Complex Variables

Analytic functions

A function f(z) = u(x, y) + iv(x, y) of z = x + iy is analytic in a region if it has a derivative at every point there. A necessary condition is the Cauchy-Riemann equations:

u_x = v_y and u_y = −v_x

If the partial derivatives are continuous and satisfy these equations, f is analytic, and f′(z) = u_x + iv_x.

• Polynomials, eᶻ, sin z and cos z are analytic everywhere (entire). 1/z is analytic except at z = 0.
• z̄ (the conjugate) and |z|² are not analytic anywhere except possibly at single points: for z̄, u_x = 1 but v_y = −1.
• The real and imaginary parts of an analytic function are harmonic: u_xx + u_yy = 0. Given u, its harmonic conjugate v follows from the Cauchy-Riemann equations.

Example. Is u = x² − y² the real part of an analytic function? u_xx + u_yy = 2 − 2 = 0, so yes: v = 2xy, and f(z) = z².

Complex integration

Cauchy's integral theorem. If f is analytic inside and on a simple closed curve C, then ∮_C f(z) dz = 0.

Cauchy's integral formula. If f is analytic inside and on C, and a is inside C:

∮_C f(z)/(z − a) dz = 2πi f(a), and ∮_C f(z)/(z − a)ⁿ⁺¹ dz = (2πi/n!) f⁽ⁿ⁾(a).

A basic result: ∮ dz/(z − a) = 2πi around a, and ∮ (z − a)ⁿ dz = 0 for every other integer n.

Series and singularities

• Taylor series about a point where f is analytic: f(z) = Σ f⁽ⁿ⁾(a)(z − a)ⁿ/n!.
• Laurent series in an annulus around a singularity: f(z) = Σ aₙ(z − a)ⁿ over all integers n. The negative powers form the principal part.
• A singularity is removable if the principal part is empty (sin z/z at 0), a pole of order m if it ends at (z − a)^−m, and essential if it is infinite (e^1/z at 0).

Residues

The residue of f at a is the coefficient a₋₁ of 1/(z − a) in the Laurent series.

• Simple pole: Res = lim_z→a (z − a)f(z). For f = p/q with a simple zero of q: Res = p(a)/q′(a).
• Pole of order m: Res = (1/(m − 1)!) lim_z→a d^m−1/dz^m−1 [(z − a)ᵐ f(z)].

Residue theorem. ∮_C f(z) dz = 2πi × (sum of the residues inside C).

Example. ∮ dz/(z(z − 2)) around |z| = 1: only z = 0 is inside, with residue 1/(0 − 2) = −1/2, so the integral is 2πi(−1/2) = −πi. Around |z| = 3 both poles are inside; the residues −1/2 and 1/2 cancel and the integral is 0.

Worked example

Check the Cauchy-Riemann equations for f(z) = z² numerically, and compute ∮ dz/(z(z − 2)) around |z| = 1 by direct numerical integration.

import cmath, math

f = lambda z: z * z
x, y, h = 1.3, -0.7, 1e-6
u = lambda x, y: f(complex(x, y)).real
v = lambda x, y: f(complex(x, y)).imag
ux = (u(x + h, y) - u(x - h, y)) / (2 * h)
vy = (v(x, y + h) - v(x, y - h)) / (2 * h)
uy = (u(x, y + h) - u(x, y - h)) / (2 * h)
vx = (v(x + h, y) - v(x - h, y)) / (2 * h)
print(f"ux = {ux:.4f}, vy = {vy:.4f}, uy = {uy:.4f}, -vx = {-vx:.4f}")

def contour(g, r, n=200000):
    total = 0
    for k in range(n):
        t = 2 * math.pi * (k + 0.5) / n
        z = r * cmath.exp(1j * t)
        total += g(z) * 1j * z * (2 * math.pi / n)
    return total

g = lambda z: 1 / (z * (z - 2))
I1 = contour(g, 1)
I3 = contour(g, 3)
print(f"around |z| = 1: real part {abs(I1.real):.6f}, imaginary part {I1.imag:.6f} (expected -pi = {-math.pi:.6f})")
print(f"around |z| = 3: magnitude {abs(I3):.6f} (expected 0)")

ux = 2.6000, vy = 2.6000, uy = 1.4000, -vx = 1.4000
around |z| = 1: real part 0.000000, imaginary part -3.141593 (expected -pi = -3.141593)
around |z| = 3: magnitude 0.000000 (expected 0)

Practice questions