A friend of mine is taking a course on analytic number theory in the spring and needs to learn basic complex analysis in a couple of weeks.  I decided to do a post (self-contained, except for Stokes’ formula) on deducing the Cauchy theorems and their applications from Stokes’ theorem now instead of later–when I’ll talk about several complex variables.  It might be objected that Stokes’ theorem is just Green’s theorem for $n=2$, commonly used in undergraduate treatments, but my goal was to take an expository challenge: write something rigorous on complex variables in as short a space as possible without sacrificing readability.  So Stokes’ theorem for manifolds is preferable to Green’s theorem as stated in a vague way about “insides of a curve” (before, say, the Jordan curve theorem is proved) and the traditional proof of Green’s theorem via rectangular decompositions.

So, let’s consider an open set ${O \subset \mathbb{C}}$, and a ${C^2}$ function ${f: O \rightarrow \mathbb{C}}$. We can consider the differential

$\displaystyle df := f_x dx + f_y dy$

which is a complex-valued 1-form on ${O}$. It is also convenient to write the differential using the ${z}$ and ${\bar{z}}$-derivatives I talked about earlier, i.e.

$\displaystyle f_z := \frac{1}{2}\left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) f, \quad f_{\bar{z}} := \frac{1}{2}\left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) f.$

The reason these are important is that if ${w_0 \in O}$, we can choose ${A,B \in \mathbb{C}}$ with

$\displaystyle f(w_0+h) = f(w_0) + Ah + B \bar{h} + o(|h|), \ h \in \mathbb{C}$

by differentiability, and it is easy to check that ${A=f_z(w_0), B=f_{\bar{z}}(w_0)}$. So we can define a function ${f}$ to be holomorphic if it satisfies the differential equation

$\displaystyle f_{\bar{z}} = 0,$

which is equivalent to being able to write

$\displaystyle f(w_0 + h) =f(w_0) + Ah + o(|h|)$

for each ${w_0 \in O}$ and a suitable ${A \in \mathbb{C}}$. In particular, it is equivalent to a difference quotient definition. The derivative ${f_z}$ of a holomorphic function thus satisfies all the usual algebraic rules, under which holomorphic functions are closed.

Recall now that ${f_{z\bar{z}}}$ is a multiple of the Laplacian. In particular, it follows that a holomorphic function is harmonic, and thus infinitely differentiable.

The other reason we care about the ${\bar{z}}$-derivative is that we can write, for any ${f}$,

$\displaystyle df = f_z dz + f_{\bar{z}} d \bar{z}$

where naturally enough ${dz = dx + idy, d\bar{z} = dx - idy}$. It is easy to check that ${dz, d \bar{z}}$ and ${\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar{z}}}$ are dual to each other as complex-valued elements of the cotangent and tangent spaces, respectively. So, as a result, we can compute on (complex-valued) 1-forms:

$\displaystyle \boxed{d(f(z) dz ) = df \wedge dz = (f_z dz + f_{\bar{z}} d \bar{z}) \wedge dz = - f_{\bar{z}} dz \wedge d \bar{z}.}$

This will lead to the Cauchy theorem. Note that ${dz \wedge d \bar{z} = 2i dx \wedge dy}$ is a multiple of the usual area element.

Complex integration

Cauchy’s theorem is now a simple corollary of Stokes’ theorem. Given a 1-dimensional manifold ${\gamma \subset O}$, we can consider the integral of a complex-valued 1-form

$\displaystyle \int_{\gamma} g(z) dz,$

which, given a parametrization, can of course be computed using the change-of-variables formula, and by taking real and imaginary parts. More interestingly, when ${\gamma}$ is the boundary of a compact 2-dimensional submanifold-with-boundary ${X}$ we can compute it using Stokes theorem:

$\displaystyle \int_{\gamma} g(z) dz = \iint_X d( g(z) dz) = - \iint_X g_{\bar{z}} dz \wedge d \bar{z}.$

In particular, if ${g}$ is holomorphic on ${O}$, we find:

Theorem 1 (Cauchy)

For ${\gamma}$ the boundary of ${X \subset O}$, we have$\displaystyle \int_{\gamma} g(z) dz = 0.$

Using a similar approach, we can establish the following:

Theorem 2 (Cauchy formula)

If ${w}$ is in the interior of ${X}$ and ${g}$ is holomorphic in ${O}$, then$\displaystyle \frac{1}{2 \pi i} \int_{\gamma} \frac{g(z) }{z-w} dz = g(w).$

In the proof, we will actually show something more general. So let us not assume ${g}$ holomorphic, only differentiable.

To prove this, let’s consider the manifold-with-boundary ${Y:=X - N_r(w)}$ for ${r>0}$ very small, as in the figure.

The boundary ${\partial Y}$ is the disjoint union of ${\partial X}$ with the usual orientation and the circle ${C_r(w)}$ with the opposite orientation. Also, ${\frac{g(z)}{z-w}}$ is holomorphic in a neighborhood of ${Y}$. This is the key fact, and we will use this together with a limiting argument on ${r}$. So

$\displaystyle \int_{\partial Y} = \int_\gamma \frac{g(z)}{z-w} dz - \int_{C_r(w)} \frac{g(z)}{z-w} dz = \iint_{X - N_r(w)} d\left( \frac{g(z)}{z-w} dz\right).$

First, we can compute the term ${\int_{C_r(w)} \frac{g(z)}{z-w} dz}$ by using the parametrization ${t \rightarrow w + re^{it}, 0 \leq t < 2 \pi}$ of the circle. Then ${dz = d(re^{it}) = ir e^{it} dt}$, so

$\displaystyle \int_{C_r(w)} \frac{g(z)}{z-w} dz = \int_{0}^{2 \pi} \frac{g(w+re^{it})}{re^{it}} i re^{it}dt = i \int_{0}^{2\pi} g(w+re^{it}) dt$

and as ${r \rightarrow 0}$, this approaches ${2 \pi i g(w)}$. The integrand in the second term becomes ${-dz \wedge d \bar{z}}$ times

$\displaystyle \frac{\partial}{\partial \bar{z}} \left( \frac{g(z)}{z-w} \right) = -g_{\bar{z}} \frac{1}{z-w} + g(z) \frac{\partial}{\partial \bar{z}} \left( \frac{1}{z-w}\right) = g_{\bar{z}} \frac{1}{z-w}$

because the ${\bar{z}}$-derivative clearly satisfies the product rule and ${\frac{1}{z-w}}$ is holomorphic (by the quotient rule).

So, when we let ${r \rightarrow 0}$, using the integrability of ${\frac{1}{z-w}}$, we find:

$\displaystyle \boxed{ \int_{\gamma} \frac{g(z)}{z-w} dz = 2 \pi i g(w) - \iint_X \frac{g_{\bar{z}}}{z-w} dz \wedge d \bar{z}.}$

This is a more general analog of the Cauchy formula, and it clearly implies the usual one when ${g}$ is holomorphic.

Representability as power series

Fix a holomorphic function ${f}$ on ${O}$. In a neighborhood ${N_r(z_0)}$ with ${\overline{N_r(x_0)} \subset O}$; then for any ${z \in N_r(z_0)}$ we have

$\displaystyle f(z) = \frac{1}{2\pi i } \int_{C_r(z_0)} \frac{ f(w)}{w-z} dw$

In particular, we have represented ${f(z)}$ as a weighted average of the functions ${g_{(w)}(z) := \frac{1}{w-z}}$ for ${w}$ ranging over the circle ${C_r(z_0)}$. Each function ${g_{(w)}=\frac{1}{w-z}}$ is represented by a power series in ${N_r(z_0)}$, the geometric series. Thus, the weighted average ${f}$ has the same property. We now prove this rigorously; by a translation assume ${z_0=0}$. So, writing out the geometric series:

$\displaystyle f(z) = \sum_n \frac{1}{2\pi i } \int_{C_r(0)} \left( \frac{z}{w}\right)^n \frac{ f(w)}{w} dw = \sum_n c_n z^n$

where ${c_n}$ is the constant ${\frac{1}{2\pi i } \int_{C_r(0)} \left( \frac{1}{w}\right)^n \frac{ f(w)}{w} dw}$. It is immediate that ${c_n = O(r^{-n})}$, so convergence questions don’t arise when ${|z|, and the interchange of summation and integration was justified. In fact the convergence is uniform on compact subsets (where ${|z|}$ is bounded by something less than ${r}$) by the same argument. We have thus proved that ${f}$ can be represented as a power series converging in the whole disk. Differentiation of this power series can be done term-by-term as one can check directly, though the justification is slightly technical.

Laurent series

The Taylor series just discussed works for functions in a disk. In an annulus, we can’t necessarily do that, but we end up with terms of negative order. As usual, let ${f: O \rightarrow \mathbb{C}}$ be holomorphic. Suppose the closed annulus ${\overline{A_r^R(0)} \subset O}$, where ${A_r^R(0) = \{ z: r<|z|. Then by Cauchy’s formula applied to ${X= \overline{A_r^R(0)}}$, we have for ${z \in A_r^R(0)}$,

$\displaystyle 2 \pi i f(z) = \left( \int_{C_R} - \int_{C_r} \right) \frac{f(w)}{w-z} dw = I_1(z)- I_2(z).$

Now ${I_1}$ is holomorphic in ${N_R(0)}$ so can be represented as a power series as before. ${I_2}$ is holomorphic in the “disk at infinity” ${\{ z: |z|>r \}}$, and we can write for such ${z}$

$\displaystyle I_2(z) = \frac{1}{z} \int_{C_r} \frac{f(w)}{w/z-1} dw = \sum_{n} \frac{1}{z^{n+1}} \int_{C_r} f(w) w^n dw .$

The convergence issues can be handled as before because the integrands are ${O(r^n)}$ and ${|z|>r}$.

In particular, we can represent ${f}$ in the annulus as

$\displaystyle f(z) = \sum_{n \in \mathbb{Z}} c_n z^n ,$

where the ${c_n}$ are uniquely determined constants. The uniqueness is seen because one can choose ${r'}$ between ${r,R}$, and get

$\displaystyle \int_{C_{r'}} f(z) z^{-k} dz = \sum_{n \in \mathbb{Z}} \int_{C_{r'}} c_n z^{n-k} dz = 2 \pi c_k,$

because all but one of the integrals vanish by direct computation. So ${f}$ determines the ${c_k}$.