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Complex analysis

lesson · about 3 minutes

Calculus with complex numbers: residues and contour integrals.

Pen and paper is fine · no calculator needed why?

Opens at level 60. You're level 1. You can read and practice here now.

the idea

Complex analysis is calculus for functions of z = x + iy. A function is analytic on a region when it has a complex derivative at every point there. Polynomials in z, eᶻ, sin z and cos z are analytic everywhere.

The residue theorem says an integral once counterclockwise around a closed loop depends only on the poles inside it, the points where the function blows up. Each pole adds 2πi times its residue.

techniques

Cover up the pole

A simple pole z = a, from a factor (z − a) on the bottom.

Cover up (z − a) and put z = a into what is left.
That value is the residue.

worked example

Example: What is the residue of f(z) = 2/((z − 1)(z + 3)) at z = 1? Give an exact number or fraction.

Cover up (z − 1). What is left is 2/(z + 3).
Put in z = 1: 2/(1 + 3) = 2/4 = 1/2.

Answer: 1/2

Add the residues inside

An integral around a circle |z| = R, counterclockwise.

Keep only the poles inside the circle, closer to 0 than R.
The integral is 2πi × (the sum of their residues). For an answer kπi, k is twice that sum.

worked example

Example: The integral of f(z) = 2/((z − 1)(z + 3)) around the circle |z| = 2 (counterclockwise) equals kπi. What is k?

Poles: 1 and −3. Only 1 is inside the circle of radius 2.
Its residue is 2/(1 + 3) = 1/2.
So k is twice 1/2, which is 1.

Answer: 1

Let the roots of unity cancel

The n roots of unity, the solutions of zⁿ = 1: their sum, product or powers.

Evenly spread around the unit circle, they balance out: for n ≥ 2, the sum is 0.
Product: each non-real root times its mirror image is 1. Left over: 1, and also −1 if n is even.
kth powers: if n divides k, each is 1 and the sum is n. Otherwise it is 0.

worked example

Example: What is the product of all 4th roots of unity?

The 4th roots of unity are 1, i, −1 and −i.
i and −i are a mirror pair with product 1.
That leaves 1 × (−1), so the product is −1.

Answer: −1

watch out for

Counting a pole outside the circle, which adds nothing to the integral.
Flipping a residue’s sign. For 1/((z − 1)(z − 3)) at z = 1, the residue is 1/(1 − 3), which is −1/2.
Calling |z|² or conj(z) analytic because it looks smooth. Neither has a complex derivative on any open region.

practice

Roots of unity

worked example

What is the sum of the 12th powers of all 12th roots of unity?

Answer: 12

Each of the 12 roots ω has ω¹² = 1.
So the sum is 12 × 1 = 12.

Residues

worked example

What is the residue of f(z) = 1/((z − 5)(z + 4)) at z = −4? Give an exact number or fraction.

Answer: -1/9

z = −4 is a simple pole: f(z) = g(z)/(z + 4) with g(z) = 1/(z − 5).
The residue is g(−4) = 1/(−4 − 5) = −1/9.

Contour integrals

worked example

The integral of f(z) = (z² + 3z + 4)/z around the circle |z| = 3 (counterclockwise) equals kπi. What is k?

Answer: 8

The only pole, z = 0, lies inside |z| = 3.
Its residue is the numerator at z = 0, which is its constant term: 4.
Residue theorem: 2πi × 4 = 8πi, so k = 8.

Analytic or not?

worked example

Three of these functions of z = x + iy are analytic (complex-differentiable) everywhere. Which one is not?

conj(z)
z²
z² + 3z
z³

Answer: conj(z)

conj(z) = x − iy: ∂u/∂x = 1 but ∂v/∂y = −1, so Cauchy–Riemann fails everywhere.
The other three, z³, z² + 3z and z², are analytic everywhere: polynomials in z, or built from eᶻ.