Zero Staircase Integral Condition for Primitive

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 closed staircase contours $C$ in $D$.

Then $f$ has a primitive.

Proof
Choose any $z_0 \in D$, and define $F: D \to \C$ by:


 * $\displaystyle F \left({w}\right) = \int_{C_w} f \left({z}\right) \ \mathrm dz$

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

From Connected Domain is Connected by Staircase Contours, it follows that such a contour exists in $D$.

If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \left({- C_w}\right)$ 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 $B_r \left({w}\right) \subseteq D$, and for all $z \in B_r \left({w}\right)$:


 * $\left\vert{f \left({z}\right) - f \left({w}\right) }\right\vert < \dfrac \epsilon 2$

Let $h = x+iy \in \C \setminus \left\{ {0}\right\}$ with $x, y \in \R$ such that $\left\vert{h}\right\vert < r$.

Let $\mathcal L$ 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 $w + x, w + h \in B_r \left({w}\right)$, it follows from Open Ball is Convex Set that $\mathcal L$ is a contour in $B_r \left({w}\right)$.

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

From Derivative of Complex Polynomial, it follows that $\dfrac{ \mathrm d}{\mathrm dz} f \left({w}\right) z = f \left({w}\right)$, so:

We can now show that $F' \left({w}\right) = f \left({w}\right)$, as:

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