Holomorphic Functions

A function that is complex-differentiable at a single point is interesting; a function that is differentiable on a whole open region is the central object of complex analysis. We give it a name.

A function f is holomorphic (equivalently, analytic) on an open set if it is complex-differentiable at every point of that set. If f is holomorphic on all of \mathbb{C}, it is called entire. Polynomials and the exponential e^z are entire; \bar{z} is holomorphic nowhere.

By the previous page, being holomorphic is the same as satisfying the Cauchy–Riemann equations (with continuous partials) throughout the region. From those two equations alone, something beautiful falls out.

The headline: holomorphic ⟹ harmonic

We will show that if f = u + iv is holomorphic, then both u and v solve Laplace's equation

u_{xx} + u_{yy} = 0,

the defining equation of a harmonic function. The proof is two differentiations and an addition.

Step 1 — start from Cauchy–Riemann. Holomorphy gives

u_x = v_y, \qquad u_y = -v_x.

Step 2 — differentiate the first equation by x. Both sides:

u_{xx} = v_{yx}.

Step 3 — differentiate the second equation by y. Both sides:

u_{yy} = -v_{xy}.

Step 4 — add them. The mixed partials of v are equal, v_{yx} = v_{xy} (equality of mixed partials), so they cancel:

u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0.

So u is harmonic. Step 5 — repeat for v: differentiate u_x = v_y by y and u_y = -v_x by x, then subtract, and the u terms cancel:

v_{xx} + v_{yy} = 0.

Both parts of a holomorphic function are harmonic. We call them harmonic conjugates of one another.

Seeing it for z²

Take f(z) = z^2, so u = x^2 - y^2 and v = 2xy. Check u:

u_{xx} = 2, \qquad u_{yy} = -2, \qquad u_{xx} + u_{yy} = 2 + (-2) = 0.

And v = 2xy has v_{xx} = 0 and v_{yy} = 0, so v_{xx} + v_{yy} = 0 too. Both harmonic, exactly as the theorem promised.

A function f holomorphic on an open set:

is complex-differentiable at every point of that set; if that set is all of \mathbb{C}, f is entire (e.g. polynomials, e^z);
satisfies the Cauchy–Riemann equations throughout, which forces its parts u and v to be harmonic: u_{xx} + u_{yy} = 0 and v_{xx} + v_{yy} = 0;
so u and v are harmonic conjugates; their level curves meet at right angles.

Laplace's equation u_{xx} + u_{yy} = 0 is everywhere in physics: it governs steady-state heat, electrostatic potential, and ideal fluid flow. The fact that the real and imaginary parts of any holomorphic function automatically solve it makes complex analysis a factory for solutions to these equations. Even better, the level curves of u and of v always cross at right angles — one family is the equipotentials, the other the flow lines. A single holomorphic function hands you both at once.

Being holomorphic means complex-differentiable in a whole neighbourhood — not just at one point — and that is an astonishingly strong demand with no parallel in real calculus. On the real line a function can be differentiable once but not twice (think of a function whose derivative has a kink). Nothing like that exists in the complex world. A function holomorphic on a region is automatically infinitely differentiable there, and it equals its own Taylor series — that is, it is analytic. So "holomorphic" and "analytic", which sound like two different properties, turn out to describe the exact same class of functions. Complex differentiability, once, on an open set, quietly hands you everything else for free.

The two families are perpendicular

For f(z) = z^2 the level curves u = x^2 - y^2 = c and v = 2xy = k are two families of hyperbolas. Slide the level values and watch where a curve from each family crosses: they always intersect at right angles — a visible fingerprint of Cauchy–Riemann. (This orthogonality holds for every holomorphic function.)