Zero Simple Staircase Integral Condition for Primitive/Lemma

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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:

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

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 $\map {\gamma_k} {b_k}$.


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 $k \in \O$, 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 $k \in \set {1, \ldots, n}$ such that the intersection of the images of $C_k$ and $C_{k + 1}$ contain more elements than their common end point $\map {\gamma_k} {b_k}$, otherwise we can put $C' = C$.

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

$\Img {C_k} \cap \Img {C_{k + 1} } = \gamma_k \sqbrk {\closedint {a'} {b_k} } \cup \gamma_{k + 1} \sqbrk {\closedint {a_{k + 1} } {b'} }$

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 $\map {\gamma_k} {a_k}$ to $\map {\gamma_k} {a'}$, and then to $\map {\gamma_{k + 1} } {b_{k + 1} }$.

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 $\gamma_k \restriction_{\closedint {a'} {b_k} }$, and $\tilde C_{k + 1}$ as the contour with the parameterization $\gamma_{k + 1} \restriction_{\closedint {a_{k + 1} } {b'} }$.

Here, $\gamma_k \restriction_{\closedint {a'} {b_k} }$ denotes the restriction of $\gamma_k$ to $\closedint {a'} {b_k}$.

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

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

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

where $C'$ has the desired property.

$\blacksquare$