# Divergence, curl, and gradient of a complex function

From an answer here I got Green's theorem for functions in the complex plane

$$\oint f(z) \, dz = i \iint \left( \nabla f \right) \, dx \, dy = i \iint \left( 1 {\partial f \over \partial x} + i {\partial f \over \partial y} \right) \, dx \, dy$$

Which leads to the well known Cauchy's integral theorem

$$\oint f(z) \, dz = \iint \left( \frac{- \partial f_x}{\partial y} + \frac{- \partial f_y}{\partial x} \right)+ i \left( \frac{\partial f_x}{\partial x} + \frac{- \partial f_y}{\partial y} \right) \, dx \, dy$$ From which I then get $$\oint f(z) \, dz = \iint \left( \nabla \times f + i \nabla \cdot f \right) \, dx \, dy$$ I'm hoping someone here can tell me whether I'm on the right track or not.

Keep in mind that $$\nabla = 1 {\partial \over \partial x} + i {\partial \over \partial y}$$

• You can use \mathbf{f} $\mathbf{f}$ or \vec{f} $\vec{f}$ for vectors. Nov 23, 2017 at 4:57
• There are no vectors. The complex numbers are being treated as though they were vectors though Nov 23, 2017 at 5:03
• Bro, it’s just a MathJax tip; no need to get defensive about it. And by the way, I would strongly contend that divergence and curl only make sense as operators on vectors and that vectors within the complex plane are very much actual vectors. Nov 23, 2017 at 5:13
• I didn't get defensive about it. Why would you think I was getting defensive about it? Nov 23, 2017 at 5:17
• Well it took me 3 days to do what should have taken 30 minutes but I finally did it. Its all based on solitaryroad.com/c606.html Nov 23, 2017 at 8:49

I'm not a physicist, but I think that gradient, curl, and divergence are strictly for a real $d$-dimensional environment, in particular for $d=2$ and $d=3$. I have never met your strange complex definition of $\nabla$.
On the other hand it is of course possible to prove the Cauchy integral formula using Green's theorem in the form $$\int_{\partial \Omega}\bigl(P(x,y)\>dx+Q(x,y)\>dy\bigr)=\int_\Omega(Q_x-P_y)\>{\rm d}(x,y)\ .\tag{1}$$ Write your analytic $f$ in the form $f=u+ iv$ as well as $dz$ in the form $dz=dx+i dy$. Then by definition of complex line integrals you have $$\int_{\partial\Omega}f(z)\>dz=\int_{\partial\Omega}(u\>dx-v\>dy)+i\int_{\partial\Omega}(v\>dx+ u\>dy)\ ,$$ to which you can apply $(1)$ separately. Finally the CR equations will come to your rescue.