# Zero Simple Staircase Integral Condition for Primitive

## Theorem

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

Let $\ds \oint_C \map f z \rd z = 0$ 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 $\ds \oint_C \map f z \rd z = 0$, 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 $\gamma: \closedint a b \to \C$, where $\closedint a b$ is a closed real interval.

Denote the parameterization of $C_k$ as $\gamma_k: \closedint {a_k} {b_k} \to \C$.

### Lemma

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

Let $C$ be a closed staircase contour in $D$.

Then there exists a contour $C'$ such that:

$\ds \oint_C \map f z \rd z = \oint_{C'} \map f z \rd z$

This contour $C'$ has the property that:

for all $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$.

### Splitting up the Contour

The lemma shows that given a staircase contour $C$, we can assume that for $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$.

This means that in order to intersect itself, $C$ must be a concatenation of at least $4$ directed smooth curves.

Now, we prove the main requirement for Zero Staircase Integral Condition for Primitive, that $\ds \oint_C \map f z \rd z = 0$.

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:

 $\ds \oint_C \map f z \rd z$ $=$ $\ds \int_a^b \map f {\map \gamma t} \map {\gamma'} t \rd t$ Definition of Complex Contour Integral $\ds$ $=$ $\ds 0$ Derivative of Complex Polynomial: $\gamma$ is constant

For $n = 4$, $C$ can only be a closed staircase contour if $C$ is a simple contour.

Then by hypothesis:

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

This is the basis for the induction.

### Induction Hypothesis

For $N \in \N$, if $C$ is a closed staircase contour that is a concatenation of $n$ directed smooth curves with $n \le N$, then:

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

This is the induction hypothesis.

### Induction Step

This is the induction step:

Suppose that $C$ is a closed staircase contour that is a concatenation of $n + 1$ directed smooth curves.

If $C$ is a simple contour, the induction hypothesis is true by the original assumption of this theorem.

Otherwise, define $t_0 = a$, and $t_3 = b$.

Define $t_1 \in \closedint a b$ as the infimum of all $t \in \closedint a b$ for which $\gamma$ intersects itself.

Then define $t_2 \in \hointl {t_1} b$ as the infimum of all $t \in \hointl {t_1} b$ for which $\map \gamma t = \map \gamma {t_1}$.

For $k \in \set {1, \ldots, 3}$, define $\tilde C_k$ as the staircase contour with parameterization $\gamma \restriction {\closedint {t_{k - 1} } {t_k} }$.

Then $\tilde C_2$ is a closed staircase contour that is a concatenation of at least $4$ directed smooth curves.

Then both $\tilde C_1 \cup \tilde C_3$ and $\tilde C_2$ are a concatenation of fewer than $n + 1$ directed smooth curves, so:

 $\ds \oint_C \map f z \rd z$ $=$ $\ds \oint_{\tilde C_1 \cup \tilde C_2 \cup \tilde C_3} \map f z \rd z$ $\ds$ $=$ $\ds \oint_{\tilde C_1 \cup \tilde C_3} \map f z \rd z + \oint_{\tilde C_2} \map f z \rd z$ Contour Integral of Concatenation of Contours $\ds$ $=$ $\ds 0$ Induction Hypothesis

$\blacksquare$