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Jeffrey Hellrung Thursday, April 27, 2006 Math 246A, Homework 03
- (a) Consider the standard model of the Riemann sphere in R^3. Show that all rotations of the sphere correspond to linear fractional transformations under the stereographic projection. (b) Prove that stereographic projection provides a bijection between the set of circles on the Riemann sphere and the set of circles and lines on the complex plane. Solution
(a) All rotations of S^2 = {x ∈ R^3 | |x| = 1} are compositions of rotations about the 3 coordinate axes:
A 1 =
0 cos θ − sin θ 0 sin θ cos θ
, A 2 =
cos θ 0 sin θ 0 1 0 − sin θ 0 cos θ
, A 3 =
cos θ − sin θ 0 sin θ cos θ 0 0 0 1
(Actually, we only need 2 of the above rotation matrices, but we’ll use all 3 for completeness). We show that each of these rotations corresponds to a linear fractional transformation of C under stereographic projection; the claim then follows, since linear fractional transformations are closed under compositions. Let π : S^2 → C denote the stereographic projection, where we have
π(x 1 , x 2 , x 3 ) =
x 1 + ix 2 x 3
and π−^1 (z) =
2 ℜ(z) |z|^2 + 1
2 ℑ(z) |z|^2 + 1
|z|^2 − 1 |z|^2 + 1
Now to show that each Ai corresponds to a linear fractional transformation, we will make some observations on where choice points are mapped to, construct the corresponding linear fractional transformation, then show (using Mathematica to simplify the algebra) that it coincides with the rotation map. We begin by noticing that, under the map A 1 ,
π−^1 (1) = (1, 0 , 0) 7 → (1, 0 , 0) = π−^1 (1) π−^1
−i sin θ 1+cos θ
= (0, − sin θ, − cos θ) 7 → (0, 0 , −1) = π−^1 (0) π−^1
i sin θ 1 −cos θ
= (0, sin θ, cos θ) 7 → (0, 0 , 1) = π−^1 (∞)
thus we consider the linear fractional transformation T 1 where 1, (^) 1+cos−i^ sin^ θθ , 1 i−^ sincos^ θ θ 7 → 1 , 0 , ∞, respec- tively, and this is given by
T 1 z =
z + (^) 1+cosi^ sin^ θ θ
1 − 1 i−^ sincos^ θ θ
z − 1 i−^ sincos^ θ θ
1 + (^) 1+cosi^ sin^ θ θ
Mathematica simplifies this by the command T1[z_] = FullSimplify[ ((z + ISin[t]/(1 + Cos[t]))(1 - ISin[t]/(1 - Cos[t])))/ ((z - ISin[t]/(1 - Cos[t]))(1 + ISin[t]/(1 + Cos[t])))]
to
T 1 z =
z cos(θ/2) + i sin(θ/2) cos(θ/2) + iz sin(θ/2)
We now just need to verify that π−^1 (T 1 z) = A 1 π−^1 (z).
We use Mathematica for this verification. Start by defining π−^1 and A 1 :
piinv[z_] = {(z + Conjugate[z]) /( zConjugate[z] + 1 ), (z - Conjugate[z]) /(I(zConjugate[z] + 1)), (zConjugate[z] - 1)/( z*Conjugate[z] + 1 )} A1 = {{1, 0, 0}, {0, Cos[t], -Sin[t]}, {0, Sin[t], Cos[t]}}
and then simplify π−^1 (T 1 z) and A 1 π−^1 (z):
FullSimplify[piinv[T1[z]], Element[t, Reals]] FullSimplify[A1. piinv[z], Element[t, Reals]]
In both cases,
π−^1 (T 1 z) =
2 ℜ(z) |z|^2 + 1 , cos θ
2 ℑ(z) |z|^2 + 1 − sin θ
|z|^2 − 1 |z|^2 + 1 , sin θ
2 ℑ(z) |z|^2 + 1
|z|^2 − 1 |z|^2 + 1
= A 1 π−^1 (z).
Thus π−^1 (T 1 z) = A 1 π−^1 (z), as desired.
Under the map A 2 ,
π−^1
cos θ 1 −sin θ
= (cos θ, 0 , sin θ) 7 → (1, 0 , 0) = π−^1 (1) π−^1
sin θ 1+cos θ
= (sin θ, 0 , − cos θ) 7 → (0, 0 , −1) = π−^1 (0) π−^1
− sin θ 1 −cos θ
= (− sin θ, 0 , cos θ) 7 → (0, 0 , 1) = π−^1 (∞)
thus we consider the linear fractional transformation
T 2 z =
z − (^) 1+cossin^ θ θ
cos θ 1 −sin θ +^
sin θ 1 −cos θ
z + (^1) −sincos^ θ θ
cos θ 1 −sin θ −^
sin θ 1+cos θ
Mathematic simplifies this by the command
T2[z_] = FullSimplify[ ((z - Sin[t]/(1 + Cos[t]))* (Cos[t]/(1 - Sin[t]) + Sin[t]/(1 - Cos[t])))/ ((z + Sin[t]/(1 - Cos[t]))* (Cos[t]/(1 - Sin[t]) - Sin[t]/(1 + Cos[t])))]
to
T 2 z =
z cos(θ/2) − sin(θ/2) cos(θ/2) + z sin(θ/2)
We now define A 2 :
A2 = {{Cos[t], 0, Sin[t]}, {0, 1, 0}, {-Sin[t], 0, Cos[t]}}
and then simplify π−^1 (T 2 z) and A 2 π−^1 (z):
FullSimplify[piinv[T2[z]], Element[t, Reals]] FullSimplify[A2. piinv[z], Element[t, Reals]]
- Find a conformal map from the quarter disc {z | |z| < 1 , ℜ(z) > 0 , ℑ(z) > 0 } to the disc {z | |z| < 1 }.
Hint: It can be constructed as a composition of linear fractional transformations and the function z 7 → z^2 (used twice). Show that the map you construct is conformal. Why is z 7 → z^4 not a solution to the problem? Solution We use the composition of the following conformal mappings.
- z 7 → z^2 maps the quarter disc {|z| < 1 , ℜ(z) > 0 , ℑ(z) > 0 } to the half disc {|z| < 1 , ℑ(z) > 0 }, where 0 7 → 0, 1 7 → 1, i 7 → −1.
- z 7 → 1+ 1 −zz maps the half disc {|z| < 1 , ℑ(z) > 0 } to the quarter plane {ℜ(z) > 0 , ℑ(z) > 0 }, where − 1 7 → 0, 0 7 → 1, i 7 → i, 1 7 → ∞.
- z 7 → z^2 maps the quarter plane {ℜ(z) > 0 , ℑ(z) > 0 } to the half plane {ℑ(z) > 0 }, where 0 7 → 0, 1 7 → 1, i 7 → −1.
- z 7 → z z−+ii maps the half plane {ℑ(z) > 0 } to the disc {|z| < 1 }, where − 1 7 → i, 0 7 → −1, 1 7 → −i, ∞ 7 → 1.
The composition is conformal since each of the components are conformal. z 7 → z^4 maps the quarter disc {|z| < 1 , ℜ(z) > 0 , ℑ(z) > 0 } to the slitted disc {|z| < 1 , ℑ(z) < 0 or ℜ(z) 6 = 0} 6 = {|z| < 1 }.
- (a) We use standard x, y coordinates for the complex plane. Consider the partial differential operators
∂ ∂z
∂x
− i
∂y
and ∂ ∂z
∂x
∂y
and prove that for any i, j ≥ 0, ∂ ∂z zj^ zk^ = jzj−^1 zk
and similarly for the other operator. (b) Let N ≥ 0 be an integer and let aj,k, bj,k be complex numbers for j, k = 0,... , N. Prove that if
∑^ N
j,k=
aj,kzj^ zk^ =
∑^ N
j,k=
bj,kzj^ zk
then aj,k = bj,k for all j, k = 0,... , N. Solution
(a) For z = x + iy,
∂ ∂x z
j (^) zk (^) = ∂ ∂x (x^ +^ iy)
j (^) (x − iy)k = j(x + iy)j−^1 (x − iy)k^ + k(x + iy)j^ (x − iy)k−^1 = jzj−^1 zk^ + kzj^ zk−^1
∂ ∂y z
j (^) zk^ = ∂ ∂y (x^ +^ iy)
j (^) (x − iy)k = ij(x + iy)j−^1 − ik(x + iy)j^ (x − iy)k−^1 = ijzj−^1 zk^ − ikzj^ zk−^1
so ∂ ∂z
zj^ zk^ =
∂x
− i
∂y
zj^ zk^ = jzj−^1 zk
and ∂ ∂z
zj^ zk^ =
∂x
∂y
zj^ zk^ = kzj^ zk−^1.
(b) By subtracting each side, we can assume that bj,k = 0. We prove the claim by induction on j + k. For j + k = 2N , we know that j = k = N , and differentiating N times each by z and z allows us to conclude that aj,k = aN,N = 0. Now supposing the claim for all j + k > l, let j, k be such that j + k = l. Then differentiating j times by z and k times by z leaves only the aj,k term (all other terms “surviving” the differentiation must have the sum of the indices greater than l, hence are zero), so aj,k = 0, and the claim is proved by induction.
- The following is a qualifying exam problem from Fall 2003.
“Let f (z) = u(x, y)+iv(x, y) denote a non-constant holomorphic function on some open domain D ⊂ C. Show that at each point of D, the “level curves” u(x, y) = constant and v(x, y) = constant intersect at right angles.” This problem is slightly ill-posed. Give a counterexample to the problem. Then minimally modify the problem so that it becomes a good qual problem and solve it. Solution As a counterexample, we need only consider f (z) = z^2 = x^2 − y^2 + i 2 xy and D = C, whose levels curves of the real and imaginary parts at z = 0 do not intersect at right angles. Note that f ′(0) = 0, hence we only need to modify the problem to stipulate that f be conformal in D. Given f conformal, we know that neither ∇u nor ∇v can vanish. Thus, at a point (x, y) ∈ C, the level curves of u and v interesect at right angles if and only if ∇u ⊥ ∇v, and
∇u · ∇v =
∂u ∂x
∂u ∂y
∂v ∂x
∂v ∂y
by the Cauchy-Riemann equations.
- Let f : C → C be a holomorphic function. Assume |f (z)| = c is constant. Prove that f is constant.
Solution Let f (z) = u(x, y) + iv(x, y). Then we are given that u^2 + v^2 = c^2 , hence
u
∂u ∂x
∂v ∂x
and u
∂u ∂y
∂v ∂y
The Cauchy-Riemann equations allow us to write the latter equation as
−u ∂v ∂x
and adding u times the first equation to v times the third equation gives
( u^2 + v^2
) (^) ∂u ∂x
