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Complex analysis concepts, including Möbius transformations and the definition and properties of the cross ratio of four complex numbers. It covers topics such as the definition of a Möbius map, the preservation of the cross ratio under Möbius transformations, and the calculation of cross ratios for specific examples.
Typology: Study Guides, Projects, Research
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1.1. A little history. The study complex numbers arose from try to find solutions to polynomial equations. Al-Khwarizmi (780-850) in his book Algebra had solutions to quadratic equations of various types. Under the caliph Al-Mamun (reigned 813-833) Al-Khwarizmi became a member of the House of Wisdom in Bagdad, The first to solve the polynomial equation x^3 + px = q was Scipione del Ferro (1465-1526). On his deathbed, del Ferro confided the formula to his pupil Antonio Maria Fiore, who subsequently challenged another mathematician Nicola “Tartaglia” Fontana (1500-1557) to a mathematical contest on solving cubics. (The name Tartaglia means “stammerer” a symptom of injuries acquired aged 12 dur- ing the french attack on his home town of Bresca). The night before the contest, Tartaglia rediscovered the formula and won the contest. Tartaglia in turn told the formula (but not the proof) to an influential mathematician Gerolamo Cardano (1501-1576), provided he signed an oath to secrecy. However, from a knowledge of the formula, Cardano was able to reconstruct the proof. Later, Cardano learned that del Ferro, not Tartaglia, had originally solved the problem and then, feeling under no further obligation towards Tartaglia, proceeded to publish the result in his Ars Magna (1545). Cardano was also the first to introduce complex numbers a +
b into algebra, but had misgivings about it. In the Ars magna he observed, for example, that the problem of finding two numbers that add to 10 and multiply to 40 was satisfied by 5 +
−15 and 5 −
−15 but regarded the solution as both absurd and useless. Cardan was also said to have correctly predicted the exact date of his own death (but it has also been claimed that he achieved this by committing suicide!). Rene Descartes (1596-1650), the mathematician and philospher coined the term imaginary: “For any equation one can imagine as many roots [as its degree would suggest], but in many cases no quantity exists which corresponds to what one imag- ines.” Abraham de Moivre (1667-1754), a protestant, left France to seek religious refuge in London at eighteen years of age. There he befriended Isaac Newton. In 1698 he mentions that Newton knew, as early as 1676 of an equivalent expression to what is today known as de Moivres theorem (and is probably one of the best known formulae) which states that:
(cos(θ) + i sin(θ))n^ = cos(nθ) + i sin(nθ)
where n is an integer. (De Moivre, like Cardan, is famed for predicting the day of his own death. He found that he was sleeping 15 minutes longer each night and summing the arithmetic progression, calculated that he would die on the day that he slept for 24 hours.) Leonhard Euler (1707-1783) introduced the notation i =
−1 in his book Introductio in analysin innitorum in 1748, and visualized complex numbers as points with rectangular coordinates, but did not give a satisfactory foundation for complex numbers. In contrast, there are indications that Carl Friedrich Gauss (1777-1855). had been in possession of the geometric representation of complex numbers since
Exercise 1.1. Show that the map φ : C → M 2 (R) (= 2 × 2 real matrices)
φ : x + iy 7 →
x y −y x
is a monomorphism.
1.3. Some useful reference books. (1) R. Churchill and J. Brown, Complex Variables and Applications (ISBN 0-07-010905-2). This is a fairly readable account including much of the material in the course. (2) I. Stewart and D. Tall, Complex Analysis (ISBN 0-52-128763-4). This is a popular and accessible book. (3) L. Alhfors, Complex Analysis: an Introduction to the Theory of Analytic Functions of One Complex Variable (ISBN 0-07-000657-1). This is a classic textbook, which contains much more material than included in the course and the treatment is fairly advanced. (4) S. Krantz and R. Greene, Function Theory of One Complex Variable (ISBN 0-82-183962-4). This is a nice textbook, which contains much more material than included in the course. (5) S. Krantz, Complex Analysis: The Geometric Viewpoint (0-88-385035-4). The first chapter gives a nice summary of some of the ideas in the course. The rest of the book is very interesting, but too geometric for this course.
We denote by Ĉ = C ∪ {∞} the Riemann sphere. There is a natural “stereo- graphic” projection between the sphere (minus the “north pole” (1, 0 , 0)) and the complex plane
π : {(x 1 , x 2 , x 3 ) ∈ R^3 : x^2 + x^22 + x^23 = 1} − {(1, 0 , 0)} → C
defined by
π(x 1 , x 2 , x 3 ) = z :=
x 1 1 − x 3
+i
x 2 1 − x 3
In particular, 0 it the image of the south pole (0, 0 , −1), and the unit circle |z| = 1 is the image of the equator x 3 = 0.
Definition 2.1. A circle on the sphere corresponds to the intersection of x^21 + x^22 + x^23 = 1 with a plane ax + by + cz = d.
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The following is easy to prove. Lemma 2.2. The stereographic projection of a circle on the sphere is either a circle or a line in C.
Proof. The image of the intersection under the projection can be written as a(z + z) − ib(z − z) + c(|z|^2 − 1) = d(|z|^2 + 1)
If we write z = x + iy then
(d − c)(x^2 + y^2 ) − 2 ax − 2 by + (d + c) = 0.
Case I: If c = d then this is the equation of a straight line.
Case II: If c 6 = d then
x^2 + y^2 −
2 ax d − c
2 by d − c
(d + c) d − c
which we can rearrange as ( x −
a d − c
y −
b d − c
a^2 + b^2 + (c^2 − d^2 ) (d − c)^2
It only remains to show that a^2 + b^2 + c^2 − d^2 > 0 to see this is the equation of a circle. However,
|α 0 | = |ax + by + cz| ≤
x^2 + y^2 + z^2 ︸ ︷︷ ︸ =
a^2 + b^2 + c^2
by the usual Cauchy-Schrwartz inequality and we are done.
Note that in the proof there is an equality in the last line only when (a, b, c) = λ(x 1 , x 2 , x 3 ) for some λ, i.e., the plane is tangent to the sphere.
Remark 2.3. The inverse images of a points z ∈ C is a triple (x 1 , x 2 , x 3 ) lying on the sphere and satisfying |z|^2 = (x^21 + x^22 )/(1 − x 3 )^2 = (1 + x 3 )/(1 − x 3 ) (since x^21 + x^22 + x^23 = 1). We can then write
x 1 =
z + z 1 + |z^2 |
since 1 − x 3 =
|z|^2 + 1|
Similarly,
x 2 =
z − z 1 + |z^2 |
and x 3 =
|z|^2 − 1 1 + |z^2 |
We deduce that
(x 1 , x 2 , x 3 ) =
z + z 1 + |z^2 |
z − z 1 + |z^2 |
|z|^2 − 1 1 + |z^2 |
This completes the proof.
Exercise 2.4. Prove the converse, i.e., the preimage of a circle or a straight line in C is a circle on the Riemann sphere.
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Example 2.11. There are four fundamental examples of M¨obius transforma- tions f : ̂C → Ĉ. a) Translations: z 7 → z + b = (^01) .z.z++1b where b ∈ C. These are often called parabolic transformations. b) Rotations: z 7 → az =
√a.z+ 0 .z+√a−^1 where^ |a|^ = 1.^ These are often called elliptic transformations.
c) Expansions (and Contractions): z 7 → λz =
√ λ.z+ 0 .z+ √ λ−^1 with^ λ >^ 1 (or 0 < λ < 1) and λ real. These are often called hyperbolic transformations. d) Inversions: z 7 → (^1) z.
We can use this to deduce the following.
Lemma 2.12. Every M¨obius map can be written as a composition of M¨obius maps of the above type.
Proof. Every Mobius map is a combination of three types of maps f (z) = Az, f (z) = z + B and f (z) = 1/z, where A, B ∈ C, since we can write
az + b cz + d
a c (cz^ +^ d) +^
b − adc
cz + d
a c
b − adc
cz + d
which is a composition of there
z 7 → cz 7 → cz + d 7 →
cz + d
b −
ad c
cz + d
b −
ad c
cz + d
a c
The follow result is very useful. Consider circles in the complex plane of the form C := {z ∈ C : |z − z 0 | = r} where z 0 ∈ C and r > 0.
Theorem 2.13. The image of a circle or straight line under a M¨obius trans- formation is a circle or straight line.
Proof. It suffices to consider three different types of M¨obius transfomations: (1) If f (z) = Az where A ∈ C with A 6 = 0 then the image f (C) is a circle. (2) If f (z) = z + B where B ∈ C then the image f (C) is a circle. (3) If f (z) = 1/z then the image f (C) is a circle. We first observe if z = x+iy lies on a circle centred at z 0 = x 0 + iy 0 of radius r > 0 then
(x − x 0 )^2 + (y − y 0 )^2 = r^2 (1)
Consider the set of z such that |z − p| |z − q|
= k (2)
where p, q ∈ C and k > 0. If p = u + iv and q = s + it then this becomes:
(x − u)^2 + (y − v)^2 = k^2
(x − s)^2 + (y − t)^2
In particular, we can rewrite (1) in terms of (3) by choosing p, q, k such that
x 0 =
(u − k^2 s) 1 − k^2
y 0 =
v − k^2 t 1 − k^2
r^2 = −
u^2 + v^2 − k^2 (s^2 + t^2 ) 1 − k^2
u − sk^2 1 − k^2
v − tk^2 1 − k^2
Finally we can see that if z satisfies (2) then
|z − p| |z − q|
= k ⇐⇒
∣ (^1) z − (^1) p
∣ (^1) z −^1 q
q p
k (4)
i.e., a version of (3) (with p replaced by (^1) p , q replaced by (^1) q and k replaced by (^) pq k). In particlar, the image of the circle given by (2) is a circle given by (3). Theorem 2.14. For distinct z 1 , z 2 , z 3 ∈ Ĉ and distinct w 1 , w 2 , w 3 ∈ Ĉ there exists a unique M¨obius map f : Ĉ → Ĉ such that f (zi) = wi, for i = 1, 2 , 3.
Proof. We first prove existence and then uniqueness.
Existence. Consider the case that z 1 , z 2 , z 3 6 = ∞. Let
S(z) =
(z − z 2 )(z 1 − z 3 ) (z − z 3 )(z 1 − z 2 )
then we see that S(z 1 ) = 1, S(z 2 ) = 0, S(z 3 ) = ∞. Consider
T (z) =
(z − w 2 )(z 1 − w 3 ) (z − w 3 )(z 1 − w 2 )
then we see that T (w 1 ) = 1, T (w 2 ) = 0, T (w 3 ) = ∞. If we define f (z) := S−^1 ◦T (z) then we can then observe that f (zi) := T −^1 ◦ S(zi) = wi (i = 1, 2 , 3). In the case that z 1 = ∞ then we let S(z) = z z−−zz^23 and similarly for T. In the case that z 2 = ∞ then we let S(z) = z z^1 −−zz 33 and similarly for T. In the case that z 3 = ∞ then we let S(z) = (^) zz 1 −−zz^22 and similarly for T.
Uniqueness. We can assume without loss of generality that w 1 = 1, w 2 = 0, w 3 = ∞. (Otherwise we can additionally compose with a M¨obius map g; ̂C → Ĉ taking w 1 , w 2 , w 3 to 1, 0 , ∞, respectively, and then we can apply the following argument to show that f 1 ◦ g−^1 = f 2 ◦ g−^1 , which therefore gives us f 1 = f 2 ). Assume that fj : ̂C → Ĉ are two M¨obius maps j = 1, 2 such that fj (1) = z 1 , fj (0) = z 2 , fj (∞) = z 3. Since S(z) = f 1 − 1 ◦ f 2 (z) is a Mobius transformation we can write
S(z) =
az + b cz + d
Moreover, f 1 − 1 ◦ f 2 : Ĉ → Ĉ fixes the three points 1, 0 , ∞. In particular, we see that
S(0) = b/d = 0 =⇒ b = 0 S(∞) = a/c = ∞ =⇒ c = 0, and S(1) = (a + b)/(c + d) = a/d = 1
from which we deduce that S(z) = z for all z, i.e., f 1 = f 2.
Let us consider a could of examples of this result. Example 2.15. Find the M¨obius transformation f which maps − 1 , 0 , 1 to the points −i, 1 , i. Assume that f (z) = azcz++db. Since f (0) = bd = 1 we have b = d and f (z) = azcz++bb.
Similarly, since f (−1) = − −iaic++bb = 1 =⇒ ic − ib = −a + b and f (1) = iaic++bb = i =⇒ ic + ib = a + b. Adding the last two equations gives c = −ib and subtracting gives a = ib. Thus
f (z) = ibz + b −ibz + b
iz + 1 −iz + 1
i − z i + z
(Formally, we should also multiply the coefficients by constant to get ad − bc = 1)
Step 2: We can trivially translate C 4 between the two parallel lines L 1 , L 2 to precisely two more circles C 5 and C 6 of the same radius such that: C 5 and C 6 are both tangent to both L 1 and L 2 and C 3.
Step 3: We then define C = f −^1 (C 5 ) and C′^ = f −^1 (C 6 ). Since again M¨obius transformations take circles to circles (where straight lines are understood as circles passing through infinity) we deduce that C, C′^ are the only two such circles tangent to C 1 , C 2 , C 3.
Theorem 2.20. Descartes Theorem: Given four mutually tangent circles C 1 , C 2 , C 3 and C 4 whose radii are r 1 , r 2 , r 3 , r 4 then F (r 1 , r 2 , r 3 , r 4 ) = 0 where
F (r 1 , r 2 , r 3 , r 4 ) = 2(r− 1 2 + r 2 − 2 + r− 3 2 + r− 4 2 ) − (r− 1 1 + r− 2 1 + r− 3 1 + r− 4 1 )^2 (1)
Proof. We begin with two simple estimates. Consider a circle E of radius k. (1) We can reflect in it a second circle C of radius r whose centres are separated by d > r + k. In particular, the circles are disjoint and have disjoint interiors. The image circle f (E′) with have radius k^2 r/(d^2 − r^2 ). (2) Consider next a straight line L at a distance b > k from the centre of E. We can reflect it in the circle E and the image is a circle of radius k^2 / 2 b.
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Step 1. Let C 1 , C 2 , C 3 and C 4 be four mutually tangent circles. Choose E to be a (large) circle centred at the tangency point ξ = (x 0 , y 0 ) between C 1 and C 2. We can invert the four circles in E by f , say, to arrive at a configuration (after some rotation and translation) with: L 1 = f (C 1 ) a line given by y = 1; L 2 = f (C 2 ) a line given by y = −1; and C 3 ′ = f (C 3 ) and C 4 ′ = f (C 4 ) being circles of radius 1 with centres (− 1 , 0) and (1, 0), respectively, i.e., the straight lines are “circles” with radius r 1 = r 2 = ∞ and the other two circles have radius r 3 = r 4 = 1. In particular, we observe that it satisfies (1).
Step 2. We can apply the two estimates to the lines L 1 , L 2 and the circles C 3 ′, C′ 4 to obtain
r(C 3 ) =
k^2 x^20 − 2 x 0 + y^20 , r(C 4 ) =
k^2 x^20 + 2x 0 + y^20 , r(C 2 ) =
k^2 2(y 0 − 1) and r(C 1 ) =
k^2 2(y 0 + 1)
The result follows by substituting.
Proceeding inductively we can construct circle packings.
Exercise 2.21. If a M¨obius map f (z) = azcz++bd maps real numbers to real num- bers then show that a, b, c, d are all real numbers. Show that f : H → H is a bijection.
Application 2.22 (Hyperbolic Half-plane). The upper half plane H has the Poincar´e metric written as (dx^2 + dy^2 )/y^2 , i.e., it is similar to the usual Euclidean
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Lemma 2.26. The value tr(f ) is preserved by conjugacy, i.e., if f 1 , f 2 are con- jugate then tr(f 1 ) = tr(f 2 )
Proof. Since every M¨obius map g can be written as a composition of basic M¨obius transformations (translation, inversion and rotation) it suffices to show the result where g is of each type.
Translations: if g(z) = z + β then we can explicitly write
f 2 (z) = g−^1 f 1 g(z) =
a(z − β) + b c(z − β) + d
a(z − β) + b + β(c(z − β) + d)) c(z − β) + d
=
z(a + βc) + (b + βd − βa − β^2 c) cz + (d − βc)
and we see that tr(f 2 ) = (a + βc) + (d − βc) = a + d = tr(f 1 ).
Inversions: If g(z) = 1/z then
f 2 (z) = g−^1 f 1 g(z) =
a/z + b c/z + d
=
a + bz c + dz
and we see that tr(f 2 ) = d + a = tr(f 1 ).
Rotations: If g(z) = αz then
f 2 (z) = g−^1 f 1 g(z) =
α
aαz + b cαz + d
az + b/α cαz + d
and we see that tr(f 2 ) = a + d = tr(f 1 ).
Definition 2.27. We say that a rational map is parabolic it it has precisely one fixed point in Ĉ
In particular, being parabolic is easily seen to be a conjugacy invariant (i.e., if f 1 has a single fixed point z 0 , say, then f 2 has a single fixed point g−^1 z 0. In particular, if f 1 has a fixed point f 1 (z 0 ) = z 0 we can choose g so that g(∞) = z 0 and then the conjugate map f := f 2 fixes ∞.
Lemma 2.28. Let f be a M¨obius map with f (∞) = ∞. (1) Then f is of the form f (z) = αz + β; (2) f has a second distinct fixed point (i.e., it is not parabolic) if and only if α 6 = 1; (3) if the second fixed point is 0 (i.e., f (0) = 0) then f (z) = αz.
Proof. (1) Since f (∞) = ∞ this immediately implies that f 2 (z) = αz + β, for some α, β ∈ C.
(2) If f has a second point z 0 , say, (other than ∞) then f (z 0 ) = αz 0 + β = z 0 , i.e, z 0 = (^) αβ− 1 with α 6 = 1. In particular, if α 6 = 1 then f has a second fixed point (in addition to ∞). Conversely, if α = 1 then f has no other fixed points except ∞ since f (z) = z + β is a straightforward translation. (In the degenerate case β = 0 this is just the identity.)
(3) Since f (0) = 0 we see that β = 0 and the result follows.
Remark 2.29. In fact M¨obius transformations f 1 , f 2 are conjugate if and only if tr(f 1 )^2 = tr(f 2 )^2
The non-identity M¨obius transformations are commonly classified into three types:
(1) parabolic (conjugate to z 7 → z + β with a single fixed point ∞); (2) elliptic (conjugate to z 7 → αz with |α| = 1 with points 0, ∞); and (3) loxodromic (everythingelse). The hyperbolic transformations are those conjugate to the expanding and con- tracting maps z 7 → λz (with 0 < λ < 1 or 1 < λ < +∞) being a subclass of the loxodromic ones
Lemma 2.30. Let f : Ĉ → ̂C be a non-trivial M¨obius transformation. It is (1) a parabolic M¨obius transformation if and only if tr(f ) = ± 2 ; (2) an elliptic M¨obius transformation if and only if − 2 < tr(f ) < 2 ; (3) a hyperbolic M¨obius transformation if and only if tr(f ) > 2. Proof. Assume that f (∞) = ∞ (else replace by a conjugate map with this property).
Case I: f is parabolic. By definition, in this case f has no other fixed points and we see from the lemma see that f (z) = z + β. This always corresponds to tr(g) = 1 + 1 = 2 (or −2 since we recall that multiplying a, b, c, d by −1 gives the same map).
Case II: f is not parabolic. In this case, f has a second fixed point. Assume that f (0) = 0 (else we can replace f by a conjugate map with this property) and then by the lemma we can write f (z) = αz. Then we require a = d−^1 =
α (to satisfy ad − bc = ad = 1). We can now consider the different values of the trace tr(f ) = a + (^1) a. (1) If tr(f ) = ±2 then a + 1/a = ±2 and so a(a^2 ± 2 a + 1) = (a ± 1)^2 = 0. But a = ±1 which means that f (z) = z, the identity map, which is excluded by the non-triviality assumption. The proofs of parts (2) and (3) are slightly jumbled up. (a) More generally, if a = reiθ^ then
tr(f ) = a +
a
= reiθ^ + e−iθ r = (r + 1/r) cos θ + i(r − 1 /r) sin θ
If we assume that the trace in (1) is real (i.e., Im(tr(f )) = (r − 1 /r) sin θ = 0) then this is equivalent to having either r = 1/r(= 1) or θ = 0, π or both. In the first case, if |a| = r = 1 then f is elliptic, and tr(f ) = 2 cos θ ∈ (− 2 , 2). In the second case, if θ = 0 or π then f is hyperbolic and tr(f ) = r + 1/r ≥ 2. (b) If we assume that the trace in (1) has a non-zero imaginary part (i.e., tr(f ) 6 ∈ R) then this is equivalent to requiring both r 6 = 1/r and θ 6 = 0, π. But then since |a| = r 6 = 1 this means that the map is not elliptic (as well as not being parabolic). Thus by definition it is loxodromic. Remark 2.31. A loxodromic Mobius map is a composition of a hyperbolic and elliptic map. In addition, a loxodromic Mobius map has that tr(f ) ∈ C − [− 2 , 2].
Exercise 2.32. Show that if f is hyperbolic then f n(z) converges to one of the fixed points.
Completing the square we can write ∣ ∣ ∣ ∣w^ −^
(ad − bc) ac − ac
2 = |w|^2 +
(ad − bc) ac − ac w +
(bc − ad) ac − ac w +
(ad − bc) ac − ac
2
dd − db ac − ac
(ad − bc) ac − ac
2 (= R^2 ).
In particular, this is seen to be the equation of a circle of radius R provided the right hand side is positive. We can rewrite this as
dd − db ac − ac
(ad − bc) ac − ac
(da − ca) ca − ca
|a|^2 |d|^2 + |b|^2 |c|^2 − abcd − abcd |ac − ac|
Thus it suffices to observe that we can write the numerator as
(ad − bc)(ad − bc) = |ad − bc|^2 > 0
using that ad − bc 6 = 0. In either case we see that f −^1 (R ∪ {∞}) is a circle or straight line
As an immediate corollary we have another presentation of the proof that M¨obius transformation’s preserve circles (although the proof is essentially the same as before).
Corollary 2.38. If f is a M¨obius transformation then it maps circles (or lines) to circles (or lines).
Proof. Let C be a circle (or line) determined by three distinct points z 1 , z 2 , z 3 ∈ C. Then z 0 ∈ C ⇐⇒ (z 0 , z 1 , z 2 , z 3 ) ∈ R ⇐⇒ (f (z 0 ), f (z 1 ), f (z 2 ), f (z 3 )) ∈ R ⇐⇒ (f (z 0 ), f (z 1 ), f (z 2 ), f (z 3 )) ∈ f (C) =: C′
where C′^ is a circle (or line). In particular, the circle (or line) for (z 1 , z 2 , z 3 ) is mapped to the circle (f (z 1 ), f (z 2 ), f (z 3 )).
Finally, the following is an interesting geometric interpretation of cross ratios.
Remark 2.39 (Cross ratios and hyperbolic geometry). We let H^3 = {(z, t) : z = x + iy ∈ C, t > 0 }
denote the three dimensional upper half-space with metric
ds^2 =
dx^2 + dy^2 + dt^2 t^2
We can consider a geodesic γ : R → H^3 with end points
z 1 = lim t→−∞ γ(t) := γ(−∞) and z 2 = lim t→∞ γ(t) := γ(+∞)
More precisely, for any two distinct t 1 < t 2 the curve [t 1 , t 2 ] 3 t 7 → γ(t) is the shortest path between γ(t 1 ) and γ(t 2 ) in H^2. It appears a semi-circular arc in H^3. Given z 3 , z 4 ∈ C we can consider the geodesic γ′^ with end points z 3 = γ′(−∞) and z 4 = γ′(+∞). We denote the Hausdorff distance between the curves γ and γ by d(γ, γ′) = inf t 1 ,t 2 ∈R
d(γ(t 1 ), γ′(t 2 )).
The connection between the cross ratio of the four complex numbers z 1 , z 2 , z 3 , z 4 (i.e., the end points) and the geometry is the following.
Lemma 2.40 (Fenchel and Ahlfors). |(z 1 , z 2 , z 3 , z 4 )| = tanh(d(γ, γ′))
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The argument of (z 1 , z 2 , z 3 , z 4 ) also has a geometric interpretation. It cor- responds to the change in the angle by parallel transporting a frame along the geodesic realizing this distance.
Definition 3.1. Assume that U ⊂ C is an open set, i.e., for every z 0 ∈ U there exists > 0 such that
B(z 0 , ) = {z ∈ C : |z − z 0 | < } ⊂ U.
It is usually also convenient to additionally assume that:
(1) U is path connected (i.e., for any two points z, w ∈ U we can find a continuous path γ : [0, 1] → U such that γ(0) = z and γ(1) = w).
(2) U is simply connected (i.e., any closed path γ : [0, 1] → U with γ(0) = γ(1) can be contracted to a single point or, equivalently, U is homeomorphic to the unit disk).
We will assume that U always has these properties, unless we explicitly state otherwise, and frequently call it a domain. We will present three equivalent definitions once we have introduced three new lots of notation.
3.1. Ingredients for the first definition (using power series). We want to start by recalling the more intuitive definition of analytic functions.