The formula in complex analysis is

$$\int f(\gamma(t))\cdot(\gamma'(t)dt$$

and the formula in the real variable setting, for a gradient field, is:

$$\int F\cdot dr$$ $$=\int f_x\,dx + f_y\,dy + f_z\,dz,$$

where the integrand is said to be an "exact differential" (or total differential.)

Are the formulas essentially the same thing, when we regard the complex function as a "vector field" mapping $C^2 \to C^2$?

Also, can one compute line integrals of scalar-valued functions in the real-variable setting -- or would this not make any physical sense?


  • $\begingroup$ "can one compute line integrals of scalar-valued functions in the real-variable setting" Yes, like this: $$\int\limits_C {f(x,y)ds} = \int_a^b {f\left( {x(t),y(t)} \right)\left\| {\vec r'(t)} \right\|} dt$$ $\endgroup$ – user204299 Aug 1 '15 at 22:18
  • $\begingroup$ Hi @JakeLebovic, so the integrand is a scalar times the norm of r'(t), times the time-differential. What is the norm object? Is that distance traveled? So that the scalar* distance = "total distance" ... times the time-differential, which when integrated gives the total displacement. Do I have the right idea? Thanks, $\endgroup$ – user258314 Aug 1 '15 at 22:30
  • 1
    $\begingroup$ @user258314 the integral Jake Lebovic wrote is the integral with respect to arclength. Physically, you can think of $f(x,y)$ as the mass density per unit arclength then the integral gives the net mass of the curve. Or, if you prefer, $f(x,y)>0$ is the height above the $xy$-plane and this integral is the area of the curvy curtain hung below the space curve $(x(t),y(t),f(x(t),y(t))$ where $(x(t),y(t))$ parametrizes the curve $C$ in the $xy$-plane. $\endgroup$ – James S. Cook Aug 1 '15 at 23:42

In the complex domain the theory of integration is quite different. In large part, this is due to Cauchy's integral formula. Long story short, there are many theorems in complex analysis which are not replicated in the real case. So, I would not agree these are the "same" thing.

That said, there is a fundamental theorem of calculus for both. As you say, the gradient vector field $\vec{F} = \nabla f$ has $\int_C \vec{F} \cdot d\vec{r} = f(B)-f(A)$ if the curve $C$ goes from $A$ to $B$ in a domain where the $f$ is defined near to $C$. Likewise, if $f(z) = g'(z)$ then $\int_C f(z) dz = g(z_1)-g(z_o)$ where $C$ goes from $z_o$ to $z_1$ again supposing $g$ is defined near $C$.

You mention a vector field from $\mathbb{C}^2$ to $\mathbb{C}^2$. However, the integral you wrote I think is just for functions from $\mathbb{C}$ to $\mathbb{C}$. These correspond to certain integrals of vector fields on $\mathbb{R}^2$. And, there are applications to fluid-flow or two-dimensional electrostatics. That said, I suspect the most natural justification for $\int_C f(z) dz$ is simply that it is the natural generalization of the Riemann integral if we replace real numbers and functions with their complex analogs.


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