1.6 Complex-Valued Functions as Mappings · Preliminaries to Complex Analysis

[1.6.1]Remark(Real Differentiability vs. Complex Differentiability)#

We now make the distinction between treating a complex-valued function as a function $\bR^2 \to \bR^2$ . Consider the function $f(z) = \bar{z}$ , then note

f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} = \lim_{h \to 0} \frac{\bar{h}}{h},

which does not exist (parameterizing $\bar{z}/z$ by an angle yields $\exp(-2i\theta)$ , and so the limit varies based on path chosen). Thus $f: \bC \to \bC$ is not holomorphic. But now consider $g: \bR^2 \to \bR^2$ defined by $g(x, y) = (u(x, y), v(x, y)) = (x, -y)$ . This function, as a real-valued one, is linear, infinitely differentiable, and enjoys many properties. One such property is that its total derivative (as a linear map), expressed in the standard basis of $\bR^2$ , is written as

g' = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}.

From the perspective of multivariate theory, a map $g: \bR^2 \to \bR^2$ has a derivative that is a linear map in two-dimensions, in contrast with a map $f: \bC \to \bC$ having a singular, complex number for its derivative (that is also a one-dimensional linear map).

[1.6.2]Remark(Cauchy-Riemann Equations)#

We can link real differentiability and complex differentiability, however. For a complex function $f: \Omega \to \bC$ to be differentiable, its difference function must take on the same value as $h \to 0$ through any path. Writing $h = h_1 + ih_2$ and $z = x + iy$ (and so $f(z) = f(x, y)$ ), we may choose a point $z_0 = x_0 + i y_0 \in \Omega$ for which $f'(z_0)$ exists and yield the following:

\begin{align*} f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h} &= \lim_{h_1 \to 0} \frac{f(x_0 + h_1, y_0) - f(x_0, y_0)}{h_1} = {\frac{\partial f}{\partial x}} (x_0) \quad (h_2 = 0) \\ &= \lim_{h_2 \to 0} \frac{f(x_0, y_0 + h_2) - f(x_0, y_0)}{ih_2} = {\frac{1}{i}\frac{\partial f}{\partial y}} (x_0) \quad (h_1 = 0). \\ \end{align*}

If $f$ is holomorphic, we have that $\frac{\partial f}{\partial x} = \frac{1}{i} \frac{\partial f}{\partial y} = -i\frac{\partial f}{\partial y}$ by the uniqueness of the derivative. If we write $f(z) = u(z) + iv(z)$ , then we have that the partials $u_x, u_y, v_x, v_y$ all exist, and they are such that

\frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y} \implies \left(\frac{\partial u}{\partial x}\right) + i\left(\frac{\partial v}{\partial x}\right) = i\left[ \left(\frac{\partial u}{\partial y}\right) + i \left(\frac{\partial v}{\partial y}\right) \right] = \left(-\frac{\partial v}{\partial y}\right) + i\left(\frac{\partial u}{\partial y}\right),

implying the Cauchy-Riemann equations:

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.

[1.6.3]Remark(Derivation of Wirtinger Operators)#

We will motivate the introduction of two important differential operators known as Wirtinger operators. To get started, let $z \in \bC$ be a complex variable. Noting that $\bC \cong \bR^2$ , write $z = x + iy$ and $\bar{z} = x - iy$ . We may “solve” for $x, y$ in terms of $z, \bar{z}$ as such:

x = \frac{z + \bar{z}}{2} \quad y = \frac{z - \bar{z}}{2i}.

Treating $\bar{z}$ as independent of $z$ , we can expand the differential operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ by the multivariate chain rule.

\frac{\partial}{\partial z} = \underbrace{\frac{\partial x}{\partial z}}_{1/2} \frac{\partial}{\partial x} + \underbrace{\frac{\partial y}{\partial z}}_{1/2i} \frac{\partial}{\partial y} = \frac{1}{2} \left( \frac{\partial}{\partial x} + \frac{1}{i} \frac{\partial}{\partial y} \right),

\frac{\partial}{\partial \bar{z}} = \underbrace{\frac{\partial x}{\partial \bar{z}}}_{1/2} \frac{\partial}{\partial x} + \underbrace{\frac{\partial y}{\partial \bar{z}}}_{-1/2i} \frac{\partial}{\partial y} = \frac{1}{2} \left( \frac{\partial}{\partial x} - \frac{1}{1} \frac{\partial}{\partial y} \right).

[1.6.4]Proposition#

If $f(z) = u(z) + iv(z)$ is holomorphic at $z_0$ , then

{\frac{\partial f}{\partial \bar{z}}} (z_0) = 0 \quad f'(z_0) = {\frac{\partial f}{\partial z}} (z_0) = 2 {\frac{\partial u}{\partial z}} (z_0).

Let $F$ be a mapping in terms of real variables such that $F(x, y) = f(x + iy)$ . Then we have that $F$ is differentiable, and its derivative can be represented by a Jacobian $J_F$ such that

\det(J_F(x_0, y_0)) = |f(x_0 + i y_0)|^2.

Proof.

Suppose $f$ is holomorphic at $z_0$ . Then $f$ satisfies the Cauchy-Riemann equations, meaning ${\frac{\partial f}{\partial x}} (z_0) = \frac{1}{i} \frac{\partial f}{\partial y} (z_0)$ , and so

{\frac{\partial f}{\partial \bar{z}}} (z_0) = \frac{1}{2}\left( {\frac{\partial f}{\partial x}} (z_0) - \frac{1}{i} {\frac{\partial f}{\partial y}} (z_0) \right) = 0.

From this, we have that

f'(z_0) = \frac{1}{2}\left( {\frac{\partial f}{\partial x}}(z_0) + {\frac{\partial f}{\partial x}}(z_0) \right) = \frac{1}{2}\left( {\frac{\partial f}{\partial x}}(z_0) + \frac{1}{i} {\frac{\partial f}{\partial y}}(z_0) \right) = {\frac{\partial f}{\partial z}}(z_0).

From the Cauchy-Riemann Equations, we get that $\frac{\partial f}{\partial z} = 2\frac{\partial u}{\partial z}$ for $f = u + iv$ . We have already proved that complex differentiability implies real differentiability by the Cauchy-Riemann equations. Finally, note that

J_F(x_0, y_0) = \begin{bmatrix} u_x(x_0) & u_y(y_0) \\ v_x(x_0) & v_y(y_0) \end{bmatrix},

and so $\det(J_F(x_0, y_0)) = u_x(x_0) v_y(y_0) - u_y(y_0)v_x(x_0) = (u_x(x_0))^2 + (u_y(y_0))^2 = |f'(z_0)|^2$ .

[1.6.5]Theorem(Criterion for Holomorphicity)#

A function $f(z) = u(z) + iv(z)$ is holomorphic if and only if $u, v$ have continuous, first-order partial derivatives that satisfy the Cauchy-Riemann equations.

Proof.

$(\Longrightarrow):$ This is implied from taking the limit of the difference quotient along $\bR$ and $i\bR$ .

$(\Longleftarrow):$ Suppose $u, v$ have continuous first-order partial derivatives that satisfy the Cauchy-Riemann equations. From multivariate theory, note that

\begin{align*} u(x + h, y + k) - u(x, y) &= u_x h + u_y k + o_1(h + ik),\\ v(x + h, y + k) - v ( x, y) &= v_x h + v_y k + o_2(h + ik). \end{align*}

Then note

f(z + [h + ik]) - f(z) = (u_x + iv_x)(h + ik) + [o_1(h + ik) + i o_2(h + ik)],

meaning

f'(z) = \lim_{h + ik \to 0} \frac{f(z + [h + ik]) - f(z)}{h + ik} = u_x + i v_x.

1.6Complex-Valued Functions as Mappings