User:Anghel/Sandbox

Theorem
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.

Suppose that $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = 0$ for all simple closed staircase contours $C$ in $D$.

Then $f$ has a primitive $F: D \to \C$.

Proof
Let $C$ be a closed staircase contour in $D$, not necessarily simple.

If we show that $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = 0$, then the result follows from Zero Staircase Integral Condition for Primitive.

The staircase contour $C$ is a concatenation of $C_1, \ldots, C_n$, where the image of each $C_k$ is a line segment parallel with either the real axis or the imaginary axis.

Denote the parameterization of $C$ as $\gamma: \left[{a\,.\,.\,b}\right] \to \C$, where $\left[{a\,.\,.\,b}\right]$ is a closed real interval.

The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of.

Basis for the Induction
For $n = 1$, $C$ can only be a closed staircase contour if $\gamma$ is constant, so:

Induction Hypothesis
For $n \in \N$, if $C$ is a closed staircase contour that is a concatenation of $n+1$ directed smooth curves, then:


 * $\displaystyle \oint_C f \left({z}\right) \ \mathrm dz = 0$

Induction Step
Suppose that $C$ is a closed staircase contour that is a concatenation of $n$ directed smooth curves.