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A very first theorem that is proved in the first course of Complex Analysis would be the Gousart Theorem. Here it is:

Theorem (Goursat). Let $f:U\rightarrow\mathbb{C}$ be an analytic function. Then the integral $\displaystyle\int_{\partial R}f(z)dz=0$, where $R$ is a rectangle given by {$z=x+iy : a\leq x\leq b$ and $ c\leq y\leq d$}.

A lot of books give a rather complicated proof using quite a lot of estimation. I am just wondering whether the following proof, which looks completely natural for me, makes sense.

Proof. Let $f(z)=u(x,y)+iv(x,y)$.

$\displaystyle\int_{\partial R}f(z)dz=\int_{L_1}f(z)dz+\int_{L_2}f(z)dz+\int_{L_3}f(z)dz+\int_{L_4}f(z)dz$, where $L_1,L_2,L_3,L_4$ are the four sides of the rectangle.

One can show that by explicit calculations that $\displaystyle \int_{L_1}f(z)dz+\int_{L_2}f(z)dz+\int_{L_3}f(z)dz+\int_{L_4}f(z)dz =I_1+iI_2$, where

$I_1=\displaystyle\int_{a}^b u(x,c)-u(x,d)dx-\int_{c}^d v(b,y)-v(a,y)dy$ and $I_2=\displaystyle\int_{a}^b v(x,c)-u(x,d)dx+\int_{c}^d u(b,y)-u(a,y)dy$.

By Fundamental Theorem of Calculus and Fubini's Theorem, we have

$I_1=\displaystyle\int_{a}^b\int_{c}^{d} -\dfrac{\partial u}{\partial y}-\dfrac{\partial v}{\partial x}dydx$ and $I_2=\displaystyle\int_{a}^b\int_{c}^{d} -\dfrac{\partial v}{\partial y}+\dfrac{\partial u}{\partial x}dydx$

Since $f$ is analytic, it satisfies the Cauchy-Riemann Equations: $-\dfrac{\partial u}{\partial y}-\dfrac{\partial v}{\partial x}=-\dfrac{\partial v}{\partial y}+\dfrac{\partial u}{\partial x}=0$

So $I_1=I_2=0$. We are done.

I am just wondering whether this proof is valid. But I have never seen any classics on Complex Analysis adopting this proof.

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Stokes' theorem would give a very quick proof of this, but Sam's caveat still applies. – Dylan Moreland May 23 '12 at 0:28
up vote 8 down vote accepted

Your proof is perfectly fine, IF you assume your function $u$ is $C^1$, that is differentiable with continuous derivative. In fact that's the requirement for Green's theorem (and FTC) to hold.

If you want to get nitpicky on the otherhand, then you have to do those more complicated estimates that you mentioned (proofs on a square, proofs on polygons, etc). The issue is that when we start the groundwork for complex analysis, we first define analyticity as $f$ satisfying the Cauchy Riemann equations, in that $f$ is complex differentiable. We later find out, (in fact using stuff like the more general Cauchy's Theorem!) that analyticity implies the $f$ is complex differentiable of all orders so that will imply $f$ has continuous first derivative and therefore all that vector calculus mumbo jumbo applies.

So when you first encounter the Cauchy Goursat theorem, you would have to "cheat" and assume continuity of the first derivatives to apply the vector calculus machinery. If you want the formal proof, you have to put in some sweat!

You might now be wondering why you need continuity of partials. A typical example is the following: take $H(x)=1$ for $x>0$ and 0 otherwise, in other words the Heaviside step function. Then $H'(x)=0$ everywhere except $x=0$ where it's $\infty$. But $\int_0^1 H'(x)=0$ yet $H(1)-H(0)=1$, so the FTC does not hold. The moral here is that the condition for a complex function to be analytic is extremely strong, and again as you later find out in complex analysis, if $f(x)$ is complex differentiable then it is infinitely differentiable.

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That's wonderful! Thank you. It seems that I have omitted the crucial bit on continuity of partial derivatives. – iloveinna May 23 '12 at 0:37
The reason Goursat's name is attached to this theorem is that he showed that the assumption of continuity of the partial derivatives is not needed. Otherwise it's just Cauchy's theorem. – Robert Israel May 23 '12 at 0:49

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