14. Complex Analysis (복소해석)¶

Learning Objectives¶

Understand the differentiability of complex functions and the Cauchy-Riemann conditions
Calculate complex integrals using Cauchy's integral theorem and integral formulas
Expand complex functions into series through Taylor and Laurent series
Efficiently solve real integral problems using the residue theorem
Understand the concept of conformal mapping and apply it to physics problems

Importance in Physics: Complex analysis is a core tool across all of physics, including wave functions in quantum mechanics, potential theory in electrodynamics, stream functions in fluid mechanics, and frequency analysis in signal processing. In particular, integral calculations via the residue theorem are among the most frequently used techniques in theoretical physics.

1. Analytic Functions (해석함수)¶

1.1 Complex Differentiation and the Cauchy-Riemann Conditions¶

For a complex function $f(z) = u(x, y) + iv(x, y)$ to be differentiable at a point $z_0$, the limit

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

must converge to the same value regardless of the direction from which $\Delta z$ approaches. This leads to the Cauchy-Riemann equations:

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

A function that is differentiable at every point in a region is called analytic in that region.

Polar form ($z = re^{i\theta}$):

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

import numpy as np
import matplotlib.pyplot as plt

def check_cauchy_riemann(u_func, v_func, x, y, h=1e-7):
    """코시-리만 조건을 수치적으로 검증"""
    du_dx = (u_func(x + h, y) - u_func(x - h, y)) / (2 * h)
    du_dy = (u_func(x, y + h) - u_func(x, y - h)) / (2 * h)
    dv_dx = (v_func(x + h, y) - v_func(x - h, y)) / (2 * h)
    dv_dy = (v_func(x, y + h) - v_func(x, y - h)) / (2 * h)

    cond1 = np.abs(du_dx - dv_dy)   # ∂u/∂x = ∂v/∂y
    cond2 = np.abs(du_dy + dv_dx)   # ∂u/∂y = -∂v/∂x

    print(f"점 ({x}, {y}): |∂u/∂x - ∂v/∂y| = {cond1:.2e}, "
          f"|∂u/∂y + ∂v/∂x| = {cond2:.2e}")
    return cond1 < 1e-5 and cond2 < 1e-5

# f(z) = z² = (x² - y²) + i(2xy)
u = lambda x, y: x**2 - y**2
v = lambda x, y: 2 * x * y

print("=== f(z) = z² 코시-리만 검증 ===")
for pt in [(1, 1), (2, -1), (0.5, 3)]:
    analytic = check_cauchy_riemann(u, v, *pt)
    print(f"  해석적: {analytic}")

1.2 Harmonic Functions (조화함수)¶

The real part $u$ and imaginary part $v$ of an analytic function each satisfy Laplace's equation:

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

Such functions are called harmonic functions, and $u$ and $v$ are harmonic conjugates of each other.

Physical meaning: In two-dimensional electrostatics, the potential $\phi(x,y)$ satisfies Laplace's equation and is therefore a harmonic function. The real part of an analytic function can be interpreted as the potential and the imaginary part as the electric field line function.

import sympy as sp

x, y = sp.symbols('x y', real=True)

def is_harmonic(expr):
    """라플라시안이 0인지 확인"""
    lap = sp.simplify(sp.diff(expr, x, 2) + sp.diff(expr, y, 2))
    return lap, lap == 0

# f(z) = z³의 실수부/허수부
u_expr = x**3 - 3*x*y**2      # Re(z³)
v_expr = 3*x**2*y - y**3      # Im(z³)

for name, expr in [("Re(z³)", u_expr), ("Im(z³)", v_expr)]:
    lap, harmonic = is_harmonic(expr)
    print(f"{name} = {expr}: ∇² = {lap}, 조화함수: {harmonic}")

1.3 Examples of Analytic Functions¶

Analytic functions frequently appearing in physics:

Function	Real part $u$	Imaginary part $v$	Physical application
$e^z$	$e^x \cos y$	$e^x \sin y$	Waves, damping
$\ln z$	$\ln r$	$\theta$	Line charge potential
$z^n$	$r^n \cos n\theta$	$r^n \sin n\theta$	Multipole expansion
$1/z$	$x/(x^2+y^2)$	$-y/(x^2+y^2)$	Point charge, source/sink

from matplotlib.colors import hsv_to_rgb

def domain_coloring(f, xlim=(-2, 2), ylim=(-2, 2), N=500):
    """복소함수의 도메인 컬러링 시각화 (위상→색상, 크기→명도)"""
    xv = np.linspace(*xlim, N)
    yv = np.linspace(*ylim, N)
    X, Y = np.meshgrid(xv, yv)
    Z = X + 1j * Y

    with np.errstate(divide='ignore', invalid='ignore'):
        W = f(Z)

    H = (np.angle(W) + np.pi) / (2 * np.pi)
    V = 1 - 1 / (1 + np.abs(W)**0.3)
    HSV = np.stack([H, np.ones_like(H), V], axis=-1)

    fig, ax = plt.subplots(figsize=(6, 6))
    ax.imshow(hsv_to_rgb(HSV), extent=[*xlim, *ylim], origin='lower')
    ax.set_xlabel('Re(z)'); ax.set_ylabel('Im(z)')
    plt.tight_layout(); plt.show()

# domain_coloring(lambda z: z**3)       # z³의 도메인 컬러링
# domain_coloring(lambda z: np.exp(z))  # e^z의 도메인 컬러링

2. Complex Integrals (복소 적분)¶

2.1 Contour Integrals (경로 적분)¶

Integrating a complex function along a path $C$ is called a contour integral:

$$\oint_C f(z)\, dz = \int_a^b f(z(t))\, z'(t)\, dt$$

Key result: For counterclockwise integration over $|z - z_0| = r$:

$$\oint \frac{dz}{(z - z_0)^n} = \begin{cases} 2\pi i & n = 1 \\ 0 & n \neq 1 \end{cases}$$

def contour_integral_circle(f, z0=0, r=1, N=10000):
    """원형 경로 |z-z0|=r 위에서의 경로 적분 (수치 계산)"""
    t = np.linspace(0, 2*np.pi, N, endpoint=False)
    dt = 2*np.pi / N
    z = z0 + r * np.exp(1j * t)
    dz_dt = 1j * r * np.exp(1j * t)
    return np.sum(f(z) * dz_dt) * dt

# ∮ 1/z dz = 2πi
I1 = contour_integral_circle(lambda z: 1/z)
print(f"∮ 1/z dz = {I1:.6f}  (이론값: {2*np.pi*1j:.6f})")

# ∮ 1/z² dz = 0
I2 = contour_integral_circle(lambda z: 1/z**2)
print(f"∮ 1/z² dz = {I2:.6f}  (이론값: 0)")

# ∮ e^z/z dz = 2πi (코시 공식: f(0)=e⁰=1)
I3 = contour_integral_circle(lambda z: np.exp(z)/z)
print(f"∮ e^z/z dz = {I3:.6f}  (이론값: {2*np.pi*1j:.6f})")

2.2 Cauchy's Integral Theorem (코시 적분 정리)¶

If $f(z)$ is analytic in a simply connected region $D$, then for any closed curve $C$ in $D$:

$$\oint_C f(z)\, dz = 0$$

Physical meaning: This is equivalent to the circulation of a conservative force being zero. The integral of an analytic function is path-independent.

2.3 Cauchy's Integral Formula (코시 적분 공식)¶

If $f(z)$ is analytic inside $C$ and $z_0$ is an interior point:

$$f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0}\, dz$$

Generalization (nth derivative):

$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}}\, dz$$

This formula means that analytic functions are infinitely differentiable, and boundary values determine interior values.

from math import factorial

def cauchy_derivative(f, z0, n=0, r=1, N=10000):
    """코시 적분 공식으로 f^(n)(z0) 계산"""
    t = np.linspace(0, 2*np.pi, N, endpoint=False)
    dt = 2*np.pi / N
    z = z0 + r * np.exp(1j * t)
    dz_dt = 1j * r * np.exp(1j * t)
    integrand = f(z) / (z - z0)**(n + 1) * dz_dt
    return np.sum(integrand) * dt * factorial(n) / (2*np.pi*1j)

# 검증: f(z) = sin(z)
f = lambda z: np.sin(z)
print("코시 공식 검증: f(z) = sin(z)")
print(f"  f(0)   = {cauchy_derivative(f, 0, 0):.6f}  (정확값: 0)")
print(f"  f'(0)  = {cauchy_derivative(f, 0, 1):.6f}  (정확값: 1)")
print(f"  f''(0) = {cauchy_derivative(f, 0, 2):.6f}  (정확값: 0)")
print(f"  f'''(0)= {cauchy_derivative(f, 0, 3):.6f}  (정확값: -1)")

3. Series Expansions (급수 전개)¶

3.1 Taylor Series¶

If $f(z)$ is analytic inside a circle of radius $R$ centered at $z_0$:

$$f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n, \quad a_n = \frac{f^{(n)}(z_0)}{n!}$$

The radius of convergence $R$ is the distance from $z_0$ to the nearest singularity.

Example: The radius of convergence of the Taylor series of $f(z) = 1/(1+z^2)$ around $z=0$ is $R=1$. This is because there are poles at $z = \pm i$. Although there are no singularities on the real axis, singularities in the complex plane determine the radius of convergence.

3.2 Laurent Series (로랑 급수)¶

If $f(z)$ is analytic in an annular region $r < |z - z_0| < R$:

$$f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n$$

Terms with $n < 0$ are called the principal part, and specifically $a_{-1}$ is called the residue.

z = sp.Symbol('z')

# 로랑 급수: e^z / z³ (z=0 주위)
f1 = sp.exp(z) / z**3
print("e^z/z³ 의 로랑 급수:")
print(f"  {sp.series(f1, z, 0, n=5)}")
print(f"  유수 (z⁻¹ 계수) = 1/2\n")

# 로랑 급수: 1/(z(z-1)) (z=0 주위)
f2 = 1 / (z * (z - 1))
print("1/(z(z-1)) 의 로랑 급수 (z=0 주위):")
print(f"  {sp.series(f2, z, 0, n=4)}")

# z=1 주위
w = sp.Symbol('w')
print("\n1/(z(z-1)) 의 로랑 급수 (z=1 주위, w=z-1):")
print(f"  {sp.series(f2.subs(z, w+1), w, 0, n=4)}")

3.3 Classification of Singularities (Removable, Pole, Essential)¶

Type	Number of principal part terms	Example	$\lim_{z \to z_0} f(z)$
Removable singularity	0	$\sin z / z$ at $z=0$	Finite value
Pole of order $m$	$m$	$1/z^m$ at $z=0$	$\infty$
Essential singularity	Infinitely many	$e^{1/z}$ at $z=0$	Does not exist

Casorati-Weierstrass theorem: Near an essential singularity, the function takes on almost every complex value.

# 특이점 분류 확인
cases = [
    ("sin(z)/z (z=0, 제거가능)", sp.sin(z)/z, z, 0),
    ("1/(z-1)³ (z=1, 3차 극점)", 1/(z-1)**3, z, 1),
    ("exp(1/z) (z=0, 본질적)", sp.exp(1/z), z, 0),
]

for name, expr, var, pt in cases:
    if pt != 0:
        w = sp.Symbol('w')
        s = sp.series(expr.subs(var, w + pt), w, 0, n=5)
    else:
        s = sp.series(expr, var, 0, n=5)
    print(f"{name}:\n  {s}\n")

4. Residue Theorem (유수 정리)¶

4.1 Definition and Calculation of Residues¶

The residue of $f(z)$ at the point $z_0$ is the $a_{-1}$ coefficient of the Laurent series:

$$\text{Res}_{z=z_0} f(z) = \frac{1}{2\pi i} \oint_C f(z)\, dz$$

Methods for calculating residues:

Simple pole: $\text{Res}_{z=z_0} f = \lim_{z \to z_0} (z - z_0) f(z)$
Pole of order $m$: $\text{Res}_{z=z_0} f = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z - z_0)^m f(z)]$
Form $p/q$ (simple pole): $\text{Res}_{z=z_0} \frac{p}{q} = \frac{p(z_0)}{q'(z_0)}$

z = sp.Symbol('z')

examples = [
    ("1/(z²+1)", 1/(z**2+1), sp.I),
    ("1/(z²+1)", 1/(z**2+1), -sp.I),
    ("e^z/z²", sp.exp(z)/z**2, 0),
    ("z/(z²-3z+2)", z/(z**2-3*z+2), 1),
    ("z/(z²-3z+2)", z/(z**2-3*z+2), 2),
]

print("=== 유수 계산 ===")
for name, expr, z0 in examples:
    print(f"Res[{name}, z={z0}] = {sp.residue(expr, z, z0)}")

4.2 Residue Theorem¶

If $f(z)$ is analytic inside a closed curve $C$ except for a finite number of singularities $z_1, \ldots, z_n$:

$$\oint_C f(z)\, dz = 2\pi i \sum_{k=1}^{n} \text{Res}_{z=z_k} f(z)$$

4.3 Jordan's Lemma (조르당 보조정리)¶

When calculating real integrals using the residue theorem, we must verify that the contribution from the infinite semicircular path is zero. Jordan's lemma guarantees this.

Theorem: If $f(z) \to 0$ uniformly as $|z| \to \infty$ (upper half-plane), then for $a > 0$:

$$\lim_{R \to \infty} \int_{C_R} f(z) e^{iaz}\, dz = 0$$

where $C_R$ is a semicircle of radius $R$ in the upper half-plane. The key is that $e^{iaz} = e^{ia(x+iy)} = e^{iax}e^{-ay}$, so in the upper half-plane ($y > 0$), $e^{-ay}$ decays exponentially.

Note: If $a < 0$, use the lower half-plane semicircle, and in this case the path direction is clockwise so $(-2\pi i)$ is multiplied to the residues.

4.4 Four Types of Real Integral Calculations¶

The most important application of the residue theorem is the calculation of real integrals. We systematically classify them according to the form of the integral.

Type 1: Rational functions of trigonometric functions — $\int_0^{2\pi} R(\cos\theta, \sin\theta)\, d\theta$¶

Substitution $z = e^{i\theta}$: $\cos\theta = (z + z^{-1})/2$, $\sin\theta = (z - z^{-1})/(2i)$, $d\theta = dz/(iz)$.

The integral transforms into a contour integral over the unit circle $|z| = 1$. Calculate only the residues at poles inside the unit circle.

from scipy.integrate import quad
import sympy as sp
import numpy as np

z = sp.Symbol('z')

# --- 유형 1: ∫₀²π dθ/(2 + cosθ) ---
print("=== 유형 1: ∫₀²π dθ/(2 + cosθ) ===")
integrand1 = 2 / (sp.I * (z**2 + 4*z + 1))
inner_pole = -2 + sp.sqrt(3)  # |z| < 1인 극점
res1 = sp.residue(integrand1, z, inner_pole)
result1 = sp.simplify(2 * sp.pi * sp.I * res1)
print(f"유수 정리: {result1} = {float(result1):.6f}")
num1, _ = quad(lambda t: 1/(2 + np.cos(t)), 0, 2*np.pi)
print(f"수치 검증: {num1:.6f}\n")

Type 2: Rational functions — $\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)}\, dx$¶

Condition: $\deg(Q) \geq \deg(P) + 2$ (integral converges), $Q(x) \neq 0$ on the real axis.

Use an upper half-plane semicircular path. As $R \to \infty$, the integral over the semicircle becomes 0 (since $f(z) \to 0$ sufficiently fast):

$$\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)}\, dx = 2\pi i \sum_{\text{Im}(z_k) > 0} \text{Res}_{z=z_k} \frac{P(z)}{Q(z)}$$

# --- 유형 2: ∫₋∞^∞ dx/(x²+1)² ---
print("=== 유형 2: ∫₋∞^∞ dx/(x²+1)² ===")
res2 = sp.residue(1/(z**2+1)**2, z, sp.I)
result2 = sp.simplify(2 * sp.pi * sp.I * res2)
print(f"유수 정리: {result2} = {float(result2):.6f}")
num2, _ = quad(lambda x: 1/(x**2+1)**2, -100, 100)
print(f"수치 검증: {num2:.6f}\n")

Type 3: Fourier-type integrals — $\int_{-\infty}^{\infty} f(x) e^{iax}\, dx$ ($a > 0$)¶

Apply Jordan's lemma. If $f(z) \to 0$ as $|z| \to \infty$ (first-order is sufficient), the contribution from the upper half-plane semicircle is 0:

$$\int_{-\infty}^{\infty} f(x) e^{iax}\, dx = 2\pi i \sum_{\text{Im}(z_k) > 0} \text{Res}_{z=z_k} f(z) e^{iaz}$$

Integrals containing $\cos(ax)$ or $\sin(ax)$ use $e^{iax}$ and then take the real or imaginary part.

# --- 유형 3: ∫₋∞^∞ cos(x)/(x²+1) dx = π/e ---
print("=== 유형 3: ∫₋∞^∞ cos(x)/(x²+1) dx ===")
f3 = sp.exp(sp.I*z) / (z**2 + 1)
res3 = sp.residue(f3, z, sp.I)
result3 = sp.simplify(2 * sp.pi * sp.I * res3)
print(f"유수 정리: Re({result3}) = π/e = {float(sp.pi/sp.E):.6f}")
num3, _ = quad(lambda x: np.cos(x)/(x**2+1), -100, 100)
print(f"수치 검증: {num3:.6f}\n")

Type 4: Branch cut integrals — $\int_0^{\infty} x^{a-1} f(x)\, dx$ ($0 < a < 1$)¶

If the integrand contains $x^a$ (where $a$ is not an integer), a branch cut is needed. Use a keyhole contour.

Representative example: $\int_0^{\infty} \frac{x^{a-1}}{1+x}\, dx = \frac{\pi}{\sin(\pi a)}$ ($0 < a < 1$)

Solution strategy: 1. Choose the positive real axis as the branch cut for $f(z) = z^{a-1}/(1+z)$ 2. Keyhole contour: above branch cut → large circle → below branch cut → small circle 3. Below the branch cut, $z^{a-1} = |z|^{a-1} e^{2\pi i(a-1)}$ 4. Residue at $z = -1$: $e^{i\pi(a-1)} = -e^{i\pi a}$

$$\int_0^{\infty} \frac{x^{a-1}}{1+x} dx - e^{2\pi i(a-1)} \int_0^{\infty} \frac{x^{a-1}}{1+x} dx = 2\pi i \cdot (-e^{i\pi a})$$

$$(1 - e^{2\pi i(a-1)}) I = -2\pi i e^{i\pi a} \implies I = \frac{\pi}{\sin(\pi a)}$$

# --- 유형 4: ∫₀^∞ x^{a-1}/(1+x) dx = π/sin(πa) ---
print("=== 유형 4: ∫₀^∞ x^{a-1}/(1+x) dx ===")
for a in [0.25, 0.5, 0.75]:
    theory = np.pi / np.sin(np.pi * a)
    numerical, _ = quad(lambda x: x**(a-1)/(1+x), 0, np.inf)
    print(f"  a = {a}: π/sin(πa) = {theory:.6f}, 수치 = {numerical:.6f}")

5. Conformal Mapping (등각사상)¶

5.1 Definition and Properties of Conformal Mapping¶

An analytic function $w = f(z)$ with $f'(z_0) \neq 0$ is conformal in the neighborhood of $z_0$:

Angle preservation: The angle of intersection of two curves is preserved after mapping
Laplace equation invariant: Harmonic functions remain harmonic after mapping
Boundary condition preservation: Physical boundary conditions remain valid after mapping

5.2 Möbius Transformations¶

$$w = \frac{az + b}{cz + d}, \quad ad - bc \neq 0$$

Properties: Maps circles and lines to circles and lines, uniquely determined by three points.

def mobius_transform(z, a, b, c, d):
    """뫼비우스 변환 w = (az+b)/(cz+d)"""
    return (a*z + b) / (c*z + d)

def plot_mobius(a, b, c, d, title="뫼비우스 변환"):
    """격자선의 뫼비우스 변환 시각화"""
    fig, axes = plt.subplots(1, 2, figsize=(12, 5))
    t = np.linspace(-2, 2, 300)

    for x0 in np.linspace(-2, 2, 9):
        zv = x0 + 1j*t; wv = mobius_transform(zv, a, b, c, d)
        axes[0].plot(zv.real, zv.imag, 'b-', alpha=0.4, lw=0.7)
        m = np.abs(wv) < 10
        axes[1].plot(wv.real[m], wv.imag[m], 'b-', alpha=0.4, lw=0.7)

    for y0 in np.linspace(-2, 2, 9):
        zv = t + 1j*y0; wv = mobius_transform(zv, a, b, c, d)
        axes[0].plot(zv.real, zv.imag, 'r-', alpha=0.4, lw=0.7)
        m = np.abs(wv) < 10
        axes[1].plot(wv.real[m], wv.imag[m], 'r-', alpha=0.4, lw=0.7)

    for ax, lab in zip(axes, ['z-평면', 'w-평면']):
        ax.set_xlim(-4,4); ax.set_ylim(-4,4)
        ax.set_aspect('equal'); ax.grid(True, alpha=0.3); ax.set_title(lab)
    fig.suptitle(f'{title}: w=({a}z+{b})/({c}z+{d})')
    plt.tight_layout(); plt.show()

# 케일리 변환 (상반면→단위원): w = (z-i)/(z+i)
# plot_mobius(1, -1j, 1, 1j, "케일리 변환")

5.3 Physics Applications (Fluid Mechanics, Electric Field)¶

Complex potential: In two-dimensional incompressible irrotational flow:

$$W(z) = \phi(x, y) + i\psi(x, y), \quad \frac{dW}{dz} = v_x - iv_y$$

$\phi$ is the velocity potential, $\psi$ is the streamline function.

Flow	$W(z)$	Physical meaning
Uniform flow	$Uz$	Velocity $U$
Source/sink	$(Q/2\pi)\ln z$	Strength $Q$
Vortex	$(-i\Gamma/2\pi)\ln z$	Circulation $\Gamma$
Doublet	$\mu/z$	Dipole
Around cylinder	$U(z + a^2/z)$	Radius $a$

def plot_flow(W_func, xlim=(-3,3), ylim=(-3,3), N=400, title="유선도"):
    """복소 포텐셜로부터 유선도와 등포텐셜선 시각화"""
    xv = np.linspace(*xlim, N); yv = np.linspace(*ylim, N)
    X, Y = np.meshgrid(xv, yv); Z = X + 1j*Y

    with np.errstate(divide='ignore', invalid='ignore'):
        W = W_func(Z)
    phi, psi = np.real(W), np.imag(W)
    mask = np.abs(W) > 50; phi[mask] = np.nan; psi[mask] = np.nan

    fig, axes = plt.subplots(1, 2, figsize=(12, 5))
    axes[0].contour(X, Y, psi, levels=30, colors='blue', linewidths=0.8)
    axes[0].set_title(f'{title} - 유선 (ψ=const)')
    axes[1].contour(X, Y, phi, levels=30, colors='red', linewidths=0.8)
    axes[1].set_title(f'{title} - 등포텐셜선 (φ=const)')
    for ax in axes:
        ax.set_aspect('equal'); ax.set_xlabel('x'); ax.set_ylabel('y')
    plt.tight_layout(); plt.show()

# 순환 있는 원주 주위 흐름
U, a, Gamma = 1.0, 1.0, 2*np.pi
W_cyl = lambda z: U*(z + a**2/z) - 1j*Gamma/(2*np.pi)*np.log(z)
# plot_flow(W_cyl, title="순환 있는 원주 주위 흐름")

# 전기 쌍극자 (+q at z=d, -q at z=-d)
W_dipole = lambda z: -1/(2*np.pi) * (np.log(z-0.5) - np.log(z+0.5))
# plot_flow(W_dipole, title="전기 쌍극자 전위")

Joukowski transformation and lift: $w = z + c^2/z$ maps a circle to an airfoil. Lift per unit length by the Kutta-Joukowski theorem:

$$L = \rho U \Gamma$$

This result is elegantly derived from the residue theorem.

5.4 Schwarz-Christoffel Mapping (슈바르츠-크리스토펠 사상)¶

Problem: Find a conformal mapping that maps the upper half-plane to a polygonal region.

Schwarz-Christoffel formula: If points $x_1, x_2, \ldots, x_n$ on the real axis of the upper half-plane map to vertices $w_1, w_2, \ldots, w_n$ of a polygon, and the interior angle at each vertex is $\alpha_k \pi$:

$$\frac{dw}{dz} = A \prod_{k=1}^{n} (z - x_k)^{\alpha_k - 1}$$

where $A$ is a complex constant.

Example: The mapping from the upper half-plane to a rectangle is expressed in terms of elliptic integrals and is used in electrostatics to calculate the fringing field of parallel plate capacitors.

6. Analytic Continuation (해석적 연속)¶

6.1 Basic Concept¶

When a function $f(z)$ is defined in a region $D_1$, if an analytic function $g(z)$ in a larger region $D_2 \supset D_1$ agrees with $f$ in $D_1$, then $g$ is called an analytic continuation of $f$.

Uniqueness: If an analytic continuation exists, it is unique (by the identity theorem).

6.2 Physics Applications¶

Gamma function: Originally $\Gamma(z) = \int_0^{\infty} t^{z-1} e^{-t} dt$ is only defined for $\text{Re}(z) > 0$, but using the recurrence relation $\Gamma(z) = \Gamma(z+1)/z$, it can be analytically continued to the entire complex plane except for negative integers.

Riemann zeta function: $\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$ converges for $\text{Re}(s) > 1$, but through analytic continuation it extends to the entire complex plane except $s = 1$. This is used in physics for zeta function regularization (Casimir effect, etc.).

# 감마 함수의 해석적 연속 시각화
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import gamma

x = np.linspace(-4.5, 5, 2000)
y = np.array([gamma(xi) if abs(xi - round(xi)) > 0.02 or xi > 0.5
              else np.nan for xi in x])

plt.figure(figsize=(10, 6))
plt.plot(x, y, 'b-', linewidth=1.5)
plt.ylim(-10, 10)
for n in range(0, -5, -1):
    plt.axvline(x=n, color='red', linewidth=0.5, linestyle='--', alpha=0.5)

plt.xlabel('z')
plt.ylabel('Γ(z)')
plt.title('감마 함수의 해석적 연속 (음의 정수에서 극점)')
plt.grid(True, alpha=0.3)
plt.axhline(0, color='k', linewidth=0.5)
plt.tight_layout()
plt.show()

Exercises (연습 문제)¶

Basic Problems¶

Determine whether the following functions are analytic using the Cauchy-Riemann conditions:
(a) $f(z) = z^3$
(b) $f(z) = |z|^2$
(c) $f(z) = \bar{z}$
(d) $f(z) = e^{-z}\sin z$
Find the harmonic conjugate $v(x, y)$ of $u(x, y) = x^3 - 3xy^2 + 2x$.
Calculate the following integrals ($C$: circle of radius 2 centered at origin):
(a) $\oint_C e^z/z^2\, dz$
(b) $\oint_C \cos z/z^3\, dz$
(c) $\oint_C z^2/((z-1)(z+2))\, dz$

Intermediate Problems¶

Calculate using the residue theorem:
(a) $\displaystyle\int_0^{2\pi} \frac{d\theta}{5 + 4\cos\theta}$
(b) $\displaystyle\int_0^{\infty} \frac{x^2}{(x^2+1)(x^2+4)}\, dx$
(c) $\displaystyle\int_0^{\infty} \frac{\cos 3x}{x^2 + 1}\, dx$
Find the residues of $f(z) = z/((z-1)^2(z+2))$ at all singularities, and find the integral value over $|z|=3$.

Advanced Problems¶

Fluid mechanics: When there is a source at $z=a$ and a sink at $z=-a$, and the $x$-axis is a wall, find the streamline function in the upper half-plane using the method of images.
Quantum mechanics: Derive the density of states $\rho(E) = -\text{Im}\, G(E)/\pi$ from the Green's function $G(E) = (E - H + i\epsilon)^{-1}$ using the residue theorem.
Solution (Problem 4a):

z = sp.Symbol('z')
# ∫₀²π dθ/(5+4cosθ) → ∮ 1/(i(2z²+5z+2)) dz
integrand = 1 / (sp.I * (2*z**2 + 5*z + 2))
poles = sp.solve(2*z**2 + 5*z + 2, z)
print(f"극점: {poles}")  # z=-1/2 (내부), z=-2 (외부)
res = sp.residue(integrand, z, sp.Rational(-1, 2))
print(f"적분값: {sp.simplify(2*sp.pi*sp.I*res)}")  # 2π/3

References (참고 자료)¶

Textbooks¶

Boas, M. L. Mathematical Methods in the Physical Sciences, 3rd ed., Ch. 14
Arfken, Weber Mathematical Methods for Physicists, Ch. 6-7
Churchill, Brown Complex Variables and Applications

Supplementary Materials¶

Needham, T. Visual Complex Analysis - Geometric intuition
Ablowitz, Fokas Complex Variables: Introduction and Applications - Physics applications

Summary of Key Formulas¶

Formula	Condition
Cauchy-Riemann: $u_x = v_y$, $u_y = -v_x$	Differentiable
Cauchy's theorem: $\oint_C f\, dz = 0$	Analytic in simply connected region
Cauchy's formula: $f(z_0) = \frac{1}{2\pi i}\oint \frac{f}{z-z_0} dz$	$f$ analytic, $z_0$ interior
Residue theorem: $\oint f\, dz = 2\pi i \sum \text{Res}$	Finite singularities

Next Lesson¶

In 15. Laplace Transform (라플라스 변환), we will cover the definition and properties of the Laplace transform, inverse transforms, and applications to differential equations.