Complex Algebra 9, Exercises - Mathematics, Exercises of Algebra

Harmonic functions, linear equations,logarithm of our conformal map, fractional linear transformations,complex conjugation.

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Math 213a: Complex analysis
Problem Set #8 (12 November 2003):
Harmonic functions and their uses, cont’d
First, an observation on the coefficients of the linear equations used to determine
the logarithm of our conformal map of a finitely connected region (Ahlfors,
V.3.1, Ex.2 (p.204)):
1. Let Cbe a region whose boundary consists of the simple closed curves
C1, . . . , Cn. As in Ahlfors V.3.1, for j, k {1, . . . , n}let αjk =HCjk,
where ωkis the harmonic measure of Ckwith respect to Ω, and Cjis
traversed in the positive direction relative to Ω. Prove that αjk =αkj .
Next, some problems on the nice special case of a doubly-connected region.
We begin with an elementary problem (that is, a problem not requiring the
machinery of (sub)harmonic functions from Chapter V) that picks up a thread
from the fourth problem set, where we studied the action of PGL2(C) on pairs
of disjoint circles.
2. If is the doubly connected region bounded by two disjoint circles, show
that the conformal bijections of to an annulus are fractional linear
transformations. Deduce that the subgroup of PGL2(C) that preserves
two disjoint circles is conjugate in PGL2(C) to the group of fractional
linear transformations of the form z7→ cz for some cCwith |c|= 1.
Obtain a similar description of the stabilizer in PGL2(C) of two circles
that are tangent or meet in two points.
3. Let be the annulus {zC:r < |z|< R}. Prove that if uis a bounded
harmonic function on then, for any ρ(r, R),
log R
rI|z|=ρ
du 2πsup
z
u(z)inf
zu(z),
with equality if and only if there exist C0, C1R, with C10, such that
u(z) = C0+C1log |z|for all zΩ. Conclude that the only conformal
bijections of are z7→ cz and z7→ cRr/z (|c|= 1).
4. Suppose and 0Care doubly connected regions conformally equivalent
to the annuli r < |z|< R and r0<|z|< R0. Let their complements in
P1(C) have components E1, E2and E0
1, E0
2, with E1, E0
1containing .
Prove that if EjE0
jfor j= 1,2 (whence 0) then R/r R0/r0,
with equality only when = 0.
5. Suppose is a simply-connected bounded subset of C, and fix z0Ω. For
each r > 0 such that contains the closed disc {z:|zz0| r}, the region
r:= {z : |zz0|> r}is doubly connected, and thus conformal with
an annulus Arwith outer radius 1. Let ρ(r) be the inner radius of Ar. By
pf2

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Math 213a: Complex analysis Problem Set #8 (12 November 2003): Harmonic functions and their uses, cont’d

First, an observation on the coefficients of the linear equations used to determine the logarithm of our conformal map of a finitely connected region (Ahlfors, V.3.1, Ex.2 (p.204)):

  1. Let Ω ⊂ C be a region whose boundary consists of the simple closed curves C 1 ,... , Cn. As in Ahlfors V.3.1, for j, k ∈ { 1 ,... , n} let αjk =

Cj ∗dωk, where ωk is the harmonic measure of Ck with respect to Ω, and Cj is traversed in the positive direction relative to Ω. Prove that αjk = αkj.

Next, some problems on the nice special case of a doubly-connected region. We begin with an elementary problem (that is, a problem not requiring the machinery of (sub)harmonic functions from Chapter V) that picks up a thread from the fourth problem set, where we studied the action of PGL 2 (C) on pairs of disjoint circles.

  1. If Ω is the doubly connected region bounded by two disjoint circles, show that the conformal bijections of Ω to an annulus are fractional linear transformations. Deduce that the subgroup of PGL 2 (C) that preserves two disjoint circles is conjugate in PGL 2 (C) to the group of fractional linear transformations of the form z 7 → cz for some c ∈ C∗^ with |c| = 1. Obtain a similar description of the stabilizer in PGL 2 (C) of two circles that are tangent or meet in two points.
  2. Let Ω be the annulus {z ∈ C : r < |z| < R}. Prove that if u is a bounded harmonic function on Ω then, for any ρ ∈ (r, R), ( log

R

r

|z|=ρ

∗du ≤ 2 π

sup z

u(z) − inf z u(z)

with equality if and only if there exist C 0 , C 1 ∈ R, with C 1 ≥ 0, such that u(z) = C 0 + C 1 log |z| for all z ∈ Ω. Conclude that the only conformal bijections of Ω are z 7 → cz and z 7 → cRr/z (|c| = 1).

  1. Suppose Ω and Ω′^ ∈ C are doubly connected regions conformally equivalent to the annuli r < |z| < R and r′^ < |z| < R′. Let their complements in P^1 (C) have components E 1 , E 2 and E′ 1 , E 2 ′, with E 1 , E′ 1 containing ∞. Prove that if Ej ⊇ E′ j for j = 1, 2 (whence Ω ⊆ Ω′) then R/r ≤ R′/r′, with equality only when Ω = Ω′.
  2. Suppose Ω is a simply-connected bounded subset of C, and fix z 0 ∈ Ω. For each r > 0 such that Ω contains the closed disc {z : |z−z 0 | ≤ r}, the region Ωr := {z ∈ Ω : |z − z 0 | > r} is doubly connected, and thus conformal with an annulus Ar with outer radius 1. Let ρ(r) be the inner radius of Ar. By

the previous problem, ρ is a decreasing function of r. Prove that ρ(r) → 0 as r → 0. What happens to the conformal bijections from Ωr to Ar as r → 0?

Finally:

  1. Let Ω ∈ C be a region of connectivity n. i) If n ≤ 3, prove that Ω is conformally equivalent to its image under complex conjugation. Do this even in the (easier) case that the complement of Ω in P^1 (C) has one or more components that reduce to a point. ii∗) When n = 4, construct or prove the existence of an Ω not confor- mally equivalent to its complex conjugate, with none of the components of P^1 (C) − Ω reducing to a point.

This problem set is due Wednesday, November 19, at the beginning of class.