Lecture 14 – Conformal Mapping 1 Conformality

Lecture 14 – Conformal Mapping MATH-GA 2451.001 Complex Variables

1 Conformality 1.1 Preservation of angle 0 The open mapping theorem tells us that an analytic function such that f (z0) 6= 0 maps a small neighborhood of z0 onto a neighborhood of f(z0) in a one-to-one fashion. In particular, f maps continuously diﬀerentiable arcs through z0 onto continuously diﬀerentiable arcs through f(z0). We now show that f preserves angles between two such arcs.

Suppose f is a complex function (not necessarily analytic) deﬁned on a neighborhood of z0 and such that iϕ f(z) 6= f(z0) for all z 6= z0 in that neighborhood. If there exists w = e ∈ C, with ϕ ∈ R such that for all ∗ θ ∈ R and r ∈ R+, iθ f(z0 + re ) − f(z0) iϕ iθ iθ −−−−→ e e |f(z0 + re ) − f(z0)| r→0+ then we say that f preserves angles at z0.

0 Theorem: Suppose f is analytic at z0. Then f preserves angles at z0 iﬀ f (z0) 6= 0.

0 Proof : Let f be analytic in a neighborhood of z0, and such that f (z0) 6= 0.

iθ iθ f(z0+re )−f(z0) 0 f(z + re ) − f(z ) iθ f (z ) ∀(r, θ) ∈ ∗ × , lim 0 0 = eiθ lim re = eiθ 0 R+ R iθ |f(z +reiθ )−f(z )| 0 r→0+ |f(z0 + re ) − f(z0)| r→0+ 0 0 |f (z0)| r

0 0 So we see that if w exists, w = f (z0)/|f (z0)|.

0 Conversely, suppose that f (z0) = 0. If f is constant, f does not preserve angle. If f is not constant, there ∗ N exists N ∈ N such that f(z) = f(z0) + (z − z0) g(z), with g analytic at z0, and g(z0) 6= 0. In that case,

We see that f increases angle by a factor of N, so f does not preserve angles Deﬁnition: A function which is analytic on an open connected set Ω and has a nonvanishing derivative is called a conformal map.

Examples: • exp maps any arbitrary vertical line {z : <(z) = x0 ∈ R} onto the circle with center 0 and x0 iy0 radius e . exp maps any horizontal line {z : =(s) = y0 ∈ R onto the open ray from 0 through e . We see that exp preserves the orthogonality of these curves.

• The function f(z) = z2 maps two curves crossing at 0 with angle α into two curves crossing at 0 with angle 2α. It is not conformal at z = 0.

Note: The deﬁnition of angle preservation contains the preservation of the magnitude of the angle as well as its orientation. You can easily convince yourself that the mapping by the complex conjugate of an analytic function whose derivative does not vanish preserves the magnitude of the angle, but reverses the orientation. It is called an indirectly conformal map.

1.2 Length and area 0 • Consider an analytic function f such that f (z0) 6= 0.

|f(z) − f(z0)| 0 lim = |f (z0)| z→z0 |z − z0|

0 Any small line segment with one end point at z0 is expanded by an amount |f (z0)|. This expansion is independent of the direction of the line segment.

1 • Consider a set Ω in R2. Its area is given by ZZ A(Ω) = dxdy Ω The mapped set Ω0 = f(Ω), where f = (u(x, y), v(x, y)) is a bijective differentiable mapping, has an area given by ZZ ZZ 0 A(Ω ) = dudv = |uxvy − uyvx|dxdy Ω0 Ω Now, if f(z) = u(x, y) + iv(x, y) is a conformal mapping on an open set containing Ω, f satisfies the Cauchy- Riemann relations, so that 0 2 |uxvy − uyvx| = |f (z)| and therefore ZZ A(Ω0) = |f 0(z)|2dxdy Ω Infinitesimal areas are expanded by the factor |f 0(z)|2.

2 Linear fractional transformations 2.1 M¨obiustransformation As we have already seen in Lecture III, M¨obiustransformations, also called linear fractional transformations, are maps of the form az + b S(z) = cz + d with (a, b, c, d) ∈ C4 such that ad − bc 6= 0 in order to avoid the situation in which S is a constant function. d The domain of S is C \ {− c } if c 6= 0, and C if c = 0.

One often deﬁnes S on the extended complex plane Cˆ by setting d a S − = ∞ ,S(∞) = if c 6= 0 ,S(∞) = ∞ if c = 0 c c

d For all z ∈ C \ {− c }, ad − bc S0(z) = 6= 0 (cz + d)2 so S is a conformal map on its domain.

Lastly, we have already seen that ∀ (a, b, c, d) ∈ C4 such that ad − bc 6= 0, S has an inverse: dw − b z = S−1(w) = −cw + a

2.2 The linear group Theorem: The set M of M¨obiustransformations is a group under composition. Any M¨obius transformation is a composition of the following maps:

(1) Translation: z 7→ z + a, with a ∈ C constant (2) Scaling and rotation: z 7→ kz, k ∈ C∗ constant 1 (3) Inversion: z 7→ z

2 • Before we discuss the group structure, let us prove the last point of the theorem. If c 6= 0 az + b bc − ad a = + cz + d 2 d c c (z + c ) which is the composition of a translation, an inversion, a scaling and rotation, and another translation. a b If c = 0, S(z) = d z + d is the composition of a scaling and rotation and a translation. • Regarding the group structure of M, we have already done most of the work:

• S(z) = z is the identity • Any S has an inverse S−1

a1z+b1 a2z+b2 • If S1(z) = ∈ M and S2(z) = ∈ M, then c1z+d1 c2z+d2 Az + B S (S (z)) = 1 2 Cz + D with AB a b a b = 1 1 2 2 CD c1 d1 c2 d2 which demonstrates that the composition is also associative.

2.3 The cross ratio ˆ Let z2, z3, and z4 be distinct points in C. To these points, we associate the M¨obiustransformation

 z−z3  z−z4 if (z , z , z ) ∈ 3  z2−z3 2 3 4 C  z2−z4  z−z3 if z2 = ∞ S(z) = z−z4 z2−z4  if z3 = ∞  z−z4  z−z3  if z4 = ∞ z2−z3

ˆ 3 Observe that ∀(z2, z3, z4) ∈ C , S(z2) = 1, S(z3) = 0, S(z4) = ∞. We now show that S is the unique such linear fractional transformation, with the following general result. For any two sets of distinct complex numbers {z2, z3, z4} and {w2, w3, w4} in the extended complex plane, there exists a unique M¨obiustransform taking zn to wn, for n ∈ {2, 3, 4}. To prove this, we use the lemma below.

Lemma: A M¨obiustransform can have at most two ﬁxed points, unless S is the identity map.

Proof of the lemma: A ﬁxed point z0 of S is such that

az0 + b 2 = z0 ⇔ cz0 + (d − a)z0 − b = 0 cz0 + d This polynomial equation has at most two solutions, which concludes our proof.

Proof of the existence and uniqueness of the Möbiustransformation: Let us call Sz2z3z4 the Möbiustransform defined on the previous page. Sz2z3z4 maps {z2, z3, z4} to {1, 0, ∞}. The inverse map of Sw2,w3,w4 takes {1, 0, ∞} to {w , w , w }. Hence, S−1 ◦ S takes {z , z , z } to {w , w , w }. 2 3 4 w2,w3,w4 z2z3z4 2 3 4 2 3 4 Now, let us imagine that there are two linear fractional transformations S and T sending {z2, z3, z4} to −1 −1 {w2, w3, w4}. Then, ∀n ∈ {2, 3, 4}, S ◦ T (wn) = S(zn) = wn. S ◦ T has three fixed points, so S = T ˆ Definition: The cross ratio (z1, z2, z3, z4) is the image of z1 ∈ C under the unique Möbiustransformation which maps {z2, z3, z4} to {1, 0, ∞}

3 2.4 Circlelines

Proposition: Let r and s be real numbers, and k ∈ C. The equation r|z|2 + kz + kz + s = 0 • represents a line if r = 0 and k 6= 0

• represents a circle if r 6= 0 and |k|2 ≥ rs, with equation

k 1p z + = |k|2 − rs r r This result can be easily proved by writing z = x + iy and expanding in x and y.

Deﬁnition: The locus of the points of r|z|2 + kz + kz + s = 0, if non-empty, is called a circleline.

Note: The definition above may feel a bit odd at first, in the sense that it tries to combine under the same term two different objects: lines and cirles. The reason why the definition makes sense in our context is that both circles and extended lines in the complex plane correspond to circles on the Riemann sphere, as discussed in Lecture I. Some authors, including Ahlfors, do not even use the term circleline, and call the locus of the points above a circle.

2 Lemma: Let r ∈ R, and (z1, z2) ∈ C , with z1 6= z2. The locus of the equation |z − z1| = r|z − z2| represents a circle if r 6= 1, and a line if r = 1, namely the line that is perpendicular to the line segment [z1, z2] and passes through its midpoint.

Theorem: A M¨obiustransformation maps a circleline to a circleline.

Proof : Since any M¨obiustransformation is the composition of a translation, a scaling and rotation, and an inversion, we just have to show that the theorem holds independently for each of these transformations.

• The image of r|z|2 + kz + kz + s = 0 under the translation z 7→ w = z + a is

r|w − a|2 + k(w − a) + kw − a + s = 0 ⇔ r|w|2 + (k − a)w + (k − a)w + r|a|2 − (ka + ka) + s = 0

The last three terms are real numbers, so this is indeed the equation of a circleline.

• The result is immediate for a scaling and rotation, which corresponds to multiplication by a nonzero complex number

• For the inversion, we distinguish the case in which the circleline is a circle, and the case in which the circleline is a line (i) Say the circleline is a circle, with equation |z − a| = r. If a 6= 0, the image of this circle under inversion is

1 1 r − a = r ⇔ w − = |w| w a |a| and from the previous lemma we know that the equation on the right-hand side is the equation of a circleline. If a = 0, the image of |z| = r under inversion is the circle |w| = 1/r. (ii) Let us now consider the line kz + kz + s = 0. Under inversion, this becomes

k k + + s = 0 ⇔ s|w|2 + kw + kw = 0 w w

which is a circleline ˆ Theorem: Let z1, z2, z3, z4 be distinct points in C. The cross ratio (z1, z2, z3, z4) is a real number iﬀ the four points lie in a circleline.

4 Proof : If (z , z , z , z ) ∈ , S maps {z , z , z , z } to points on the extended x-axis. S−1 sends 1 2 3 4 R z2z3z4 1 2 3 4 z2z3z4 {(z1, z2, z3, z4), 1, 0, ∞} to {z1, z2, z3, z4}. A M¨obiustransformation takes the extended line to a circleline, so {z1, z2, z3, z4} lie in a circleline.

Conversely, if the four points lie in a circleline, Sz2z3z4 sends the circleline to a circleline, which is the x-axis. So (z1, z2, z3, z4) ∈ R .