# Zero Simple Staircase Integral Condition for Primitive/Lemma

## 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:

$\displaystyle \oint_C f \left({z}\right) \rd z = \displaystyle \oint_{C'} f \left({z}\right) \rd z$

This contour $C'$ has the property that for all $k \in \left\{ {1, \ldots, n-1} \right\}$, the intersection of the images of $C_k$ and $C_{k+1}$ is equal to their common end point $\gamma_k \left({b_k}\right)$.

## Proof

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

As induction basis, if $C$ consists of just one directed smooth curve, we can simply put $C' = C$.

Then, we have to check a statement for all $k \in \varnothing$, so it is trivially true.

### Induction Hypothesis

As induction hypothesis, suppose that if $C$ is a concatenation of $n$ directed smooth curves, we can find a contour $C'$ with the properties above.

### Induction Step

In the induction step, let $C$ be a concatenation of $n + 1$ directed smooth curves.

Suppose that there exists $k \in \left\{ {1, \ldots, n}\right\}$ such that the intersection of the images of $C_k$ and $C_{k + 1}$ contain more elements than their common end point $\gamma_k \left({b_k}\right)$, otherwise we can put $C' = C$.

Then, one of the two line segments must contain the other line segment, so:

$\operatorname{Im} \left({C_k}\right) \cap \operatorname{Im} \left({C_{k + 1} }\right) = \gamma_k \left({\left[{a' \,.\,.\, b_k}\right]}\right) \cup \gamma_{k + 1} \left({\left[{a_{k + 1} \,.\,.\, b'}\right]}\right)$

where either $a' = a_k$, or $b' = b_{k + 1}$.

Define $C_k'$ as the contour that is parameterized as a line segment that goes from $\gamma_k \left({a_k}\right)$ to $\gamma_k \left({a'}\right)$, and then to $\gamma_{k + 1} \left({b_{k + 1} }\right)$.

Then, the image of $C_k'$ is equal to the part of the line segments that does not overlap.

Define $\tilde C_k$ as the contour with the parameterization $\gamma_k \restriction_{\left[{a' \,.\,.\, b_k}\right]}$, and $\tilde C_{k + 1}$ as the contour with the parameterization $\gamma_{k + 1} \restriction_{\left[{a_{k+1} \,.\,.\, b'}\right]}$.

Here, $\gamma_k \restriction_{ \left[{a'\,.\,.\,b_k }\right] }$ denotes the restriction of $\gamma_k$ to $\left[{a' \,.\,.\, b_k}\right]$.

Now $\tilde C_k$ is the reversed contour of $\tilde C_{k + 1}$, so:

 $\displaystyle \oint_C f \left({z}\right) \rd z$ $=$ $\displaystyle \oint_{C_1 \cup \ldots \cup C_{k - 1} } f \left({z}\right) \rd z + \oint_{C_k} f \left({z}\right) \rd z + \oint_{C_{k + 1} } f \left({z}\right) \rd z + \oint_{C_{k+2} \cup \ldots \cup C_{n + 1} } f \left({z}\right) \rd z$ Contour Integral of Concatenation of Contours $\displaystyle$ $=$ $\displaystyle \oint_{C_1 \cup \ldots \cup C_{k - 1} } f \left({z}\right) \rd z + \oint_{C_k'} f \left({z}\right) \rd z + \oint_{\tilde C_k } f \left({z}\right) \rd z + \oint_{\tilde C_{k + 1} } f \left({z}\right) \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n+1} } f \left({z}\right) \rd z$ Contour Integral of Concatenation of Contours $\displaystyle$ $=$ $\displaystyle \oint_{C_1 \cup \ldots \cup C_{k - 1} } f \left({z}\right) \rd z + \oint_{C_k'} f \left({z}\right) \rd z + \oint_{\tilde C_k } f \left({z}\right) \rd z - \oint_{\tilde C_k } f \left({z}\right) \rd z + \oint_{C_{k + 2} \cup \ldots \cup C_{n + 1} } f \left({z}\right) \rd z$ Contour Integral along Reversed Contour $\displaystyle$ $=$ $\displaystyle \oint_{C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1} } f \left({z}\right) \rd z$ Contour Integral of Concatenation of Contours

which shows that the contour integral of $f$ along $C$ is equal to the contour integral of $f$ along $C'' = C_1 \cup \ldots \cup C_{k - 1} \cup \tilde C_k \cup C_{k + 2} \ldots \cup C_{n + 1}$.

This new contour $C''$ is a staircase contour which is a concatenation of $n$ directed smooth curves, so by the induction hypothesis, there exists a contour $C'$ such that:

$\displaystyle \oint_{C'} f \left({z}\right) \rd z = \oint_{C''} f \left({z}\right) \rd z$

where $C'$ has the desired property.

$\blacksquare$