Contour Integration I: Cauchy and Morera's Integral Theorems

We're interested to find out if there exists an integral form of the Cauchy-Riemann equations. On one hand, this sounds absurd -- this is asking if there's an "integral form" of complex differentiability. On the other, the Cauchy-Riemann equations are just partial differential equations.

The standard relationship between differential and integral formulations of things is Stoke's theorem -- the theorem that tells you that adding things on a lot of tiny curves gives you a thing on a big curve. So let's see what a complex integral on a tiny square looks like.

Observe that the integral on AB is (using the midpoint as the partition tag) is $\varepsilon$ times the midpoint of $f(A)f(B)$, while the integral on CD is $-\varepsilon$ times the midpoint of $f(C)f(D)$. The sum of these is $\varepsilon$ times the line connecting these midpoints (the red arrow in the diagram below). Similarly, the sum of the other two parts of the integral is $i$ times the blue arrow in the diagram below.

Because a holomorphic function preserves squares and their orientation, these cancel out, and the integral gives zero. One can then use Green's theorem to show that the integral of a holomorphic (on $D$) function $f$ on the closed curve $\partial D$ is zero. (If you wanted to be completely formal, the equivalent would be to just apply Green's theorem and note that the local integral is zero, which is what the geometry above shows).

$$\oint_\gamma f(z)dz=0$$
Alternatively, one may write, for a simply-connected region $D$: if $f$ is holomorphic on $D$, the integral of $f$ on all closed curves contained in $D$ is zero. This is known as Cauchy's Integral theorem (or the Cauchy-Goursat theorem).

One also immediately sees that the converse holds -- if the function weren't holomorphic, the blue arrow would not be a right-angle rotation of the red one, and you could construct closed curves on which this cancellation doesn't occur. This converse -- if the integral of a continuous function $f$ on all closed curves contained in an open region $D$ are zero, then $f$ is holomorphic -- is called Morera's Integral theorem.

(The "openness" requirement in Morera's theorem is important because we want to ensure the integral is an actual global property -- that it's across some amount of "space".)

Think about how surprising this is for a moment.

• Cauchy's theorem tells us that for a simply-connected region, existence of a derivative implies existence of a primitive. 
• Morera's theorem tells us that for a continuous function, existence of a primitive implies existence of a derivative.

Morera's theorem does not show that a holomorphic function is infinitely-differentiable. Do you see why?

The Cauchy Riemann Equations: what do they really mean?

Question: Geometrical Interpretation of Cauchy Riemann equations?

One might think that being differentiable on $\mathbb{R}^2$ is sufficient for differentiability on $\mathbb{C}$. But the Jacobian of an arbitrary such function doesn't have a natural complex number representation.

$$
\left[ {\begin{array}{*{20}{c}}
{\partial u/\partial x} & {\partial u/\partial y} \\
{\partial v/\partial x} & {\partial v/\partial y}
\end{array}} \right]
$$
Another way of putting this is that no complex-valued derivative (see below for an example, known as the Wirtinger derivative) you can define for an arbitrary function fully captures the local behaviour of the function that is represented by the Jacobian.

$$
\frac{df}{dz} = \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right) + i\left(\frac{\partial v}{\partial x}-\frac{\partial v}{\partial y}\right)
$$
The idea is that we should be able to define a complex-valued derivative "purely" for the value $z$, without considering directions, i.e. we want to consider $\mathbb{C}$ one-dimensional in some sense (the sense being "as a vector space"). More precisely, the derivative in some direction in $\mathbb{C}$ should determine the derivative in all other directions in a natural manner -- whereas on $\mathbb{R}^2$, the derivatives in *two* directions (i.e. the gradient) determines the directional derivatives in all directions.

If you think about it, this is quite a reasonable idea -- it's analogous to how not every linear transformation on $\mathbb{R}^2$ is a linear transformation on $\mathbb{C}$ -- only spiral transformations are.

$$
\left[ {\begin{array}{*{20}{c}}
{a} & {-b} \\
{b} & {a}
\end{array}} \right]
$$
How would we generalise differentiability to an arbitrary manifold? Here's an idea: a function is differentiable if it is locally a linear transformation. So on $\mathbb{R}^2$, any Jacobian matrix is a linear transformation. But on $\mathbb{C}$, only Jacobians of the above form are linear transformations -- i.e. the only linear transformation on $\mathbb{C}$ is multiplication by a complex number, i.e. a spiral/amplitwist. So a complex differentiable function is one that is locally an amplitwist (geometrically), which can be stated in terms of the components of the Jacobian as:

$$
\begin{align}
\frac{\partial u}{\partial x} & = \frac{\partial v}{\partial y} \\
\frac{\partial u}{\partial y} & = - \frac{\partial v}{\partial x} \\
\end{align}
$$
This is precisely why you shouldn't (and can't) view complex differentiability as some basic first-degree smoothness -- there is a much richer structure to these functions, and it's better to think of them via the transformations they have on grids.

One might observe that the Cauchy-Riemann equations can also be restated as: $\frac{\partial f(z)}{\partial\bar{z}}=0$, where the derivative is the Wirtinger derivative defined above. This seems completely bizarre to me, but apparently it makes sense with some differential geometry, and I'm given the keyword "complexifying the tangent bundle".