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.
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.)