Geometric Interpretation of Antiderivative?

Question

Could someone please give me a geometric interpretation of the above theorem?

Answer 1

It's terribly hard to explain geometry without having any pictures on hand, but I'll try my best. I'll be working from the perspective of vector fields ($\mathbb{R}^2 \to \mathbb{R}^2$): specifically, we're going to be working with the vector field $\overline{f} = u-iv$.

Wait, you say! Why work with the conjugate instead of the original $f=u+iv$? We're going to see that $f$ being analytic is equivalent to the vector field $\overline{f}$ having certain very nice properties. Namely, $\overline{f}$ is "curl"-free and divergence-free. This in turn will allow us, on a simply-connected domain, to construct a "potential function" $\Phi$ and a "stream function" $\Psi$ for $\overline{f}$. Finally, when we put together $\Phi + i\Psi$, we'll turn out to have an antiderivative of $f$!

Some notation to start with: Given $z = a+bi$ and $w = c+di$ ($a,b,c,d$ real), write $z \cdot w = ac+bd$ (the dot product) and $z \times w = ad - bc$ (the "cross product", or rather, the 3rd component of the cross product of $(a,b,0)$ and $(c,d,0)$).

First we look at the "curl". By the Cauchy-Riemann equations, $$ \nabla \times \overline{f} = - \partial_x v - \partial_y u = 0. $$ So $\overline{f}$ is a conservative (irrotational) vector field, which means that it has the property that the "work" (line integral) of $\overline{f}$ along any closed curve $\Gamma$ is $0$. (We need the simple-connectedness to make this conclusion!) To put it another way, the work along any path $\gamma$ from $a$ to $b$ depends only on the endpoints $a$ and $b$, not the path taken. With a fixed base point $z_0$, we can define a potential function \begin{align} \Phi(Z) &= (\text{Work of }\overline{f} \text{ along any path from }z_0\text{ to }Z) \\ &= \int_{z_0}^Z \overline{f} \cdot dz = \int_{z_0}^Z (u - iv) \cdot (dx + i\,dy) = \int_{z_0}^Z u\,dx - v\,dy. \end{align} Note that the gradient of $\Phi$ is $\overline{f}$.

Next we look at the divergence. By the Cauchy-Riemann equations again, $$ \nabla \cdot \overline{f} = \partial_x u - \partial_y v = 0. $$ Making use of the simple-connectedness again, $\overline{f}$ is being divergence-free, and so has the property that the "flux" of $\overline{f}$ across any closed curve $\Gamma$ is $0$. Just like the work, the flux across any path $\gamma$ from $a$ to $b$ depends only on the endpoints $a$ and $b$, not the path taken. We can then define a "stream function" \begin{align} \Psi(Z) &= (\text{Flux of }\overline{f} \text{ across any path from }z_0\text{ to }Z) \\ &= \int_{z_0}^Z \overline{f} \cdot (-i\,dz) = \int_{z_0}^Z (u - iv) \cdot (dy - i\,dx) = \int_{z_0}^Z u\,dy + v\,dx. \end{align} (Minor note: I've implicitly made a choice of orientation for the flux.) This is also equal to $$ \int_{z_0}^Z (i\overline{f}) \cdot dz; $$ in particular, the gradient of $\Psi$ is $i\overline{f}$.

---

Now we finally get to the geometry. When we make a plot of the vector field $\overline{f}$, we can draw a family of equipotentials: level curves of the potential function $\Phi$. We can also draw a family of streamlines: level curves of the stream function $\Psi$.

Choose the levels to be the multiples of some very small constant $k > 0$. If you plot these equipotentials and streamlines together on diagram, you'll see that the equipotentials and streamlines

1. intersect at right angles,

and even more importantly,

2. form a grid of small (approximate-)squares.

Why? Because the gradients $\nabla \Phi = \overline{f}$ and $\nabla \Psi = i\overline{f}$ are (1) perpendicular and (2) have the same magnitude.

Now define $F = \Phi + i\Psi$. Geometrically, $F$ is the mapping which sends the equipotentials and streamlines of the vector field $\overline{f}$ to vertical and horizontal lines (respectively). Of course, the horizontal and vertical lines are all equally spaced (with spacing $k$). So they, too,

1. intersect at right angles
2. form a grid of small (actual) squares.

The function $F$ maps a grid of small "squares" to a grid of small squares. That means it's analytic.

Let's find its derivative. The gradient of $\Phi$ at a point $z$ is $\overline{f(z)}$; let's call that $Re^{-i\phi}$. Traveling a distance of $k/R$ in the gradient direction $e^{-i\phi}$ (i.e. the direction perpendicular to the equipotential through $z$) increases the potential from $\Phi(z)$ to $\Phi(z)+k$. On the other hand, the direction of $\nabla\Phi$ is the direction perpendicular to $\nabla\Psi$. So traveling $(k/R)e^{-i\phi}$ won't change $\Psi$ - it's in the direction along the streamline through $z$.

Putting these together, changing $z$ by $(k/R)e^{-i\phi}$ causes $F$ to change by $k$. Remember, this is in the limit as $k \to 0$. So we've determined the value of $F'(z)$: the derivative of $F$ at $z$ is $Re^{i\phi}=f(z)$. Hence, $F$ is indeed an antiderivative of $f$.

Finally, we'll relate $F$ to the formula you're probably used to: \begin{align} \int_{z_0}^z f(z) \, dz &= \int_{z_0}^z (u+iv)(dx+i\,dy) \\ &= \int_{z_0}^z (u\,dx - v\,dy) + i(u\,dy + v\,dx) \\ &= \Phi(z) + i\Psi(z) \\ &= F(z). \end{align} Now you see the innocent looking formula $F(z)=\int_{z_0}^z f(z)\,dz$ hides a great deal of physical and geometric content. It all has to do with the vector field $\overline{f}$.

---

For a less condensed and more picture-ful exposition (you need to see the figures! you need to see the figures!), please pick up a copy of Needham's Visual Complex Analysis. This post is greatly indebted to the book, and almost entirely based on the chapter "Vector Fields and Complex Integration".

Answer 2

Assume that $f:\>\Omega\to{\mathbb C}$ is an analytic function on some domain (connected open set) $\Omega\subset{\mathbb C}$, and that $z_0\in\Omega$ is an arbitrary point in this domain. Then there is a (maybe small) disk $D(z_0)$ centered at $z_0$ which is completely contained in $\Omega$. Using the Taylor development of $f$ at $z_0$, or suitable line integrals together with the CR equations, one can show that the restriction of $f$ to this disk has a primitive, valid in this disk only. This means that there is an analytic function $F_{[z_0]}:\>D(z_0)\to{\mathbb C}$ such that $$f(z)={F_{[z_0]}}'(z)\qquad\forall z\in D(z_0)\ .$$ Such a local primitive $F_{[z_0]}$ is not uniquely determined. In fact it is determined exactly up to an additive constant.

Assume that such a construction has been performed at all points $z_0\in\Omega$.

The geometric content of Theorem 18.3.3 above is the following: When the domain $\Omega$ is simply connected then one can adjust the integration constants of all these local primitives in such a way that a globally defined function $$F:\quad\Omega\to{\mathbb C}$$ emerges, which is then a global primitive of $f$ on $\Omega$: $$f(z)=F'(z)\qquad \forall z\in\Omega\ .$$ When $\Omega$ is not simply connected then one cannot guarantee this for every $f:\>\Omega\to{\mathbb C}$. The most prominent example is $\Omega:={\mathbb C}\setminus\{0\}$ (the punctured plane) and $$f(z):={1\over z}\qquad(z\in\Omega)\ .$$ Here the functions $$F_{[z_0]}(z):={\rm Log}{z\over z_0}$$ can serve as local primitives near their $z_0$, but it is impossible to concoct a global primitive of $f$ out of these.