**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?
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