\ifdefined\inmaster\else\def\subonly{\jobname}\input{complex}\fi

\chapter{Complex integration of differential forms}

Let us now consider integration of complex-valued differential forms.
The most interesting case is that of analytic 1-forms.

\section{Line integrals of complex 1-forms}
\label{sec:complex-line-integrals}

The definition of the integral of a 1-form over an oriented curve works just as well when the 1-form is complex-valued.
That is, given a 1-form $f(x,y)\,\dx + g(x,y)\,\dy$ and a curve $\curve$ with parameter $t\in [a,b]$, we can substitute $x = x(t)$, $y=y(t)$, $\dx = x'(t)\,\dt$, and $\dy = y'(t)\,\dt$ to get
\[ \lint{\curve} \Big(f(x,y)\,\dx + g(x,y)\,\dy\Big) = \itint t a b {\Big( f\big(x(t),y(t)\big)\, x'(t) + g\big(x(t),y(t)\big)\,y'(t)\Big)}. \]
There is nothing magical and complexy about this; we can just regard it as two separate integrals of real 1-forms, one for the real part and one for the imaginary part.
On the other hand, recall that we can also write $f(x,y)\,\dx + g(x,y)\,\dy$ as
\[\half (f(x,y) -i\, g(x,y))\,\dz + \half (f(x,y) +i\, g(x,y))\,\dzbar \]
We can thus also try to evaluate the integral by substituting $x = x(t)$, $y=y(t)$, and
\begin{align*}
  \dz &= z'(t) \,\dt = (x'(t) + i\,y'(t))\,\dt\\
  \dzbar &= \zbar'(t) \,\dt = (x'(t) - i\,y'(t))\,\dt.
\end{align*}
This yields the formula
\begin{align*}
  \lint{\curve} &\Big(f(x,y)\,\dx + g(x,y)\,\dy\Big)\\
  &= \lint{\curve} \half (f(x,y) -i\, g(x,y))\,\dz + \half (f(x,y) +i\, g(x,y))\,\dzbar\\
  &= \itint t a b {\half \Big((f(x,y) -i\, g(x,y)) (x'(t) + i\,y'(t)) + (f(x,y) +i\, g(x,y))\,(x'(t) - i\,y'(t))\Big)}
\end{align*}
Multiplying this out, we of course get the same formula.
However, note that it now involves complex multiplication.

In particular, suppose we have a 1-form written as $f(z) \,\dz$, with no $\dzbar$ component.
Then this formula becomes
\[ \lint{\curve} f(z) \,\dz = \itint t a b {f(z)\, z'(t)}\]
where the right-hand side involves multiplying the complex numbers $f(z)$ and $z'(t)$.
This is the definition of a complex line integral that you may find in your textbook (though, as in one-variable calculus, it is often described as an integral of the function $f$ rather than the 1-form $f(z)\,\dz$).
The point being made here is that it really is a special case of the line integral of a (complex-valued) differential form: one of the special sort $f(z)\,\dz$.

\section{The fundamental theorem of calculus}
\label{sec:complex-ftc}

Recall that the two-dimensional fundamental theorem of calculus for line integrals says that given a function $f(x,y)$ with continuous partial derivatives, we have
\[ \lint{\curve} \df = f(\curvex(b),\curvey(b)) - f(\curvex(a),\curvey(a)). \]
In the complex plane, we have $\df = \pder f z \,\dz + \pder f \zbar \,\dzbar$, so that this becomes
\[ \lint{\curve} \pder f z \,\dz + \pder f \zbar \,\dzbar  = f(\curvez(b)) - f(\curvez(a)). \]
Now suppose that $f$ is furthermore complex-differentiable.
We have seen that this means that $\df$ can be written as $f'(z) \,\dz$; thus in this case we have
\[ \lint{\curve} f'(z)\,\dz  = f(\curvez(b)) - f(\curvez(a)). \]
This looks exactly like the fundamental theorem of calculus for complex line integrals that is given in your textbook.
We can't \emph{quite} prove that theorem in this way, because in order to use the fundamental theorem for ordinary line integrals we need the partial derivatives of $f$ to be continuous, whereas for complex line integrals we don't need to assume that.
(It is true that any complex-differentiable function automatically has continuous partial derivatives --- and indeed is infinitely differentiable --- but we haven't yet proven that.)

\section{Integrals of analytic 1-forms}
\label{sec:integrals-of-analytic-1forms}

Now suppose instead that we are given an \emph{analytic 1-form}, i.e.\ a complex-valued 1-form $f(z) \,\dz$ such that $f$ is complex-differentiable on some region $\region$.
Then as we have seen, $\ed(f(z)\,\dz) = f'(z)\,\dz \wedge\dz = 0$ on $\region$.
Therefore, as long as $f$ has continuous partial derivatives, we can apply Green's theorem to conclude that
\[ \lint{\bdry\region} f(z)\,\dz = \sint{\region} \ed(f(z)\,\dz) = 0. \]
This is essentially \emph{Cauchy's theorem} or the \emph{closed curve theorem}: if $f$ is complex-differentiable throughout a region $\region$, then its complex line integral around the boundary of that region is zero.

Now we can't quite prove Cauchy's theorem this way, because Green's theorem requires $f$ to have continuous partial derivatives, whereas Cauchy's theorem does not.
(It is true that any complex-differentiable function automatically has continuous partial derivatives --- and indeed is infinitely differentiable --- but Cauchy's theorem is used in the proof of that fact, so we can't assume it at this point.)
However, it shows that a slightly weaker form of Cauchy's theorem is really just a corollary of Green's theorem.

The point is that complex-differentiability of $f$ is a \emph{differential equation}, which in particular implies that the vector field corresponding to the 1-form $f(z)\,\dz$ is \emph{divergence-free}.
Thus, Green's theorem (which is the two-dimensional form of the divergence theorem) implies that this vector field is ``conservative''.