Closed/Exact Forms and Complex Functions If $f : D \to \Bbb{C}$ is a function, where $D$ is simply connected, what does it mean to say that $f(z)dz$ is either closed or exact? I know what it means for $Pdx + Qdy$ to be exact or closed when everything is real, but what happens when you move over to $\Bbb{C}$?
 A: There is no issue with having $i$ in the expressions. On any smooth manifold $M$ we can talk about complex-valued smooth functions $M \to \mathbb{C}$ and more generally complex-valued smooth differential $k$-forms $\Omega^k(M, \mathbb{C})$; these are exactly like real differential $k$-forms, just allowing complex coefficients, and in fact as vector spaces we have
$$\Omega^k(M, \mathbb{C}) \cong \Omega^k(M, \mathbb{R}) \oplus i \Omega^k(M, \mathbb{R})$$
meaning that a complex $k$-form splits up canonically into a real part and an imaginary part just like a complex number or complex function does. Furthermore the exterior derivative $d : \Omega^k(M, \mathbb{C}) \to \Omega^{k+1}(M, \mathbb{C})$ respects this decomposition, so a complex $k$-form is closed resp. exact iff its real and imaginary parts are closed resp. exact.
So, when $M = \mathbb{C}$ the form $f(z) \, dz$ is a complex valued $1$-form $f(z) \, dz \in \Omega^1(M, \mathbb{C})$, and if we split up $f(x + iy) = u(x, y) + i v(x, y)$ and $\, dz = dx + i \, dy$ then the real and complex parts of $f(z) \, dz$ are
$$f \, dz = \left( u \, dx - v \, dy \right) + i \left( u \, dy + v \, dx \right)$$
so to say that $f \, dz$ is closed is exactly to say that $u \, dx - v \, dy$ and $u \, dy + v \, dx$ are closed. Taking the exterior derivatives of these forms gives the $2$-forms
$$\left( - \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x} \right)\, dx \, dy, \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) \, dx \, dy$$
so the vanishing of these $2$-forms is exactly the Cauchy-Riemann equations. A similar computation shows that $f(z) \, dz$ is exact iff $f$ has an antiderivative.
Of course it's annoying and in some sense unnatural to constantly break into real and complex parts like this. Instead of working with $dx$ and $dy$ as a real basis of the space of real $1$-forms, as Ivo does it's cleaner to work with $dz = dx + i \, dy$ and $d \bar{z} = dx - i \, dy$ as a complex basis of the space of complex $1$-forms, and similarly it's cleaner to work with the dual vector fields $\frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)$ and $\frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right)$. Then you can check that the exterior derivative of a smooth complex function $g \in \Omega^0(M, \mathbb{C})$ can be written
$$dg = \frac{\partial g}{\partial x} \, dx + \frac{\partial g}{\partial y} \, dy =  \frac{\partial g}{\partial z} \, dz + \frac{\partial g}{\partial \bar{z}} d \bar{z} \in \Omega^1(M, \mathbb{C})$$
(and in particular $dz$ really is the exterior derivative of $z = x + iy \in \Omega^0(M, \mathbb{C})$ and $d \bar{z}$ really is the exterior derivative of $\bar{z} = x - iy \in \Omega^0(M, \mathbb{C})$ so all our notation is consistent), and now the computation is faster and more conceptual: we have
$$d \left( f \, dz \right) = \frac{\partial f}{\partial \bar{z}} \, d \bar{z} \, dz$$
which vanishes iff $\frac{\partial f}{\partial \bar{z}} = 0$; this is a compact restatement of the Cauchy-Riemann equations. Similarly we have $dg = f \, dz$ iff $\frac{\partial g}{\partial \bar{z}} = 0$ and $\frac{\partial g}{\partial z} = f$ which says exactly that $f$ has an antiderivative. (If $g$ is holomorphic then $\frac{\partial g}{\partial z}$ as we've defined it really does compute the holomorphic derivative in the usual sense, so again all of our notation is consistent.) This is worth stating explicitly:

Proposition: The $1$-form $f(z) \, dz \in \Omega^1(M, \mathbb{C})$ is closed iff $f$ is holomorphic and exact iff $f$ has an antiderivative.

The point of this is that it explains, from the point of view of differential geometry, the behavior of contour integrals:

*

*Because $f(z) \, dz$ is closed we get that the contour integral $\int_{\gamma} f(z) \, dz$ only depends on the homology class of $\gamma$, and in particular in a simply connected region (whose homology vanishes) we get that $\int_{\gamma} f(z) \, dz$ vanishes and in fact that $f(z) \, dz$ is exact, hence $f$ has an antiderivative.

*If $M$ is $\mathbb{C}$ minus a finite set of points $z_1, \dots z_k$ then the de Rham cohomology $H^1(M, \mathbb{C})$ has basis given by the forms $\frac{dz}{z - z_i}$, which are closed but not exact. If we write $f(z) \, dz$ as a linear combination of such forms plus an exact form (which amounts to subtracting off the residues of $f$ at each $z_i$), we get the residue theorem, and in particular Cauchy's integral formula.

A: Same thing. For a $1$-form $\alpha = f\,{\rm d}z + g\,{\rm d}\bar{z}$ we have $${\rm d}\alpha = (f_z\,{\rm d}z+f_{\overline{z}}\,{\rm d}\bar{z})\wedge {\rm d}z + (g_z\,{\rm d}z+g_{\overline{z}}\,{\rm d}\bar{z})\wedge {\rm d}\bar{z} = (g_z-f_{\bar{z}})\,{\rm d}z\wedge {\rm d}\bar{z}.$$You say that $\alpha$ is closed if ${\rm d}\alpha = 0$ and exact if $\alpha = {\rm d}h$ for some function $h$. In particular, we have that ${\rm d}(f\,{\rm d}z) = f_{\bar{z}}\,{\rm d}\bar{z}\wedge {\rm d}z$, so that $f\,{\rm d}z$ is closed if and only if $f_{\bar{z}} = 0$, which is to say that $f$ is holomorphic.
Here, of course, ${\rm d}z = {\rm d}x+{\rm i}\,{\rm d}y$, ${\rm d}\bar{z} = {\rm d}x - {\rm i}\,{\rm d}y$, as well as $2\partial_z = \partial_x - {\rm i}\,\partial_y$ and $2\partial_{\bar{z}} = \partial_x + {\rm i}\,\partial_y$.
