Zero Staircase Integral Condition for Primitive

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

Let $z_0 \in D$.

Suppose that $\ds \oint_C \map f z \rd z = 0$ for all closed staircase contours $C$ in $D$.

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


 * $\ds \map F w = \int_{C_w} \map f z \rd z$

where $C_w$ is any staircase contour in $D$ with start point $z_0$ and end point $w$.

Proof
From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C_w$ in $D$ with start point $z_0$ and end point $w$.

If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \paren {-C_w}$ is a closed staircase contour.

Then the definition of $F$ is independent of the choice of contour, as:

We now show that $F$ is the primitive of $f$.

Let $\epsilon \in \R_{>0}$.

By definition of continuity, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} w \subseteq D$, and for all $z \in \map {B_r} w$:


 * $\size {\map f z - \map f w} < \dfrac \epsilon 2$

Let $h = x+iy \in \C \setminus \set 0$ with $x, y \in \R$ such that $\size h < r$.

Let $\LL$ be the staircase contour that goes in a horizontal line from $w$ to $w + x$, and continues in a vertical line from $w + x$ to $w + h$.

As $w + x, w + h \in \map {B_r} w$, it follows from Open Ball is Convex Set that $\LL$ is a contour in $\map {B_r} w$.

Then $C_w \cup \LL$ is a staircase contour from $z_0$ to $w + h$, so:

From Derivative of Complex Polynomial, it follows that $\dfrac \rd {\rd z} \map f w z = \map f w$, so:

We can now show that $\map {F'} w = \map f w$, as:

When $h$ tends to $0$, we have $\map {F'} w = \map f w$ by definition of differentiability.