Chapter 2 • Theory & Derivations

Holomorphic Functions, Cauchy-Riemann Equations & Harmonic Conjugates

Complex differentiability, the Cauchy-Riemann equations in Cartesian and polar forms, geometric conformality, and harmonic conjugate potentials.

§2.1 Complex Differentiability, Holomorphic / Analytic Functions & Geometric Conformal Interpretation

1. Complex Differentiability

Let $f: \Omega \to \mathbb{C}$ be a function defined on an open set $\Omega \subseteq \mathbb{C}$, and let $z_0 \in \Omega$. We say that $f$ is complex differentiable at $z_0$ if the limit:

$$f'(z_0) = \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z}$$

exists. Crucially, this limit must be completely independent of the direction in which $\Delta z = \Delta x + i\Delta y \to 0$ in the 2D complex plane!

Definition 2.1: A function $f$ is called holomorphic (or analytic) on an open set $\Omega$ if it is complex differentiable at every point of $\Omega$. A function holomorphic on the entire complex plane $\mathbb{C}$ is called an entire function.

2. Geometric Meaning: Infinitesimal Conformal Invariance

Write the derivative in polar form: $f'(z_0) = r e^{i\theta} \ne 0$. For an infinitesimal displacement $dz$, the differential transformation is:

$$dw = f'(z_0) dz = (r e^{i\theta}) |dz| e^{i\phi} = (r |dz|) e^{i(\phi + \theta)}$$

This means the mapping $w = f(z)$ acts infinitesimally as a uniform scaling by factor $r = |f'(z_0)|$ and a rigid rotation by angle $\theta = \arg f'(z_0)$. Consequently, angles between curves and orientations are locally preserved (conformality).

§2.2 The Cauchy-Riemann Equations in Cartesian and Polar Forms with Sufficiency Proofs

1. Derivation of the Cauchy-Riemann Equations

Let $f(z) = u(x, y) + i v(x, y)$ where $u, v: \mathbb{R}^2 \to \mathbb{R}$ are real-valued. Approach $0$ along the real axis ($\Delta z = \Delta x$):

$$f'(z) = \lim_{\Delta x \to 0} \frac{u(x+\Delta x, y) - u(x, y) + i[v(x+\Delta x, y) - v(x, y)]}{\Delta x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$

Approach $0$ along the imaginary axis ($\Delta z = i\Delta y$):

$$f'(z) = \lim_{\Delta y \to 0} \frac{u(x, y+\Delta y) - u(x, y) + i[v(x, y+\Delta y) - v(x, y)]}{i\Delta y} = \frac{1}{i}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y}$$

Equating real and imaginary parts yields the celebrated Cauchy-Riemann Equations:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

2. The Sufficiency Theorem

Theorem 2.1 (Looman-Menchoff / Sufficiency): Let $u(x, y)$ and $v(x, y)$ have continuous first-order partial derivatives throughout an open neighborhood of $z_0 = (x_0, y_0)$. If the Cauchy-Riemann equations hold at $z_0$, then $f(z) = u + iv$ is complex differentiable at $z_0$ with derivative: $$f'(z_0) = u_x(x_0, y_0) + i v_x(x_0, y_0) = u_x - i u_y$$

3. Polar Form of Cauchy-Riemann Equations

In polar coordinates $z = r e^{i\theta}$, $f(z) = u(r, \theta) + i v(r, \theta)$:

$$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}, \qquad \frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta}$$

The derivative is given by $f'(z) = e^{-i\theta} \left( \frac{\partial u}{\partial r} + i \frac{\partial v}{\partial r} \right)$.

§2.3 Harmonic Functions, Harmonic Conjugates, Orthogonal Trajectories & Fluid Potentials

1. Harmonic Functions

Let $f = u + iv$ be holomorphic on domain $D$. If $u, v \in C^2(D)$, differentiate the C-R equations:

$$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial v}{\partial y}\right) = \frac{\partial^2 v}{\partial x \partial y}, \qquad \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}\left(-\frac{\partial v}{\partial x}\right) = -\frac{\partial^2 v}{\partial y \partial x}$$

By Schwarz's theorem on mixed partials ($v_{xy} = v_{yx}$):

$$\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \qquad \nabla^2 v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$$

Both the real and imaginary parts of a holomorphic function are harmonic functions! The function $v$ is called the harmonic conjugate of $u$.

2. Orthogonal Equipotentials and Fluid Streamlines

Compute the inner product of gradients:

$$\nabla u \cdot \nabla v = \left( \frac{\partial u}{\partial x} \right) \left( \frac{\partial v}{\partial x} \right) + \left( \frac{\partial u}{\partial y} \right) \left( \frac{\partial v}{\partial y} \right) = (u_x)(-u_y) + (u_y)(u_x) = 0$$

Hence, the level curves $u(x, y) = c_1$ and $v(x, y) = c_2$ are strictly mutually orthogonal wherever $f'(z) \ne 0$. In 2D fluid dynamics, $u$ is the velocity potential and $v$ is the stream function.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Example 2.1: Harmonic Conjugate & Construction of Holomorphic Function

Verify that $u(x, y) = x^3 - 3xy^2 + 2x$ is harmonic, and find its harmonic conjugate $v(x, y)$ such that $f(z) = u + iv$ with $f(0) = 0$.

Step 1: Check Laplace's Equation $\nabla^2 u = 0$

\frac{\partial u}{\partial x} = 3x^2 - 3y^2 + 2, \quad \frac{\partial^2 u}{\partial x^2} = 6x

\frac{\partial u}{\partial y} = -6xy, \quad \frac{\partial^2 u}{\partial y^2} = -6x

\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 6x - 6x = 0

Hence $u(x, y)$ is harmonic on $\mathbb{R}^2$.

Step 2: Apply Cauchy-Riemann Equations
From $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}$:

\frac{\partial v}{\partial y} = 3x^2 - 3y^2 + 2

Integrating with respect to $y$:

v(x, y) = \int (3x^2 - 3y^2 + 2)\,dy = 3x^2 y - y^3 + 2y + g(x)

Step 3: Differentiate with respect to $x$ and equate to $-u_y$

\frac{\partial v}{\partial x} = 6xy + g'(x) = -\frac{\partial u}{\partial y} = -(-6xy) = 6xy \implies g'(x) = 0 \implies g(x) = C

Thus $v(x, y) = 3x^2 y - y^3 + 2y + C$.

Step 4: Initial condition and $f(z)$ formulation

f(0) = u(0, 0) + i v(0, 0) = 0 + i C = 0 \implies C = 0

$$f(z) = (x^3 - 3xy^2 + 2x) + i(3x^2 y - y^3 + 2y) = (x + iy)^3 + 2(x + iy) = z^3 + 2z$$

Final Answer & Physical Insight

$v(x, y) = 3x^2 y - y^3 + 2y$ and $f(z) = z^3 + 2z$.

Tier 2 • Intermediate Exam Example 2.2: Constancy of Holomorphic Functions with Constant Modulus

Prove that if $f(z) = u(x, y) + i v(x, y)$ is holomorphic on a connected domain $D$ and $|f(z)| = c$ is constant on $D$, then $f(z)$ must be constant.

Case 1: $c = 0$
If $|f(z)| = 0$, then $f(z) = 0$ for all $z \in D$, which is identically constant.

Case 2: $c > 0$
Since $|f(z)|^2 = u^2 + v^2 = c^2$, differentiate partially with respect to $x$ and $y$:

2u \frac{\partial u}{\partial x} + 2v \frac{\partial v}{\partial x} = 0 \implies u u_x + v v_x = 0 \quad \text{--- (1)}

2u \frac{\partial u}{\partial y} + 2v \frac{\partial v}{\partial y} = 0 \implies u u_y + v v_y = 0 \quad \text{--- (2)}

Step 2: Apply Cauchy-Riemann equations
Recall $u_y = -v_x$ and $v_y = u_x$. Substituting into (2):

-u v_x + v u_x = 0 \implies v u_x - u v_x = 0 \quad \text{--- (3)}

Step 3: Linear System for $(u_x, v_x)$
From (1) and (3), we have the matrix equation:

\begin{pmatrix} u & v \\ v & -u \end{pmatrix} \begin{pmatrix} u_x \\ v_x \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

The determinant of this coefficient matrix is:

\det \begin{pmatrix} u & v \\ v & -u \end{pmatrix} = -u^2 - v^2 = -(u^2 + v^2) = -c^2 \ne 0

Since the determinant is strictly non-zero, the linear system has only the trivial solution:

u_x = 0, \quad v_x = 0

By the Cauchy-Riemann equations, $v_y = u_x = 0$ and $u_y = -v_x = 0$.
Therefore, the gradient of both $u$ and $v$ vanishes throughout $D$. Since $D$ is connected, $u$ and $v$ are constant, which proves $f(z)$ is constant. $\blacksquare$

Final Answer & Physical Insight

Proved: $f'(z) = 0$ everywhere on connected domain $D \implies f(z)$ is constant.

Tier 3 • Honors Challenge Example 2.3: Proof of Maximum Modulus Principle via Mean Value Property

Prove the Maximum Modulus Principle using the Mean Value Property for holomorphic functions.

Step 1: The Mean Value Property
Let $f(z)$ be holomorphic in a domain $D$. For any $z_0 \in D$ and circle $C_r: z = z_0 + r e^{i\theta}$ lying with its disk in $D$, Cauchy's Integral Formula gives:

f(z_0) = \frac{1}{2\pi i} \oint_{C_r} \frac{f(z)}{z - z_0}\,dz = \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z_0 + r e^{i\theta})}{r e^{i\theta}} (i r e^{i\theta})\,d\theta = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + r e^{i\theta})\,d\theta

Step 2: Triangle Inequality
Taking the modulus:

|f(z_0)| \le \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + r e^{i\theta})|\,d\theta

Step 3: Assume local maximum at interior point
Suppose $|f(z)|$ attains a local maximum at $z_0$, so $|f(z)| \le |f(z_0)|$ for all $z \in D(z_0, \delta)$. Then for any $0 < r < \delta$:

|f(z_0)| \le \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + r e^{i\theta})|\,d\theta \le \frac{1}{2\pi} \int_0^{2\pi} |f(z_0)|\,d\theta = |f(z_0)|

The two outer expressions are equal, forcing the inequality to be an equality:

\frac{1}{2\pi} \int_0^{2\pi} [|f(z_0)| - |f(z_0 + r e^{i\theta})|]\,d\theta = 0

Since the integrand is continuous and non-negative, it must vanish identically:

|f(z_0 + r e^{i\theta})| = |f(z_0)| \quad \text{for all } \theta \in [0, 2\pi]

Step 4: Extension to connected domain
This proves $|f(z)|$ is constant on $D(z_0, \delta)$. By Problem 9, a holomorphic function with constant modulus is constant. By the Identity Theorem, $f(z)$ is constant throughout the entire connected domain $D$. $\blacksquare$

Final Answer & Physical Insight

Proved: Mean value equality forces $|f(z)| \equiv |f(z_0)| \implies f(z)$ is constant.