Complex Functions | ChatWhole Learn

Why It Matters

ℹ️ Why It Matters

Complex functions are fundamentally different from real functions in ways that have profound consequences. A function that is differentiable once in the complex sense is automatically infinitely differentiable and analytic (representable by a power series) — a rigidity with no real-variable counterpart. This extraordinary property makes complex analysis far more powerful than real analysis in many contexts. Analytic functions preserve angles (conformal mappings), enabling the solution of Laplace's equation in complicated geometries used in fluid flow, electrostatics, and heat conduction. The Cauchy-Riemann equations provide a clean test for analyticity, connecting the partial derivatives of the real and imaginary parts. Understanding complex functions is essential for residue calculus, the study of Riemann surfaces, and applications in engineering and physics where frequency-domain analysis is indispensable.

Core Definitions

DfComplex Function

A complex function is a mapping $f: D \to \mathbb{C}$ where $D \subseteq \mathbb{C}$ , assigning to each $z \in D$ a value $w = f(z) \in \mathbb{C}$ . Writing $z = x + iy$ and $f(z) = u(x,y) + iv(x,y)$ , the function has real part $u$ and imaginary part $v$ , both functions of two real variables.

DfLimit of a Complex Function

We say $\lim_{z \to z_0} f(z) = L$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $|f(z) - L| < \varepsilon$ whenever $0 < |z - z_0| < \delta$ .

DfComplex Derivative

The derivative of $f$ at $z_0$ is $f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$ , provided the limit exists and is independent of the direction from which $h \to 0$ .

DfAnalytic (Holomorphic) Function

A function $f$ is analytic (holomorphic) at $z_0$ if $f'(z_0)$ exists in some open neighborhood of $z_0$ . If $f$ is analytic at every point in a domain $D$ , we say $f$ is analytic on $D$ . A function analytic on all of $\mathbb{C}$ is called entire.

DfHarmonic Function

A real-valued function $\phi(x,y)$ with continuous second partial derivatives satisfying Laplace's equation $\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$ is called harmonic. The real and imaginary parts of any analytic function are harmonic.

DfConformal Mapping

A mapping $w = f(z)$ is conformal at $z_0$ if it preserves the angle (both magnitude and orientation) between any two smooth curves passing through $z_0$ . Analytic functions with $f'(z_0) \neq 0$ are conformal at $z_0$ .

DfSingular Point

A point $z_0$ where $f$ is not analytic is called a singular point (or singularity). If $f$ is analytic in a punctured disk $0 < |z - z_0| < R$ but not at $z_0$ , then $z_0$ is an isolated singularity.

DfMeromorphic Function

A function that is analytic in a domain $D$ except for isolated poles (singularities where $f$ behaves like $1/(z-z_0)^n$ ) is called meromorphic on $D$ .

Key Formulas

Cauchy-Riemann Equations

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

Here,

$u(x,y)$ =Real part of f(z) = u + iv
$v(x,y)$ =Imaginary part of f(z) = u + iv

Derivative in Terms of Partial Derivatives

f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}

Here,

$\frac{\partial u}{\partial x}$ =Partial of real part w.r.t. x
$\frac{\partial v}{\partial x}$ =Partial of imaginary part w.r.t. x

Laplace's Equation

\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0

Here,

$\nabla^2$ =Laplacian operator
$\phi(x,y)$ =Harmonic function

Conjugate Harmonic Functions

v(x,y) = \int \frac{\partial u}{\partial y} dx + C \quad \text{or via Cauchy-Riemann}

Here,

$u$ =Known harmonic function (real part)
$v$ =Conjugate harmonic function (imaginary part)

Möbius Transformation

w = f(z) = \frac{az + b}{cz + d}, \quad ad - bc \neq 0

Here,

$a, b, c, d$ =Complex constants with ad - bc ≠ 0
$z$ =Complex variable

Polar Form of Cauchy-Riemann

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

Here,

$r$ =Radial coordinate in polar form
$\theta$ =Angular coordinate in polar form

Complex Exponential

e^z = e^{x+iy} = e^x(\cos y + i\sin y)

Here,

$z = x + iy$ =Complex variable
$e^x$ =Modulus of e^z
$y$ =Argument of e^z

Complex Trigonometric Functions

\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos z = \frac{e^{iz} + e^{-iz}}{2}

Here,

$z$ =Complex variable
$iz$ =Imaginary multiple of z

Important Theorems

ThCauchy-Riemann Equations (Necessary and Sufficient Conditions)

Let $f(z) = u(x,y) + iv(x,y)$ be defined on a domain $D$ .

Necessary condition: If $f'(z_0)$ exists at some point $z_0 = x_0 + iy_0$ , then the partial derivatives of $u$ and $v$ exist at $(x_0, y_0)$ and satisfy:

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

Sufficient condition: If the four partial derivatives of $u$ and $v$ exist, are continuous at $(x_0, y_0)$ , and satisfy the Cauchy-Riemann equations, then $f$ is analytic at $z_0$ .

Proof sketch (necessity): Consider $h \to 0$ along the real axis: $\frac{f(z_0+h) - f(z_0)}{h} = \frac{u(x_0+h,y_0) - u(x_0,y_0)}{h} + i\frac{v(x_0+h,y_0) - v(x_0,y_0)}{h} \to u_x + iv_x$ . Along the imaginary axis ( $h = ik$ ): $\frac{f(z_0+ik) - f(z_0)}{ik} = \frac{u(x_0,y_0+k) - u(x_0,y_0)}{ik} + i\frac{v(x_0,y_0+k) - v(x_0,y_0)}{ik} \to -iu_y + v_y$ . Equating real and imaginary parts gives the Cauchy-Riemann equations.

ThAnalyticity Implies Infinite Smoothness

If $f$ is analytic in a domain $D$ , then $f$ has derivatives of all orders in $D$ , and $f$ is equal to its Taylor series expanded about any point in $D$ . This is fundamentally different from real analysis, where a function can be differentiable once but not twice.

Implication: Analytic functions are extraordinarily rigid. If two analytic functions agree on any set with a limit point in $D$ , they are identical throughout $D$ (Identity Theorem).

ThHarmonic Conjugate Theorem

If $u(x,y)$ is harmonic in a simply connected domain $D$ , then there exists a unique (up to a constant) harmonic function $v(x,y)$ such that $f(z) = u + iv$ is analytic. The function $v$ is called the harmonic conjugate of $u$ .

Construction: Given $u$ , use $v_x = -u_y$ and $v_y = u_x$ , then integrate to find $v$ .

ThConformal Mapping Property

If $f$ is analytic at $z_0$ and $f'(z_0) \neq 0$ , then $f$ is conformal at $z_0$ : it preserves the angle between any two smooth curves through $z_0$ , both in magnitude and orientation.

Geometric meaning: Near $z_0$ , $f$ acts like a rotation (by $\arg f'(z_0)$ ) and a scaling (by $|f'(z_0)|$ ). Small shapes are preserved up to rotation and scaling.

ThClassification of Singularities

Let $f$ have an isolated singularity at $z_0$ . Exactly one of the following holds:

Removable singularity: $f$ can be redefined at $z_0$ to be analytic. The Laurent series has no negative-power terms.
Pole of order $n$ : $f(z) \sim c_{-n}/(z-z_0)^n$ as $z \to z_0$ , with $c_{-n} \neq 0$ . The Laurent series has finitely many negative-power terms.
Essential singularity: The Laurent series has infinitely many negative-power terms. By the Casorati-Weierstrass theorem, $f$ comes arbitrarily close to every complex value in any neighborhood of $z_0$ .

ThMaximum Modulus Principle

If $f$ is analytic and non-constant in a bounded domain $D$ and continuous on $\overline{D}$ , then $|f|$ attains its maximum on the boundary $\partial D$ , never in the interior.

Corollary: If $f$ is analytic on a bounded domain and $|f|$ has a local maximum in the interior, then $f$ is constant.

Worked Examples

📝Example 1: Testing Analyticity with Cauchy-Riemann Equations

Determine whether $f(z) = z^2$ is analytic.

Step 1: Write $z^2 = (x+iy)^2 = x^2 - y^2 + 2xyi$ , so $u = x^2 - y^2$ and $v = 2xy$ .

Step 2: Compute partial derivatives: $u_x = 2x$ , $u_y = -2y$ , $v_x = 2y$ , $v_y = 2x$ .

Step 3: Check Cauchy-Riemann: $u_x = 2x = v_y = 2x$ ✓ and $u_y = -2y = -v_x = -2y$ ✓.

The equations are satisfied everywhere, and the partials are continuous everywhere. Therefore $f(z) = z^2$ is entire (analytic on all of $\mathbb{C}$ ).

Derivative: $f'(z) = u_x + iv_x = 2x + 2yi = 2z$ .

📝Example 2: Non-Analytic Function

Show that $f(z) = \overline{z} = x - iy$ is not analytic anywhere.

Step 1: $u = x$ , $v = -y$ .

Step 2: $u_x = 1$ , $v_y = -1$ . Since $u_x \neq v_y$ ( $1 \neq -1$ ), the Cauchy-Riemann equations fail everywhere.

Conclusion: $\overline{z}$ is not analytic at any point. This function is real-differentiable everywhere but not complex-differentiable.

📝Example 3: Finding the Harmonic Conjugate

Given $u(x,y) = x^2 - y^2$ (which is harmonic since $u_{xx} + u_{yy} = 2 + (-2) = 0$ ), find $v$ such that $f(z) = u + iv$ is analytic.

Step 1: From Cauchy-Riemann: $v_y = u_x = 2x$ and $v_x = -u_y = 2y$ .

Step 2: Integrate $v_y = 2x$ with respect to $y$ : $v = 2xy + g(x)$ for some function $g(x)$ .

Step 3: Differentiate with respect to $x$ : $v_x = 2y + g'(x)$ . Set equal to $2y$ : $g'(x) = 0$ , so $g(x) = C$ .

Result: $v(x,y) = 2xy + C$ , and $f(z) = (x^2 - y^2) + i(2xy + C) = z^2 + iC$ .

📝Example 4: Conformal Mapping — Möbius Transformation

Consider $w = f(z) = \frac{z - i}{z + i}$ . Show it maps the real axis to the unit circle.

Step 1: For $z = x \in \mathbb{R}$ : $|w| = \left|\frac{x - i}{x + i}\right| = \frac{|x - i|}{|x + i|} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} = 1$ .

So the real axis maps to the unit circle $|w| = 1$ .

Step 2: Check specific points: $f(0) = \frac{-i}{i} = -1$ , $f(1) = \frac{1-i}{1+i} = \frac{(1-i)^2}{2} = \frac{-2i}{2} = -i$ , $f(-1) = \frac{-1-i}{-1+i} = \frac{(-1-i)(-1-i)}{(-1+i)(-1-i)} = \frac{1+2i-1}{2} = i$ .

Step 3: Where does $f(z) = 0$ ? When $z = i$ . Where does $f(z) = \infty$ ? When $z = -i$ . The point $i$ maps to the origin; the point $-i$ maps to infinity.

This Möbius transformation maps the upper half-plane to the interior of the unit disk.

📝Example 5: Complex Exponential and Logarithm

Compute $e^{1 + i\pi/2}$ and $\log(-1)$ .

Part (a): $e^{1 + i\pi/2} = e^1 \cdot e^{i\pi/2} = e(\cos(\pi/2) + i\sin(\pi/2)) = e(0 + i) = ei$

Part (b): $\log(z) = \ln|z| + i\arg(z)$ . For $z = -1$ : $|z| = 1$ , $\arg(-1) = \pi + 2k\pi$ .

$\log(-1) = \ln 1 + i(\pi + 2k\pi) = i(2k+1)\pi$ for any integer $k$ .

The principal value is $\operatorname{Log}(-1) = i\pi$ .

📝Example 6: Classifying Singularities

Classify the singularity of $f(z) = \frac{e^z}{z^2}$ at $z = 0$ .

Step 1: Expand $e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$

Step 2: $f(z) = \frac{1}{z^2}\left(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \cdots\right) = \frac{1}{z^2} + \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \cdots$

Step 3: The Laurent series has finitely many negative-power terms (up to $z^{-2}$ ), with $c_{-2} = 1 \neq 0$ .

Conclusion: $z = 0$ is a pole of order 2 (double pole).

Practice Problems

📝Problem 1: Cauchy-Riemann Verification

Verify that $f(z) = e^z$ is entire.

💡Solution

$f(z) = e^{x+iy} = e^x\cos y + ie^x\sin y$ , so $u = e^x\cos y$ and $v = e^x\sin y$ .

$u_x = e^x\cos y$ , $v_y = e^x\cos y$ ✓

$u_y = -e^x\sin y$ , $-v_x = -e^x\sin y$ ✓

The Cauchy-Riemann equations are satisfied everywhere, and the partials are continuous everywhere. Therefore $e^z$ is entire.

$f'(z) = u_x + iv_x = e^x\cos y + ie^x\sin y = e^z$ .

📝Problem 2: Harmonic Conjugate

Find the harmonic conjugate of $u(x,y) = e^x\cos y$ .

💡Solution

Step 1: Check harmonicity: $u_{xx} = e^x\cos y$ , $u_{yy} = -e^x\cos y$ . Sum = 0 ✓.

Step 2: From Cauchy-Riemann: $v_y = u_x = e^x\cos y$ and $v_x = -u_y = e^x\sin y$ .

Step 3: Integrate $v_y = e^x\cos y$ : $v = e^x\sin y + g(x)$ .

Step 4: Differentiate: $v_x = e^x\sin y + g'(x) = e^x\sin y$ implies $g'(x) = 0$ .

Result: $v = e^x\sin y + C$ , and $f(z) = e^x\cos y + ie^x\sin y = e^z$ (consistent with the known form of the complex exponential).

📝Problem 3: Conformal Mapping

Find the image of the line $y = 1$ under the mapping $w = z^2$ .

💡Solution

Step 1: Parameterize: $z = x + i$ , so $w = (x+i)^2 = x^2 - 1 + 2xi$ .

Step 2: Write $w = u + iv$ : $u = x^2 - 1$ and $v = 2x$ .

Step 3: Eliminate $x$ : From $v = 2x$ , we get $x = v/2$ . Substitute into $u$ :

u = (v/2)^2 - 1 = v^2/4 - 1

So $u = v^2/4 - 1$ , or equivalently $v^2 = 4(u + 1)$ .

Result: This is a parabola opening to the right with vertex at $(-1, 0)$ in the $w$ -plane.

📝Problem 4: Singularity Analysis

Find and classify all singularities of $f(z) = \frac{z}{(z-1)(z+2)}$ .

💡Solution

Step 1: The function is undefined at $z = 1$ and $z = -2$ . These are isolated singularities.

Step 2: At $z = 1$ : $\lim_{z \to 1} (z-1)f(z) = \lim_{z \to 1} \frac{z}{z+2} = \frac{1}{3} \neq 0$ . Simple pole (order 1).

Residue: $\operatorname{Res}(f, 1) = \frac{1}{3}$ .

Step 3: At $z = -2$ : $\lim_{z \to -2} (z+2)f(z) = \lim_{z \to -2} \frac{z}{z-1} = \frac{-2}{-3} = \frac{2}{3} \neq 0$ . Simple pole.

Residue: $\operatorname{Res}(f, -2) = \frac{2}{3}$ .

Verification by partial fractions: $\frac{z}{(z-1)(z+2)} = \frac{A}{z-1} + \frac{B}{z+2}$ . Solving: $A = 1/3$ , $B = 2/3$ . ✓

📝Problem 5: Laplace Equation Solution

Solve $\nabla^2 u = 0$ in the upper half-plane with boundary condition $u(x, 0) = e^{-x^2}$ .

💡Solution

Since $u$ is harmonic in the upper half-plane with given boundary data, we can find a conformal mapping to transform this to a simpler domain, or use the Poisson integral formula for the half-plane:

u(x, y) = \frac{y}{\pi}\int_{-\infty}^{\infty} \frac{e^{-t^2}}{(x-t)^2 + y^2} dt

Alternatively, note that $e^{-z^2}$ is entire, so $e^{-z^2} = e^{-(x^2-y^2)}[\cos(2xy) - i\sin(2xy)]$ is analytic. Its real part $u(x,y) = e^{-(x^2-y^2)}\cos(2xy)$ is harmonic.

Check boundary: $u(x, 0) = e^{-x^2}\cos(0) = e^{-x^2}$ ✓

So $u(x,y) = e^{-(x^2-y^2)}\cos(2xy)$ is a solution.

Common Mistakes

Mistake	Correction	Example
Assuming real differentiability implies complex differentiability	The function $\overline{z}$ is real-differentiable but not complex-differentiable	$\overline{z}$ fails Cauchy-Riemann
Forgetting continuity of partials in the sufficiency condition	Cauchy-Riemann + continuity of partials → analyticity; without continuity, C-R may hold but $f'$ may not exist	Need $u_x, u_y, v_x, v_y$ continuous
Confusing $\sin z$ with $\sin(x+iy)$ real trigonometric	$\sin(x+iy) = \sin x\cosh y + i\cos x\sinh y$ — it's unbounded!	$\\|\sin(iy)\\| = \\|\sinh y\\|$ grows without bound
Assuming $e^z$ is one-to-one	$e^z$ is periodic: $e^{z+2\pi i} = e^z$ ; it maps horizontal strips to $\mathbb{C} \setminus \{0\}$	$e^0 = e^{2\pi i} = 1$
Wrong sign in Cauchy-Riemann	It's $u_x = v_y$ and $u_y = -v_x$ , not $u_x = v_x$	Common sign error
Assuming conformality at critical points	Where $f'(z_0) = 0$ , the mapping is NOT conformal (angles may not be preserved)	$w = z^2$ at $z = 0$ : angles double
Forgetting that harmonic conjugates are unique only up to a constant	$v$ and $v + C$ are both conjugates of $u$	Different choices of $C$ give different $f$

Interview / Exam Questions

Q1: What is the relationship between analytic functions and harmonic functions?

A1: If $f(z) = u + iv$ is analytic, then both $u$ and $v$ are harmonic (satisfy Laplace's equation). Conversely, in a simply connected domain, every harmonic function $u$ has a harmonic conjugate $v$ such that $u + iv$ is analytic. This is because the Cauchy-Riemann equations $u_x = v_y$ and $u_y = -v_x$ imply $u_{xx} + u_{yy} = v_{xx} + v_{yy} = 0$ . The connection is deep: harmonic functions are the real or imaginary parts of analytic functions, and conformal mappings transform harmonic functions from one domain to another.

Q2: Why is complex differentiability so much more restrictive than real differentiability?

A2: In the real case, $f'(x) = \lim_{h \to 0} [f(x+h) - f(x)]/h$ only requires agreement from two directions (left and right). In the complex case, $f'(z) = \lim_{h \to 0} [f(z+h) - f(z)]/h$ must give the same limit from every direction in the plane — the real axis, imaginary axis, and all other approaches. This is a much stronger condition, captured precisely by the Cauchy-Riemann equations. The result is that complex differentiability implies infinite smoothness and representability by power series, which never happens in real analysis.

Q3: Describe the geometric effect of the mapping $w = z^2$ near $z_0 = 1$ .

A3: At $z_0 = 1$ , $f'(1) = 2$ . The mapping scales by $|f'(1)| = 2$ and rotates by $\arg f'(1) = 0$ . Near $z_0 = 1$ , $z^2$ acts like a linear map that doubles distances and preserves angles. Since $f'(1) \neq 0$ , the mapping is conformal at $z_0 = 1$ : angles between curves are preserved.

At $z_0 = 0$ , $f'(0) = 0$ , so the mapping is not conformal. The angle between two curves through the origin is doubled.

Q4: What are the three types of isolated singularities, and how do you distinguish them?

A4: An isolated singularity $z_0$ of $f$ is:

Removable if $f$ can be redefined at $z_0$ to be analytic. The Laurent series has no negative powers, and $\lim_{z \to z_0} (z-z_0)f(z) = 0$ .
A pole of order $n$ if $f(z) \sim c/(z-z_0)^n$ as $z \to z_0$ . The Laurent series has finitely many negative powers, and $\lim_{z \to z_0} (z-z_0)^n f(z) \neq 0$ .
Essential if the Laurent series has infinitely many negative powers. Equivalently, $\lim_{z \to z_0} (z-z_0)^n f(z) = \infty$ for all $n$ .

Q5: If $f$ is analytic and $|f|$ has a local maximum at an interior point, what can you conclude?

A5: By the Maximum Modulus Principle, $f$ must be constant on the domain. This is a powerful rigidity result: analytic functions cannot have interior local maxima of their modulus (unless they are constant). This principle is used to prove uniqueness theorems, bound analytic functions, and establish that polynomials map circles to curves that enclose the same area. It also implies that the maximum of $|f|$ on a closed bounded domain is always achieved on the boundary.

Q6: Construct the harmonic conjugate of $u = x^3 - 3xy^2$ and form the analytic function $f(z)$ .

A6: Check: $u_{xx} = 6x$ , $u_{yy} = -6x$ , so $u_{xx} + u_{yy} = 0$ ✓ (harmonic).

From C-R: $v_y = u_x = 3x^2 - 3y^2$ and $v_x = -u_y = 6xy$ .

Integrate $v_x = 6xy$ : $v = 3x^2y + h(y)$ . Differentiate: $v_y = 3x^2 + h'(y) = 3x^2 - 3y^2$ , so $h'(y) = -3y^2$ , $h(y) = -y^3 + C$ .

$v = 3x^2y - y^3 + C$ .

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

Quick Reference

📋Formula Summary

Formula	Expression	Notes
Cauchy-Riemann	$u_x = v_y$ , $u_y = -v_x$	Necessary + sufficient (with continuity)
Derivative	$f' = u_x + iv_x = v_y - iu_y$	Either form works
Complex exponential	$e^z = e^x(\cos y + i\sin y)$	Periodic in imaginary direction
Complex sine	$\sin z = \sin x\cosh y + i\cos x\sinh y$	Unbounded in complex plane
Complex cosine	$\cos z = \cos x\cosh y - i\sin x\sinh y$	Unbounded in complex plane
Complex logarithm	$\log z = \ln\\|z\\| + i\arg z$	Multi-valued; branch cuts needed
Laplace equation	$u_{xx} + u_{yy} = 0$	Real/imaginary parts of analytic $f$
Harmonic conjugate	$v_y = u_x$ , $v_x = -u_y$	Integrate to find $v$ from $u$
Möbius transform	$w = (az+b)/(cz+d)$	Maps circles to circles
Conformality condition	$f$ analytic, $f'(z_0) \neq 0$	Preserves angles at $z_0$
Maximum modulus	$\\|f\\|$ max on boundary	Non-constant analytic functions
Laurent series	$f(z) = \sum_{n=-\infty}^{\infty} c_n(z-z_0)^n$	Reveals singularity type

Cross-References

091 - Complex Numbers — The algebraic foundations (modulus, argument, polar form) underpin all function theory.
093 - Contour Integration — Cauchy's theorem and integral formula are consequences of analyticity; they are the computational engine of complex analysis.
094 - Residue Theory — Poles and essential singularities (classified here) are the objects whose residues are computed.
095 - Applications — Fourier transforms, filter design, and conformal mappings to physical domains rely on the properties of analytic functions.
Partial Differential Equations — Harmonic functions arise as steady-state solutions; conformal mappings transform Laplace's equation between domains.
Fluid Dynamics (Topic 25): Velocity potentials and stream functions are harmonic conjugates; conformal mappings solve flow problems around obstacles.