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Math 246A
Homework 7 Solutions
Due Thursday, May 24
Note the solutions are often just an outline, but I have tried to address all of the key points and you should be able to fill in the details. If you have questions about them feel free to ask me; also if there are any errors in my solutions please let me know. Of course there are usually many other correct ways to solve each problem as well.
- Prove that f(z) = − 12 (z + 1/z) is a conformal map from the half-disc {z = x + iy : |z| < 1 , y > 0 } to the upper half-plane. [Hint: The equation f(z) = w reduces to the quadratic equation z^2 + 2wz + 1 = 0, which has two distinct roots in C whenever w 6 = ±1. This is certainly the case if w ∈ H.] First f(z) maps the half-disc into H since if z = reiθ^ where r < 1 and 0 < θ < π, then Im z = − 12 sin θ(r − 1 /r), and sin θ > 0 while r − 1 /r < 0. Following the hint z^2 + 2wz + 1 = 0 has two distinct roots if w 6 = ± 1 , and as their product is 1 one must lie in D and one in C − D¯. Also if z is a root then z + 1/z = − 2 w, where w ∈ H. Since Im (z + 1/z) = (Im z)(1 − 1 / |z|) and |z| < 1 we must have Im z > 0 as well and z is in the upper half-disc. Let g(w) = −w +
w^2 − 1 where we choose a branch of square root so that
i^2 − 1 = i
- Then g(i) ∈ D, and by continuity g(w) ∈ D (and also Im g(w) > 0 ) for all w ∈ H. The functions f and g are inverses by calculation.
10 Let F : H → C be a holomorphic function that satisfies |F (z)| ≤ 1 and F (i) = 0. Prove that
|F (z)| ≤ | z − i z + i
| for all z ∈ H.
Define G : D → H by G(z) = i (^1) 1+−ww This is a conformal map from D into H and fr (w) = r ∗ (F ◦ G) for r < 1 is a holomorphic map from D into C such that fr (0) = r ∗ (F (i)) = 0 and |fr (w)| ≤ r < 1 for all w ∈ D. Then by Schwarz’s lemma |r ∗ (F ◦ G(w))| = |fr (w)| ≤ |w| for all w ∈ D. Letting r go to 1 we get |F ◦ G(w)| ≤ |w| for all w ∈ D. letting w = G−^1 (z) = i i−+zz gives |F (z)| ≤ | z z−+ii |.
13 The pseudo-hyperbolic distance between two points z, w ∈ D is defined by
ρ(z, w) =
z − w 1 − wz¯
(a) Prove that if f : D → D is holomorphic, then ρ(f(z), f(w)) ≤ ρ(z, w) for all z, w ∈ D. Moreover, prove that if f is an automorphism of D then f preserves the pseudo-hyperbolic distance ρ(f(z), f(w)) = ρ(z, w) for all z, w ∈ D. [Hint: Consider the automorphism ψα(z) = (z − α)/(1 − αz¯) and apply the Schwarz lemma to ψf (w) ◦ f ◦ ψ− w^1 .] Following the hint, if g = ψf (w) ◦ f ◦ ψ w− 1 then g : D → D is holomorphic and g(0) = 0, and so by the Schwarz lemma it follows |g(z′)| ≤ |z′| for all z′^ ∈ D. Let z′^ = ψw (z) for z ∈ D, and ρ(f(z), f(w)) ≤ ρ(z, w) follows. For f an automorphism of D, then ρ(f(z), f(w)) ≤ ρ(z, w) and ρ(z, w) = ρ(f−^1 (f(z))), f−^1 (f(w))) ≤ ρ(f(z), f(w)); therefore ρ(f(z), f(w)) = ρ(z, w) for all z, w ∈ D. (b) Prove that |f′(z)| 1 − |f(z)|^2
1 − |z|^2
for all z ∈ D
This result is called the Schwarz-Pick lemma. By part (a), for all z, w ∈ D we have
∣ 1 f−^ (zf) (−wf)^ f(w (z))
∣ 1 z−− wz¯w
∣, which implies ∣ ∣ ∣ f^ (z z)−−fw^ (w) 1 −f (w^1 )f (z)
∣ (^1) −^1 wz¯
∣ if z 6 = w. Take the limit as w → z to get the result. One can also use that |g′(0)| ≤ 1 and then the result follows from the chain rule.
14 Prove that all conformal mappings from the upper half plane H to the unit disc D take the form
eiθ^
z − β z − β ¯
, θ ∈ R and β ∈ H.
First we show that Tβ (z) = z z−−β β¯ , β ∈ H, is a conformal map from H to D. Clearly Tβ maps H into D since |z − β| <
∣z − β¯
∣ (^) for z ∈ H. The inverse function is T (^) β− 1 (z) = z^ β¯−β z− 1 which maps^ D^ into^ H^ as a calculation shows: If β = a + bi and z = x + iy where b > 0 and x^2 + y^2 < 1 , then the imaginary part of z β¯−β z− 1 is^
b(1−x^2 ) (x−1)^2 +y^2 >^0. It follows immediately that^ Tθ,β^ =^ e
iθ z−β z− β¯ is also a conformal map from^ H^ to^ D. Let T = T 0 ,i = z z−+ii and let S be some other conformal map from H to D. Then S ◦ T −^1 = eiϕ α 1 −− ¯αzz for some ϕ ∈ R and α ∈ D. It follows S = eiϕ z z((1α−− α1)+¯)+ii(1+(1+ ¯αα)) = −eiϕ^ z+i( 1+ α−α 1 ) z( α α¯−−^11 )−i( 1+ ¯ α−α 1 ) =^ −e
iϕ
α− 1 α ¯− 1
) (^) z−i ( 1+ 1 −αα ) z+i( 1+ ¯ 1 −α α¯ ) =^ e
iθ z−β z− β¯ , where eiθ^ = −eiϕ
α− 1 α ¯− 1
and β = i
1+α 1 −α
∈ H.
15 Here are two properties enjoyed by automorphisms of the upper half-plane.
(a) Suppose Φ is an automorphism of H that fixes three distinct points on the real axis. Then Φ is the identity. Let the three points be a, b, c. Then the map T = z z−−ac bb−−ac is an LFT (possibly in P SL 2 (C), but that’s okay) mapping a, b, c to 0 , 1 , ∞ respectively. It follows T −^1 ◦ Φ ◦ T fixes 0 , 1 , ∞. The only LFT that does this is the identity map z. It follows Φ(z) = z as well. (b) Suppose (x 1 , x 2 , x 3 ) and (y 1 , y 2 , y 3 ) are two pairs of three distinct points on the real axis with x 1 < x 2 < x 3 and y 1 < y 2 < y 3. Prove that there exists (a unique) automorphism Φ of H so that Φ(xj ) = yj , j = 1, 2 , 3. The same conclusion holds if y 3 < y 1 < y 2 or y 2 < y 3 < y 1. Map x 1 , x 2 , x 3 to 0 , 1 , ∞ using S = z z−−xx^13 x x^22 −−xx^31 and map y 1 , y 2 , y 3 to 0 , 1 , ∞ using T = z z−−yy^13 y y^22 −−yy^31. Note that both S and T have positive determinant due to the ordering of the points. Divide top and bottom by the square root of the determinant and call the new maps S and T ; now they both have determinant 1 and so are in P SL 2 (R). Now let Φ = T −^1 ◦ S. For the other orderings use different maps S and T mapping the points to 0 , 1 , ∞ but in a different order and still with positive determinant. Uniqueness follows since if there were two maps Φ, Ψ then Ψ−^1 ◦ Φ fixes three points and thus must be the identity by part (a), and so Φ = Ψ.
16 Let
f(z) = i − z i + z
and f−^1 (z) = i 1 − w 1 + w
(a) Given θ ∈ R, find real numbers a, b, c, d such that ad − bc = 1, and so that for any z ∈ H
az + b cz + d
= f−^1 (eiθ^ f(z)).
Note all of the maps can be thought of as fractional linear transformations so we just need to multiply some matrices in GL 2 (C) project down to PSL 2 (C) and show that the entries are real. The matrices are ( −i i 1 1
eiθ/^2 0 e−iθ/^2
− 1 i 1 i
ieiθ/^2 + ie−iθ/^2 eiθ/^2 − e−iθ/^2 −eiθ/^2 + e−iθ/^2 ieiθ/^2 + ie−iθ/^2
This matrix has determinant -4 so normalizing (by dividing each entry by 2 i) we get ( cos(θ/2) sin(θ/2) − sin(θ/2) cos(θ/2)
a b c d
(b) Given α ∈ D, find real numbers a, b, c, d such that ad − bc = 1, and so that for any z ∈ H
az + b cz + d
= f−^1 (Ψα(f(z))).
