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Complex form of Green's theorem is $\int _{\partial S}{f(z)\,dz}=i\int \int_S{\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\,dx\,dy}$. The following is just my calculation to show both sides equal.

$$LHS=\int _{\partial S}{f(z)\,dz}=\int_{\partial S}{(u+iv)(dx+i\,dy)}=\int_{\partial S}{(u\,dx-v\,dy)+i(u\,dy+v\,dx)}$$ $$RHS=i\int \int_S{\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\,dx\,dy}=i\int \int_S{\frac{\partial f}{\partial x}\,dx\,dy+\int \int_Si\frac{\partial f}{\partial y}\,dx\,dy}=i\int_S{f\,dy-\int \int_Si\frac{\partial f}{\partial y}\,dy\,dx}=i\int_S{f\,dy-\int_Sif\,dx}=i\int_S{(u+iv)\,dy-\int_Si(u+iv)\,dx}=\int_S{i(u+iv)\,dy+\int_S(u+iv)\,dx}=LHS$$ Actually what I want to ask here is do we need to add negative sign when change the order of $dx\,dy$ in complex integral?

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You are actually integrate in the sense of first kind. You just apply the Fubini's theorem and don't have to change the sign. – lee Dec 22 '12 at 15:16
What is ' integrate in the sense of first kind' ? If I do not change the sign , then I cannot get the result. – hong wai Dec 22 '12 at 15:19

Is that the complex form of Green's theorem?

Let's use a little geometric calculus to verify the result.

$$\begin{align*} \oint_{\partial S} f(r) \, dz &= \oint_{\partial S} f(r) e_x \, d\ell \\ &= \int_S \dot f(r) e_x \dot \nabla (e_{xy} \, dA) \\ &= \int_S \dot f(r) (\dot \partial_x + e_{xy} \dot \partial_y) (e_{xy} \, dA) \\ &= e_{xy} \int_S \Big ( \frac{\partial f}{\partial x} + e_{xy} \frac{\partial f}{\partial y} \Big ) \, dA \\ &= i \int_S \Big( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \Big)\, dA \end{align*}$$

Hm, so the premise of the problem itself looks right.

I think the problem is that, in your RHS calculation, you drop an $i$ on the $\partial f/\partial y$ term for no reason. It should be just $-\int_S \partial_y f \, dA$. This leaves an extra $i$ on half the terms in the rest of the calculation.

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