3. Elementary Results of Complex Analysis. Liouville's Theorem.
We assume the following results of advanced advanced calculus:
Integration of Complex Functions over Paths. If γ: [0, 1] → C is a rectifiable curve in the complex plane, we can write
γ(t) = α(t) + iβ(t)
where α, β are real functions of bounded variation. For a real valued function h defined on a region containing the image, C, of γ, we then define
∫C f dγ = ∫[0,1] h dα(t) + ∫[0,1] h dβ(t)
where the integrals on the right are Riemann-Stieltjes integrals. For a complex valued function f = u + iv, we set
∫C f dγ = ∫C u dγ + i ∫C v dγ
= [∫[0,1] u dα(t) - ∫[0,1] v dβ(t)] + i[∫[0.,1] v dα + ∫[0,1] u dβ(t)]
If γ is differentiable except on a finite number of points (piecewise differentiable), we then have
∫C f dγ = ∫[0,1] f (dγ/dt) dt
≤ sup |f| ∫[0,1] (dγ/dt) dt
≤ sup |f| ∫[0,1] |dγ/dt| dt
and we note that the last integral is actually the length of the arc γ.
An Important Integral.
| 2π | eis | |
| ∫ | ds = 2π if |z| < 1. | |
| 0 | eis - z |
Proof. Let
| φ (s,t) = | eis | for 0 ≤ t ≤ 1, 0 ≤ s ≤ 2π |
| eis - tz |
Then φ is continuously differentiable. Hence, if g(t)= ∫[0,1] φ(s,t) ds
g'(t) = (∂/∂t) ∫[0,2π] φ(s,t) ds
| = | ∂ | 2π | eis | ds |
| ∫ | ||||
| ∂t | 0 | eis - tz |
| = | 2π | z eis | ds |
| ∫ | |||
| 0 | (eis - tz) |
by the theorem on differentiation under the integral sign. Since
| ∫ | eis | ds = | i |
| (eis - tz)2 | eis - tz |
[The constant of integration is omitted here.]
and eis = 1 for s = 0 or 2π, the definite integral vanishes, so g(t) is constant. Since g(0) = 2π, we must have g(t) = 2π for all t ∈ [0.1], giving the required result.
Proposition. Let G be a region (connected set with nonempty interior) in C, let f: G → C be analytic, and suppose |z - a| implies z ∈ G. If γ(t) = a + reit, 0 ≤ t ≤ 2π, then
| f(z) = | 1 | f(w) | dw | |
| ∫ | ||||
| 2πi | γ | w - z |
Proof. By considering G1 = {(z - a)/r : z ∈ G} and the function g(z) = f(a + rz), we see that, without loss of generality, we may take a = 0 and z = 1.
We want to show that
| 2π | f(eis) eis | ds - 2πf(z) = | 2π | / | f(e | eis - f(z) | \ | dz |
| ∫ | ∫ | | | | | |||||
| 0 | eis - z | 0 | \ | eis - z | / |
Put
| φ (s, t) = | f(z + t(eis - z)) | eis - f(z) |
| eis - z |
If g(t) = ∫[0,2π] φ(s,t) ds, then
g'(t) = (∂/∂t))∫[0,2π] φ(s,t) ds
= ∫[0,2π] eis f'(z + t(eis - z)) ds
by differentiation under the integral sign. But we have the indefinite integral∫eis f'(z + t(eis - z)) ds = - it f(z + t(eis - z))
and noting that the right hand has the same value for s = 0 and s = 2π, g'(t) = 0 and g(t) is constant. Clearly| g(0) = | 2π | / | f(z) eis | - f(z) | \ | ds = 0 |
| ∫ | | | | | ||||
| 0 | \ | eis - z | / |
since we have established above that
| 2π | eis | ds = 2π | |
| ∫ | |||
| 0 | eis - z |
Theorem (Taylor series expansion). If f is analytic for |z - a| < R, then f(z) = ∑n = 0, ... , ∞ an(z - a)n for all z in this region, where an = f(n)(a)/n!
Proof.If γ(t) = a + reit, 0 ≤ t ≤ 2π where 0 ≤ r ≤ R, then by the above proposition
| f(z) = | 1 | f(w) | dw for |z - a| < r | |
| ∫ | ||||
| 2πi | γ | w - z |
We have the geometric series expansion
| 1 | = | 1 | ∞ | / | z - a | \n |
| ∑ | | | | | ||||
| w - z | (w - a) | n=0 | \ | w - a | / |
which converges uniformly for |z - a| < r. Thus we can write
| f(z) = | ∞ | ┌ | 1 | f(w) | ┐ | (z - a)n | ||
| ∑ | | | ∫ | | | |||||
| n=0 | ┕ | 2πi | γ | (w - a)n + 1 | ┙ |
Setting
| an = | 1 | f(w) | dw | |
| ∫ | ||||
| 2πi | γ | (w - a)n + 1 |
we get
f(z) = ∑n=0,...,∞ an(z - a)n
for |z - a| < r. Clearly an = f(n)(a)/n! Since r < R was chosen arbitrarily, the expansion holds for alll z with |z - a| < R.
Corollary.
| f(n)(a) = | n! | f(w) | dw | |
| ∫ | ||||
| 2πi | γ | (w - a)n + 1 |
where γ(t) = a + reit, o ≤ t ≤ 2π.
Cauchy's Estimate. Let f be analytic for |z - a| < r and suppose |f(z)| ≤ M for all such z. Then
| |f(n)(a)| ≤ | n!M |
| Rn |
Proof. By the Corollary above
| |f(n)(a)| ≤ | | | n! | | | | | f(w) | | | d|w| | |
| | | | | ∫ | | | | | ||||
| | | 2πi | | | γ | | | (w - a)n + 1 | | | ||
= (n!/2π).(M/rn+1).2πr = (n!M/rn)
The result follows since r < R is arbitrary.
Liouville's Theorem. If f is a bounded function analytic in the entire plane, then f is constant.
Proof. If |f(z)| ≤ M for all z ∈ C, then by Cauchy's estimate, since f is analytic in every disc {w: |w - z| < R}, we have |f'(z)| ≤ M/R. Since R can be made arbitrarily large, |f(z)| = 0, and the result follows.
Link. The treatment of the subject here is an extremely sketchy one, taken from John B. Conway: Functions of one Complex Variable. a detailed development of the subject is found at the online textbook
http://math.fullerton.edu/mathews/c2003/ComplexUndergradMod.html