Zero Staircase Integral Condition for Primitive

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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:

\(\ds \int_{C_w} \map f z \rd z\) \(=\) \(\ds \int_{C_w} \map f z \rd z + \int_{C_w' \cup \paren {-C_w} } \map f z \rd z\) by assumption
\(\ds \) \(=\) \(\ds \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z + \int_{-C_w} \map f z \rd z\) Contour Integral of Concatenation of Contours
\(\ds \) \(=\) \(\ds \int_{C_w} \map f z \rd z + \int_{C_w'} \map f z \rd z - \int_{C_w} \map f z \rd z\) Contour Integral along Reversed Contour
\(\ds \) \(=\) \(\ds \int_{C_w'} \map f z \rd z\)


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:

\(\ds \map F {w + h} - \map F w\) \(=\) \(\ds \int_{C_w \cup \LL} \map f z \rd z - \int_{C_w} \map f z \rd z\)
\(\ds \) \(=\) \(\ds \int_\LL \map f z \rd z\) Contour Integral of Concatenation of Contours

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

\(\ds \int_\LL \map f w \rd z\) \(=\) \(\ds \map f w \paren {w + h} - \map f w w\) Fundamental Theorem of Calculus for Contour Integrals
\(\ds \) \(=\) \(\ds h \map f w\)


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

\(\ds \size {\dfrac {\map F {w + h} - \map F w} h - \map f w}\) \(=\) \(\ds \size {\dfrac 1 h \int_\LL \map f z \rd z - \dfrac 1 h h \map f w}\) a priori
\(\ds \) \(=\) \(\ds \size {\dfrac 1 h} \size {\int_\LL \paren {\map f z - \map f w} \rd z}\) Linear Combination of Contour Integrals
\(\ds \) \(<\) \(\ds \size {\dfrac 1 h} \dfrac \epsilon 2 \map L \LL\) Estimation Lemma for Contour Integrals, as $z \in \map {B_r} w$
\(\ds \) \(=\) \(\ds \dfrac {\size x + \size y} {\size h} \dfrac \epsilon 2\) the lengths of the line segments are $\size x$ and $\size y$
\(\ds \) \(\le\) \(\ds \epsilon\) Modulus Larger than Real Part and Imaginary Part

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

$\blacksquare$


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