $\gamma: \closedint a b \to \R^2$ such that:


 * $\ds \int_C \map f z \rd z = \ds \int_\gamma \tuple {u, -v} \cdot \rd \mathbf l + i \int_\gamma \tuple {v, u} \cdot \rd \mathbf l$

Corollary
Let $f: U \to \C$ be a holomorphic function, where $U \subseteq \C$ is an open set.

Let $C$ be a simple closed contour in $U$.

Let $\Int C \subseteq U$, where $\Int C$ denotes the interior of $C$.

Then:


 * $\ds \oint_C \map f z \rd z = 0$