Holomorphic Functions

The power of complex analysis doesn’t come from complex-differentiability at a point (though that is certainly not a trivial property for a function). Its power comes from complex-differentiability in a neighborhood around that point.

Definition 1. Let $f : \mathbb C \to \mathbb C$ be differentiable at some point $z_0 \in \mathbb C$ .

$f$ is holomorphic at $z_0$ if there exists a neighborhood $U$ of $z_0$ such that $f$ is complex-differentiable on $U$ .
$f$ is holomorphic on an open set $U$ if it is holomorphic at every point $z_0 \in U$ .
$f$ is entire if it is holomorphic on $\mathbb C$ .

For simplicity, call a nonempty open subset $D \subseteq \mathbb C \cong \mathbb R^2$ a domain if and only if it is (path-)connected. Here, a path from $z_1$ to $z_2$ is the image $\gamma([0,1])$ of a continuous map $\gamma : [0, 1] \to \mathbb C$ such that $\gamma(0) = z_1$ and $\gamma(1) = z_1$ .

Let $f : \mathbb C \to \mathbb C$ is analytic on a domain $D$ containing $z_0$ .

Theorem 1. We have $f' = 0$ on $D$ if and only if $f = f(z_0)$ on $D$ .

Proof. For the direction $(\Rightarrow)$ , $z_0 = x_0 + i y_0$ and fix $z = x + iy \in D$ such that there exists a con line

$L_{z_0, z}:= \{ (1-t) \cdot z_0 + t \cdot z : t \in [0, 1]\}.$

belong to $D$ . If $f' = 0$ , then

$u_x = u_y = v_x = v_y = 0.$

Observe that the function $\gamma : [0, 1] \to \mathbb C$ defined by

$\gamma(t) := f((1-t) \cdot z_0 + t \cdot z)$

is differentiable on $L$ . By the mean value theorem, there exists $c \in [0, 1]$ such that

$\gamma(1) = \gamma(0) + \gamma'(c) \cdot (z - z_0).$

However,

$\gamma'(c) = f'((1-c) \cdot z_0 + c \cdot z) \cdot (z - z_0) = 0 \cdot (z-z_0) = 0,$

so that $f(z) = \gamma(1) = \gamma(0) = f(z_0)$ . For arbitrary $z \in D$ , connect it to $z_0$ using the polygonal lines $L_{z_0,z_1}, L_{z_1,z_2}, \dots, L_{z_{n-1},z_n}, L_{z_n,z}$ . By the previous argument,

$\gamma(z) = \gamma(z_n) = \cdots = \gamma(z_1) = \gamma(z_0).$

The direction $(\Leftarrow)$ is obvious.

Analytic functions also play a crucial role in harmonic analysis. But first, define second-order partial derivatives by $u_{x_i x_j} := (u_{x_i})_{x_j}$ whenever they exist.

Theorem 2. If the functions $u = \mathbf f_1, v = \mathbf f_2$ have continuous first- and second-order partial derivatives, then they are harmonic in $D$ in that they satisfy the following Laplace equations:

$\left\{ \begin{aligned} u_{xx} + u_{yy} &= 0, \\ v_{xx} + v_{yy} &= 0. \end{aligned} \right.$

Proof. Since $f$ is analytic, its component functions satisfy the Cauchy-Riemann equations:

$u_x = v_y,\quad u_y = -v_x.$

Taking necessary derivatives,

$u_{xx} = v_{yx},\quad u_{yy} = -v_{xy},\quad v_{xx} = -u_{yx},\quad v_{yy} = u_{xy}.$

Thus, we prove the theorem if we can prove that $u_{xy} = u_{yx}$ . This we claim holds by Clairaut’s theorem below.

Theorem 3 (Clairaut’s Theorem). If $u$ has continuous first- and second-order partial derivatives on a domain $D$ , then $u_{xy} = u_{yx}$ .

Proof. Suppose without loss of generality that $D$ is an open square containing some $(x_0, y_0)$ and $(x_0 \pm \epsilon, y_0 \pm \epsilon)$ . Define the Fréchet-differentiable functions $f,g, h : (-\epsilon, \epsilon)^2 \to D$ by

$\begin{aligned}f(s, t) &:= u(x_0 + s, y_0 + t) - u(x_0 + s, y_0), \\ g(s, t) &:= u(x_0 + s, y_0 + t) - u(x_0 , y_0+t), \\ h(s, t) &:= u(x_0 + s, y_0 + t) - u(x_0 + s, y_0) - u(x_0 , y_0+t) + u(x_0, y_0). \end{aligned}$

Now fix $(s, t) \in (-\epsilon, \epsilon)^2$ such that $s \neq 0, t \neq 0$ . Use the mean value theorem to obtain $\xi_1,\xi_2 \in (0, 1)$ such that

$\begin{aligned} h(s,t) &= f(s,t) - f(0, t) = f_x(\xi_1 s, t) \cdot s \\ &= u_{xy}(x_0 + \xi_1 s, y_0 + \xi_2 t) \cdot s \cdot t\end{aligned}$

Similarly, use the mean value theorem to obtain $\xi_3,\xi_4 \in (0, 1)$ such that

$\begin{aligned} h(s,t) &= g_y(s, \xi_3 t) \cdot t = u_{yx}(x_0 + \xi_4 s, y_0 + \xi_3 t) \cdot s \cdot t\end{aligned}$

Hence,

$u_{xy}(x_0 + \xi_1 s, y_0 + \xi_2 t) = u_{yx}(x_0 + \xi_4 s, y_0 + \xi_3 t) .$

Taking $(s, t) \to (0, 0)$ , and applying continuity, $u_{xy}(x_0, y_0) = u_{yx}(x_0, y_0)$ . Since $(x_0, y_0)$ is arbitary, we obtain $u_{xy} = u_{yx}$ , as required.

The converse of Theorem 2 holds as well.

Corollary 1. Let $f : \mathbb C \to \mathbb C$ . If $u,v$ are harmonic functions on $D$ that satisfy the Cauchy-Riemann equations, then $f$ is holomorphic on $D$ .

And just like with complex-differentiability, we obtain a family of holomorphic functions.

Theorem 4. Suppose $f, g : \mathbb C \to \mathbb C$ are holomorphic on a domain $D$ . Then $f \pm g$ , $f \cdot g$ are holomorphic on $D$ . Furthermore, if $g \neq 0$ on $D$ , then $f/g$ is holomorphic on $D$ . Finally, if $g$ is holomorphic on $f(D)$ , then $g \circ f$ is holomorphic on $D$ , yielding the usual product, quotient, and chain rules in real-variable calculus.

Particularising to $D = \mathbb C$ , we obtain families of entire functions. Polynomials are our canonical examples, but the exponential is really the star of complex analysis.

Theorem 5. The exponential function $\exp : \mathbb C \to \mathbb C$ characterised by

$\exp(x +iy ) = e^{x} \cdot (\cos(y) + i \sin(y))$

is entire.

Proof. Writing $u = e^x \cdot \cos(y)$ and $v = e^x \cdot \sin(y)$ , it is clear that

$u_x = v_y,\quad u_y = -v_x.$

Furthermore,

$u_{xx} + u_{yy} = u + (-u) = 0,\quad v_{xx} + v_{yy} = v + (-v) = 0.$

By Corollary 1, $\exp$ is holomorphic on $\mathbb C$ . Therefore, $\exp$ is entire.

Corollary 2. The complex trigonometric functions $\cos, \sin$ defined by

$\displaystyle \cos(z) := \frac{e^{iz} + e^{-iz}}{2},\quad \sin(z) := \frac{e^{iz} - e^{-iz}}{2i}$

are entire.

What about the logarithm? This is where we will need to take a pitstop and re-evaluate our life decisions. Or rather, approach it with a log more caution than the other functions so far.

—Joel Kindiak, 11 Aug 25, 2346H

KindiakMath

Holomorphic Functions

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Holomorphic Functions

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