In vector calculus we know that if $\,\,\underline{F}\,\,$ is a $\,\,\underline{conservative}\,\,$ vector field, then:

  1. $\oint_\gamma{\textbf{F}\cdot d{\textbf{r}}} = 0$, where $\gamma$ is a closed curve.
  2. $\nabla\times \textbf{F} = 0$, i.e. the Curl Vanishes
  3. $\oint_{\gamma_1}{\textbf{F}\cdot d{\textbf{r}}}= \oint_{\gamma_2}{\textbf{F}\cdot d{\textbf{r}}}$, where $\gamma_i$ starts at $A$ and ends at $B$, Path Independent
  4. $\textbf{F} = \nabla\phi$, Existence of Potential, $\phi$ is a scalar field.

However in Complex Analysis we know that if $f$ is holomorphic along and inside a simple closed contour $C$, then: $$\int_Cf(z)dz = 0$$ where $z\in\mathbb{C}$.

I wanted to know what is the relationship between these two properties in $\mathbb{R}$ and in $\mathbb{C}$. Indeed in both we have a closed path that does not intersect itself (simple). However, in $\mathbb{R}$ we require that the vector field is conservative, whereas in $\mathbb{C}$ we require that the function is holomorphic inside and into the contour.

My idea was that the property of the conservative vector field that the curl vanishes, somehow connects to $f$ being holomorphic. Indeed by the sufficiency theorem we know that $f$ is holomorphic if the four partial derivatives $u_x,u_y,v_x,v_y$ exists and are continuous on and in the contour, furthermore the Cauchy Riemann equations hold, i.e. $u_x = v_y$ and $v_x = -u_y$.

The Cauchy-Riemann equations look like they could come out of the cross product between $\nabla$ and $\mathbf{F}$ somehow. Even though I couldn't figure it out.

I tried to make some kind of function that transforms $f: \mathbb{C}\to\mathbb{C}$ to a function $g:\mathbb{R}^2\to\mathbb{R}^2$, where $u(x,y)\to x'$ and $v(x,y) \to y'$, where $g(x',y') = (u(x,y),v(x,y))$.

However I have no idea if what I am doing even makes sense. So I would like to know if you could explain to me how the conditions of a conservative vector field translate to those in $\mathbb{C}$ and vice versa. Indeed in both cases we have that the integral around the path is zero, so there could be some relationship.

Please let me know how you would go about showing the link between these two apparently separate properties.


The holomorphic function $f(z) = u(x,y) + i v(x,y)$ corresponds to the vector field ${\bf F}(x,y) = u(x,y) {\bf i} - v(x,y) {\bf j}$. Note the $-$ sign. The Cauchy-Riemann equations for $f$ then correspond to $\bf F$ being irrotational and incompressible:

EDIT: Irrotational: $\dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}$, i.e. $\nabla \times {\bf F} = 0$.

Incompressible: $\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$, i.e. $\nabla \cdot {\bf F} = 0$.

  • $\begingroup$ Could you explain the minus sign and what you mean by incompressible please? $\endgroup$ Dec 7 '16 at 18:20
  • $\begingroup$ I've seen that incompressible means that $\nabla\cdot \mathbf{F} = 0$, however I thought that $\nabla\cdot \mathbf{F} = \nabla\cdot(\nabla\phi) = \nabla^2\phi$ which is the scalar Laplacian. How do I show it is equal to zero? $\endgroup$ Dec 7 '16 at 18:31
  • $\begingroup$ thank you, now it makes sense! Just one last question: how come that incrompressibility is not mentioned in the definition or properties of a conservative vector field? Is it because it doesn't always happen for a vector field and therefore this is an additional condition that assures that the C-R equations hold? $\endgroup$ Dec 7 '16 at 19:09
  • $\begingroup$ It's not related to being conservative, it's a separate property. $\endgroup$ Dec 7 '16 at 21:34
  • $\begingroup$ So this means that the two notions are not entirely equivalent. If you start with $f$ on $\mathbb{C}$ with all the above conditions, you don't get only a conservative vector field, but in addition to irrotationality, you get incrompressibility? $\endgroup$ Dec 8 '16 at 0:23

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