. Holomorphic Harmonic Functions Basic notation. Considering C as R, with coordinates x y, z = x + iy denotes the stard complex coordinate, in the usual way. Definition.1. Let f : U C be a complex valued function defined on some open subset U C. We say that f(z) is differentiable at z 0, with derivative f (z 0 ), if f (z 0 ) = lim z z0 f(z) f(z 0 ) z z 0. We say that f is holomorphic if it is differentiable at every point of U It is easy to prove, as in the real case, that the sum, difference, product, quotient, composition of holomorphic functions is holomorphic that the usual formulas apply. The obvious first step in analyzing the condition that the derivative exists, is to see what happens if we approach z 0 along a line, in particular along a horizontal vertical line. Along a horizontal line, we fix y = y 0 vary x. Using the obvious notation z 0 = x 0 + iy 0, f(z) = f(x, y) setting z = x + iy 0 we get f f(z) f(z 0 ) (z 0 ) = lim z z0 z z 0 f(x, y 0 ) f(x 0, y 0 ) = lim x x0 x x 0 = Similarly along a vertical line, we fix x = x 0 vary y. Setting z = x 0 + iy, we get f f(z) f(z 0 ) (z 0 ) = lim z z0 z z 0 f(x 0, y) f(x 0, y 0 ) = lim y y0 i(y y 0 ) = 1 = i i (x0,y 0 ) In other words, if f is holomorphic then = i (x0,y 0 ) 1
There is another way to interpret all of this. Suppose we start with f(x, y), a function f : R R. Now we can express x y in terms of z z, x = 1 (z + z) y = 1 i (z z) = i (z z). Thus we can consider f(x, y) as a function f(z, z) of z z. Now formally differentiating formally applying the chain rule we would have z = z + z = 1 i, z = z + z = 1 + i This suggests that we introduce operators z = 1 ( i ) z = 1. ( + i ). In these terms, the condition above becomes z = 0. In other words, a function f : C C is holomorphic if only if it does not depend on z. Separating into real imaginary parts, we get the famous Cauchy- Riemann equations. Suppose that we write f = u + iv. Then Lemma.. If is a C 1 -function then = v u: U R R v =. ɛ = u(x + h, y + k) h k goes to zero faster than (h, k) goes to zero, for any (x, y) U. Proof. By assumption δ = u(x + h, y) u(x, y) h. (x,y). goes to zero faster than h. On the other h, η = u(x + h, y + k) u(x + h, y) k.
goes to zero faster than k. As the partials are continuous, the difference goes to zero as h goes to zero. So (x,y) ( ζ = (x,y) ) k goes to zero faster than k. Thus ɛ = u(x + h, y + k) u(x, y) h k = u(x + h, y + k) u(x + h, y) k + u(x + h, y) u(x, y) h = u(x + h, y + k) u(x + h, y) + ζ + δ = η + ζ + δ, which goes to zero faster than (h, k). Proposition.3. A function f(z) is holomorphic if only if its real imaginary parts u(x, y) v(x, y) are C 1 (that is, the first derivatives exist are continuous) satisfy the Cauchy-Riemann equations. Proof. We have already seen that the derivatives of u v exist satisfy the Cauchy-Riemann equations. We will see later that if f is holomorphic then the derivative of f is holomorphic (so that f is infinitely differentiatible) this will imply that u v are C 1. Suppose that u v are C 1. (.) implies that u(x + h, y + k) u(x, y) = h + k + ɛ 1 v(x + h, y + k) v(x, y) = v h + v k + ɛ 3
where ɛ 1 ɛ go to zero faster than h + ik. f(x + h, y + k) f(x, y) = (u(x + h, y + k) u(x, y)) + i(v(x + h, y + k) v(x, y)) = h + k + ɛ 1 + i v h + i v k + ɛ ( ) ( ) = + i v h + + i v k + ɛ 1 + ɛ ( ) = + i v (h + ik) + ɛ 1 + ɛ As ɛ 1 + ɛ approaches zero faster than h + ik, it follows that f is differentiable. It is interesting to further investigate the properties satisfied by u v. Recall that the Laplacian of a real function u(x, y): R R is defined to be u = u + u, that function u is said to be Harmonic if it is C the Laplacian vanishes, that is u = 0. Now suppose that f = u + iv, where f is holomorphic, so that u is the real part of a holomorphic function. Latter we will show that f is infinitely differentiable, so that in particular u v are both C. On the other h u = v = v = u, where we used the fact that if v is C then its mixed partials are equal. Thus u = u + u = 0, so u is harmonic. Similarly v is harmonic. Two functions u v that satisfy the Cauchy-Riemann equations, are called conjugate harmonic v is said to be the conjugate harmonic function of u. Note that this is a slight abuse of language, since v is only determined up to a constant. Given a harmonic function u it is instructive to try to construct the harmonic conjugate v. 4
For example u = x y is harmonic. = x = y. Thus the harmonic conjugate satisfies v = y v = x. Take the first equation integrate with respect to x. We get that v(x, y) = xy + g(y) where g is an arbitrary function of y. Now differentiate with respect to y. Then x + g (y) = x. Thus g(y) is constant. Thus the harmonic conjugate of u is v = xy. Of course in this case the corresponding holomorphic function is f(z) = u(x, y) + iv(x, y) = (x y ) + ixy = (x + iy) = z. In fact there is another way to compute harmonic conjugates, which does not depend on using integration. In other words, we will find a formula for a holomorphic function f(z) in terms of its real part u(x, y). The trick is to consider the function g(z, z) = f(z). Note that g z = ḡ z = z = 0, so that g(z, z) = g( z) is a function only of z (g is sometimes known as an antiholomorphic function). Then u(x, y) = 1 (f(x + iy) + g(x iy)). Now treat this equality formally extend it to the case where x = z/, y = z/i. Thus u(z/, z/i) = 1 (f(z) + g(0)). Now f(z) is determined up to an imaginary constant. Thus we may as well assume that f(0) is real. In this case, we may take f(0) = g(0) = u(0, 0). Thus f(z) = u(z/, z/i) u(0, 0). For example, suppose that u(x, y) = x y. Then f(z) = u(z/, z/i) u(0, 0) = (z/) (z/i) = z / + z / = z. Note that for this to work, the expression u(z/, z/i) has to make sense (for example, u is given as a power series). 5