# MA 3B80 University of Warwick Proof of The Riemann Mapping Theorem Complex Analysis

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MA3B80
THE UNIVERSITY OF WARWICK
FOURTH YEAR EXAMINATION: Summer 2020
Complex Analysis
Time Allowed: 3 hours
advance on the departmental ‘Warwick Mathematics Exams 2020’ webpage.
Calculators, wikipedia and interactive internet resources are not needed and are not
permitted in this examination. You are not allowed to confer with other people. You
may use module materials and resources from the module webpage.
ANSWER COMPULSORY QUESTION 1 AND TWO FURTHER QUESTIONS out of
the four optional questions 2, 3, 4 and 5.
PDF document if possible, although multiple files (2 or 3) are permitted. You have
an additional 45 minutes to make the upload, and instructions are available on the
departmental ‘Warwick Mathematics Exams 2020’ webpage.
You must not upload answers to more than 3 questions, including Question 1. If you
do, you will only be given credit for your Question 1 and the first two other answers.
The numbers in the margin indicate approximately how many marks are available for
each part of a question. The compulsory question is worth twice the number of marks
of each optional question. Note that the marks do not sum to 100.
COMPULSORY QUESTION
b :=
1. (a) (i) Define a Möbius transformation mapping the extended complex plane C
b
C ∪ {∞} to C.

(ii) Find a Möbius transformation that maps the left half-plane HL := {z ∈
C : R(z) < 0} onto the unit disk ∆ := {z ∈ C : |z| < 1}, and find a Möbius transformation that maps ∆ onto the right half-plane HR := {z ∈ C : R(z) > 0}.

(b) (i) Give sufficient conditions on D ⊂ C and the curve γ that ensure that for
1
Question 1 continued overleaf
MA3B80
every holomorphic function f : D → C we have
Z
(♣)
f (z) dz = 0.
γ

(ii) Give an example of an open and connected set D, a closed piecewise C 1
curve γ : [a, b] → C, and a holomorphic function f : D → C such that (♣)
in part (i) does not hold (no proof required).
(c) (i) State the Removable Singularity Theorem (no proof required).


(ii) Show that for every holomorphic function f : D → C, D ⊂ C open, and
(z0 )
every z0 ∈ D the function F (z) = f (z)−f
defined on D {z0 } can be
z−z0
extended to a holomorphic function on all of D.

(d) (i) Let f : D → C, D ⊂ C open, be holomorphic. Show that the imaginary
part I(f ) is harmonic on D.

(ii) Give an example of an open and connected set D ⊂ C and a harmonic
function h : D → C such that there is no holomorphic function f : D → C
with I(f ) = h (no proof required).

(e) (i) State Cauchy’s Integral formulas for the derivatives of holomorphic functions.

(ii) Compute
Z
∂B2 (0)
z2
1
dz,
−1
∂B2 (0) = {z ∈ C : |z| = 2}.

(f) (i) State the Riemann Mapping Theorem.

(ii) Show that there is no holomorphic and bijective mapping from ∆ := {z ∈
C : |z| < 1} onto C.  OPTIONAL QUESTIONS 2. (a) (i) Define the four elementary Möbius transformations. 2 Question 2 continued overleaf  MA3B80 (ii) Show that every Möbius transformation f , f (z) := az + b , cz + d a, b, c, d ∈ C, can be written as a composition of the four elementary Möbius transformations in (i).  (b) Denote H+ := {z = x + iy ∈ C : y > 0} the upper half-plane and ∆ := {z ∈
θ ∈ R, the transformation z 7→ reiθ z is a composition of a rotation by θ and a dilation by a factor r.
For c 6= 0 we write
az+b
cz+d
=
a
c
+
c
.
cz+d
Then this map is obtained by
1
c
c
z 7→ cz 7→ cz + d 7→
7

7→ +
.
cz + d
cz + d
c cz + d
The case where c = 0 is even easier.
(2.5.1)
We can often use this decomposition into elementary Möbius transformations in order to prove that
certain properties are preserved under general Möbius transformations: The proof is then reduced to
checking that the property is preserved for elementary Möbius transformations, as in the following.
Keeping in mind our definition of ‘circles’ in C∞ from Definition 2.3, we have
Theorem 2.10. The image of every circle in C∞ under any Möbius transformation is also a circle in
C∞ .
Proof. We only need check this property for each type of elementary Möbius transformation. The
property is obvious for translations, rotations and dilations. The property for the complex inversion
follows from its interpretation as a 180 ◦ rotation, together with the preservation of circles/lines by
stereographic projection given in Remark 2.2.
You proved essentially this theorem by direct calculation on the exercise sheets of Analysis 3.
2.6
Three points determine a Möbius transformation
VIDEO: Three points determine a Möbius transformation
In this section we will see the fundamental (and very useful) property that if we ask for three specific
distinct points in C∞ to be mapped to another three specific distinct points in C∞ , then we determine
a unique Möbius transformation. Precisely, we have:
Theorem 2.11. Given three distinct points z1 , z2 , z3 ∈ C∞ and three distinct points w1 , w2 , w3 ∈ C∞ ,
there exists a unique Möbius transformation f that satisfies f (zi ) = wi for i = 1, 2, 3.
Thus the group of Möbius transformations has three complex degrees of freedom. This illustrates
very clearly how it is a six real-parameter family.
The proof of Theorem 2.11 will be supported by a couple of sub-results. We start with an observation
about the fixed points of Möbius transformations.
17
2.6
Three points determine a Möbius map
Lemma 2.12. Every Möbius transformation other than the identity f (z) = z has at least one, but at
most two, fixed points. In particular, if f is a Möbius transformation and z1 , z2 , z3 ∈ C∞ are distinct
points such that f (zi ) = zi , then f is the identity.
Proof. As usual, we write
az + b
.
cz + d
Observe that f being the identity corresponds to the case a = d 6= 0 and b = c = 0, so if we assume
that this is not the case then we need to show that there can be at most two fixed points.
f (z) =
We split the proof into two cases: First, if c = 0, then f is just a linear transformation f (z) = ad z + db ,
where a, d 6= 0 by nondegeneracy of Möbius transformations. Such a linear transformation has a
b
if a 6= d.
fixed point at infinity, plus at most one further fixed point at z = d−a
In the remaining case that c 6= 0, we have
f (z) =
az + b
=z
cz + d
⇐⇒
(az + b) = (cz + d)z
⇐⇒
0 = cz 2 + (d − a)z − b .
The quadratic formula gives us two solutions, which might coincide.
Next, we give a special case of the desired Theorem 2.11 in which the points wi are explicit. In this
special case we also give formulae for the Möbius transformation, which will be useful later.
Proposition 2.13. Given three distinct points z1 , z2 , z3 ∈ C∞ , there exists a Möbius transformation
f that maps z1 , z2 , z3 to 1, 0, ∞ respectively. In the case that zi 6= ∞ for i = 1, 2, 3, then it is
(z − z2 )(z1 − z3 )
.
(z − z3 )(z1 − z2 )
(2.6.1)
f (z) =
z − z2
,
z − z3
(2.6.2)
f (z) =
z1 − z3
,
z − z3
(2.6.3)
f (z) =
z − z2
.
z1 − z2
(2.6.4)
f (z) :=
In the case that z1 = ∞ we set
in the case that z2 = ∞ we set
and in the case that z3 = ∞ we set
Proof. Clearly each of these functions f are Möbius transformations. By inspection, they send the
points zi to the required image points.
In the final three cases (2.6.2) to (2.6.4) in which one of the zi is ∞, the formulae for f arise from
(2.6.1) by dropping the factor on the numerator and the factor on the denominator containing that zi .
These two factors would cancel in the limit zi → ∞, so this makes sense.
We are finally in a position to prove the main result of this section.
18
2.7
The cross ratio
Proof of Theorem 2.11.
Existence: Let f1 be the function from Proposition 2.13 that sends z1 , z2 , z3 to 1, 0, ∞ respectively.
Let f2 be the function from Proposition 2.13 that sends w1 , w2 , w3 to 1, 0, ∞ respectively. The Möbius
transformation f we seek will simply be f2−1 ◦ f1 .
Uniqueness: Suppose that we have two Möbius transformations f and g, both of which send the
points zi to wi respectively. Then g −1 ◦ f is a Möbius transformation that has all three distinct points
zi as fixed points. Lemma 2.12 then tells us that g −1 ◦ f is the identity, i.e. f ≡ g.
2.7
The cross ratio
VIDEO: The cross ratio
Definition 2.14. The cross-ratio of pairwise distinct z0 , z1 , z2 , z3 ∈ C∞ , is the point in C that is the
image of z0 under the unique Möbius transformation that sends z1 , z2 , z3 to 1, 0, ∞ respectively.
Formulae for the Möbius transformation f , and hence for f (z0 ), are given by Proposition 2.13.
Theorem 2.15. The cross-ratio is invariant under Möbius transformations.
Proof. If f is any Möbius transformation, we have to show that the cross-ratio of (z0 , z1 , z2 , z3 ) is the
same as that of (f (z0 ), f (z1 ), f (z2 ), f (z3 )). If we denote by g the unique Möbius transformation that
sends z1 , z2 , z3 to 1, 0, ∞ respectively, then this cross ratio is g(z0 ) by definition. But then the the
unique Möbius transformation that sends f (z1 ), f (z2 ), f (z3 ) to 1, 0, ∞ respectively is g ◦ f −1 , and so
the cross-ratio of (f (z0 ), f (z1 ), f (z2 ), f (z3 )) is g ◦ f −1 (f (z0 )) = g(z0 ), which is the same.
Theorem 2.16. The cross ratio of (z0 , z1 , z2 , z3 ) is real-valued if and only if z0 , z1 , z2 , z3 all lie on a
common circle in C∞ .
Recall from Definition 2.3 that circles in C∞ restrict to circles and lines in C.
Proof. Since both the cross-ratio and the property of lying on a circle are invariant under Möbius
transformations, we may assume that z1 , z2 , z3 equal 1, 0, ∞ respectively. The cross ratio of (z0 , 1, 0, ∞)
is z0 by definition. But the circle passing through 1, 0 and ∞ is the real line.
Let’s digest a consequence of what we have proved. If z0 , z1 , z2 , z3 ∈ C, then they all lie on a common
line or circle in the plane if and only if
(z0 − z2 )(z1 − z3 )
∈ R.
(z0 − z3 )(z1 − z2 )
Amazing! Try it out on a few examples.
19
2.8
Examples of Möbius transformations
2.8
Examples and special classes of Möbius transformations
VIDEO: Examples of Möbius transformations
Around 23:10 when I said ‘reflection’, I meant ‘rotation’. I’ll explain more in the live lecture.
There are numerous distinguished examples and subclasses of Möbius transformations, of which we
give a few prominent examples.
Example 2.17 (Möbius transformation that gives a bijection from a disc to a half-space). We are
looking for a Möbius transformation that provides a bijection from the disc D := {z ∈ C : |z| < 1} to the upper half-plane H=>0 := {z ∈ C : =(z) > 0}. Since Möbius transformations are
homeomorphisms from C∞ to itself, this will map the boundary of D, i.e. the unit circle, to the
boundary of H=>0 , i.e. the real axis. There are many maps with this property. By Theorem 2.11 we
can pick any three distinct points on the unit circle and map them to 1, 0 and ∞ respectively. For
example, we could pick 1, −i and i on the unit circle, in which case the map is given by (2.6.1) as
f (z) =
z+i
(z − (−i))(1 − i)
=
.
(z − i)(1 − (−i))
iz + 1
By Theorem 2.10, this Möbius transformation will map the entire unit circle ∂D to the entire real
axis. Because f is a homeomorphism, this map must send the disc D either to the upper half-plane
H=>0 , or the lower half-plane. But since it sends 0 to i, it must be the former case, as required.
Example 2.18 (Möbius transformations that give bijections from the disc to itself). Consider the
Möbius transformations of the form
z−w
for w ∈ C, with |w| < 1. (2.8.1) f (z) = w̄z − 1 We claim that Möbius transformations of this form map D onto itself, and map the boundary of D to itself. To see this, we use the easily-checked identity |z − w|2 = |w̄z − 1|2 − (1 − |z|2 )(1 − |w|2 ) to compute |f (z)|2 = |z − w|2 (1 − |z|2 )(1 − |w|2 ) = 1 − . |w̄z − 1|2 |w̄z − 1|2 Because we are assuming that 1 − |w|2 > 0, we see that |f (z)| < 1 if and only if |z| < 1, and |f (z)| = 1 if and only if |z| = 1. This argument also implies that f maps onto D, since f is a bijection from C∞ to itself. An alternative argument to obtain the surjectivity would be to observe that the inverse of f is f itself! Remark 2.19. We can slightly generalise the class of Möbius transformations from Example 2.18 that map the disc D to itself, by composing with a rotation about the origin, giving maps of the form z−w iθ f (z) = e , (2.8.2) w̄z − 1 20 2.8 Examples of Möbius transformations still with w ∈ D, and now with θ ∈ (−π, π]. A stunning fact that will follow from the so-called Schwarz lemma, Theorem 7.20, is that every holomorphic map D 7→ D that is a bijection is of the form (2.8.2). We are not starting with the assumption that the map is a Möbius transformation here. This is true in far greater generality! Think how different this is to what you have seen before. Just imagine if there were only a finite-dimensional family of real differentiable bijective functions from (−1, 1) to itself! Example 2.20 (Möbius transformations that give bijections from H=>0 to itself). A Möbius transformation g(z) from D to itself can be converted into a Möbius transformation h := f ◦ g ◦ f −1 from
H=>0 to itself, where f : D → H=>0 is the Möbius transformation from Example 2.17. One can also
conjugate in the other direction to give g = f −1 ◦ h ◦ f . Because of this, it may seem pointless to consider Möbius transformations from H=>0 to itself after already considering Möbius transformations
from D to itself in Example 2.18.
However, working in the H=>0 viewpoint has some significant advantages in certain situations. To
see one, we return to the isomorphism between the group of Möbius transformations and the group
P SL(2, C) from Section 2.4. The key observation is that the subgroup
P SL(2, R) := SL(2, R)/{±I},
essentially restricting from complex matrices in SL(2, C) to real matrices in SL(2, R), but as before
identifying each pair A and −A, corresponds to an interesting subgroup of Möbius transformations.
Indeed, we claim that they send the upper half plane H=>0 to itself. To see this, we rewrite
f (z) =
(az + b)(cz̄ + d)
ac|z|2 + ad z + bc z̄ + bd
az + b
=
=
.
cz + d
(cz + d)(cz̄ + d)
|cz + d|2
Therefore, keeping in mind that =(z̄) = −=(z) and ad − bc = 1, we have
=(f (z)) =
=(z)
=
=
2
2
|cz + d|
|cz + d|
|cz + d|2
In particular, if z ∈ H=>0 , equivalently =(z) > 0, if and only if =(f (z)) > 0, equivalently f (z) ∈
H=>0 . Thus f maps H=>0 to itself.
Similarly to before, we see that f also maps H=>0 onto itself. For example, one can observe that the
inverse of f is another Möbius transformation of the same form, and thus maps H=>0 to itself.
One can check that every Möbius transformation that maps H=>0 bijectively to itself arises in this
way. Indeed, any such map must send the real line to itself, or to ∞, and if we add the point ∞ ∈ C∞
to the real line to give a circle C, then the Möbius transformation will map this circle bijectively to
itself. If we now take the points x1 , x2 , x3 ∈ C that map to the points 1, 0, ∞ respectively, then we
can construct a Möbius transformation in P SL(2, R) using Proposition 2.13 that has the same effect
on these three points. By Theorem 2.11 itself, the two Möbius transformations must then coincide.
Example 2.21. Nonexaminable example: The rotations of the unit sphere S 2 ,→ R3 are Möbius
transformations when we identify S 2 and C∞ as above. By rotations, we mean elements of SO(3).
Since the group of Möbius transformations is isomorphic to P SL(2, C), by Lemma 2.7, it is natural
21
2.9
Conformal maps
to ask to which subgroup of P SL(2, C) these rotations correspond. It turns out that it is the subgroup
P SU (2). The corresponding Möbius transformations can be written
az − c
z 7→
,
cz + a
for a, c ∈ C with |a|2 + |c|2 = 1.
2.9
Conformal maps
VIDEO: Conformal maps
A general principle in mathematics is that one defines an object with some structure, for example a
vector space with its linear structure, and then one considers bijective maps between different objects
that preserves this structure, for example a bijection between vector spaces that preserves the linear
structure. Such bijections are typically called isomorphisms, and intuitively we view two objects that
are isomorphic as being the same. You will have seen many other examples of this viewpoint, for example isometries between metric spaces, isomorphisms between groups, homeomorphisms between
topological spaces or maybe diffeomorphisms between manifolds (if you have studied manifolds).
What is the right notion of equivalence for domains in C?
(A domain is an open and connected subset.)
Definition 2.22. Two domains Ω1 and Ω2 in C are said to be conformally equivalent if there exists a
bijective holomorphic function ϕ : Ω1 → Ω2 such that the inverse ϕ−1 is also holomorphic.
Such a map ϕ is called a conformal or biholomorphic map.
In due course, we will establish that the inverse ϕ−1 is automatically holomorphic for a bijective
holomorphic function ϕ between domains, but there is no point in fretting about that now.
This notion of equivalence gives us an equivalence relation. In particular, if Ω1 and Ω2 are conformally
equivalent via the conformal map ϕ : Ω1 → Ω2 , and Ω2 and Ω3 are conformally equivalent via the
conformal map ψ : Ω2 → Ω3 , then Ω1 and Ω3 are conformally equivalent via the conformal map
ψ ◦ ϕ : Ω1 → Ω3 . We are implicitly using the chain rule to be sure that the composition of these
conformal maps is conformal. Thus the notion of being conformally equivalent is an equivalence
relation.
The key property that is preserved by this notion of equivalence is the concept of a function on the
domain being holomorphic. More precisely, by the chain rule, a function f : Ω2 → C is holomorphic
if and only if the composition f ◦ ϕ : Ω1 → C is holomorphic.
For the rest of this section we try to get a feeling for which domains are conformally equivalent to
which other domains. Our knowledge of Möbius transformations will help. Throughout the discussion we write the unit disc as
D := {z ∈ C : |z| < 1}. 22 2.9 Conformal maps Before we begin, let’s recall that we already showed in Example 2.17 that the upper half-space is conformally equivalent to the disc D. Let’s find some more such domains. Watch the video for pictures! Example 2.23. We claim that the upper right quarter of the complex plane Q := {z ∈ C | 0 and =(z) > 0}
is conformally equivalent to the disc D.
To see this, we first observe that Q is conformally equivalent to the upper half plane H=>0 by virtue
of the conformal map z 7→ z 2 .
The upper half plane is then conformally equivalent to D thanks to Example 2.17.
Example 2.24. We claim that the upper half disc
D=>0 := {z ∈ C : |z| < 1 and =(z) > 0}
is conformally equivalent to the whole disc D.
Danger: It is very tempting to try to use the conformal map z 7→ z 2 to do this job, but this actually
shows the quite different fact that D=>0 is conformally equivalent to the disc D with the real interval
[0, 1) removed!
Instead, we notice that D=>0 is conformally equivalent to Q via the unique Möbius transformation
that sends −1, 0 and 1 to 0, 1 and ∞ respectively. One could compute this explicitly, but we can argue
more geometrically as follows: First, we know from Theorem 2.11 that this Möbius transformation
exists. Second, by the preservation of circles in C∞ from Theorem 2.10 it sends the interval [−1, 1]
to the interval [0, ∞]. Third, by Theorem 2.10 again it must send the semicircle {z ∈ C : |z| =
1 and =(z) ≥ 0} to a half line starting at 0 and going in some direction off to infinity. Finally, the
right-angle in the boundary of D=>0 at −1 must be reflected in a right angle in the boundary of
the image of D=>0 under this Möbius transformation, because the Möbius transformation preserves
angles, and so the half line must be the positive imaginary axis.
Now we have shown that D=>0 is conformally equivalent to Q, the claim follows by Example 2.23.
By this point you may be getting the false impression that every domain is conformally equivalent to
D. But this is not true. One type of counterexample would be to take the domain that is the whole of
C. If we could find a conformal map from C to D then this would be a bounded holomorphic function
on C, and Liouville’s theorem that was mentioned in Analysis 3 would tell us that it would have to
be constant, and certainly not then surjective onto D. (We will revisit Liouville’s theorem later, in
Corollary 6.8, once we have rigorously proved Cauchy’s theorem.)
Another sort of domain that would certainly fail to be conformal to the disc D would be a domain
‘with holes’ such as an annulus {z ∈ C : a < |z| < b}, where 0 < a < b < ∞. This domain is not even homeomorphic to D so it is a bit much to ask for it to be homeomorphic via a homeomorphism that is additionally a conformal map! To see that they are not homeomorphic, it suffices to notice 23 2.9 Conformal maps that one is simply connected while the other is not. The notion of simply connected has been briefly mentioned in Analysis 3. Informally it means that every loop in the space can be deformed to a point. In order to make this precise we need to define the notion of homotopy. You may have seen homotopies in greater generality in the course Introduction to Topology. Definition 2.25 (Homotopic). Let Ω ⊂ C be open and let γ1 , γ2 : [a, b] → Ω be two continuous curves (i.e. continuous maps from an interval) with the same endpoints γ1 (a) = γ2 (a) and γ1 (b) = γ2 (b). Then γ1 and γ2 are called homotopic if there exists a continuous mapping h : [0, 1] × [a, b] → Ω such that for all t ∈ [a, b] and all s ∈ [0, 1] we have h(0, t) = γ1 (t), h(1, t) = γ2 (t), h(s, a) = γ1 (a), and h(s, b) = γ1 (b). (2.9.1) Such a mapping h is called a homotopy. The notion of Ω being simply connected is intuitively that it does not contain any holes. Definition 2.26 (Simply connected). An open set Ω ⊂ C is said to be simply connected if it is connected and every closed1 continuous curve γ : [a, b] → Ω is homotopic to the constant curve γ̃ : [a, b] → Ω defined by γ̃(t) = γ(a) = γ(b). In more general situations the definition of simply connected would ask for path connectedness, but path connectedness is equivalent to connectedness when considering an open set. In this very special situation of open sets Ω in C, the notion of being simply connected is equivalent to both Ω being connected, and its complement in the Riemann sphere C∞ being connected. We will not explicitly use this formulation. At this point we can look forward to the end of the course when we prove one of the greatest theorems in the subject, namely the Riemann mapping theorem. That theorem will tell us that every simply connected domain Ω ⊂ C other than Ω = C is conformally equivalent to the disc D. See Theorem 11.1. 1 γ : [a, b] → C is said to be closed if γ(a) = γ(b), i.e. the curve closes up. 24 1 1 b a r Figure 2: Stereographic projection, cross-section 2.10 Exercises 2.1. By considering a triangle similar to the red triangle in Figure 2, prove that r = formula (2.2.1) for stereographic projection. a . 1−b Deduce the Think of Figure 2 as a cross-section in the picture describing stereographic projection. 2.2. Invert formula (2.2.1) to give (2.2.2). Hint: Solve first for x3 . Then get x1 and x2 . 2.3. We claimed in Remark 2.2 that straight lines in C are in one-to-one correspondence with circles in S 2 that pass through the north pole N , with the correspondence being given by stereographic projection π. By considering appropriate planes in R3 that pass through N , give a geometric justification of this fact. You could also check this correspondence by solving equations. See the next question. 2.4. We claimed in Remark 2.2 that circles in C are in one-to-one correspondence with circles in S 2 that don’t pass through the north pole N , with the correspondence being given by stereographic projection π. Verify this by writing down the equations of the appropriate circles. You might also try to give a geometric proof of this, but it’s trickier than the previous question! Hint: Let’s remember some basic geometric facts from school mathematics. If n := (a, b, c) is a unit vector, then the plane through the origin with normal vector n is given by (x1 , x2 , x3 ).n = 0. More generally if we shift that plane in the direction n by a distance d ∈ (−1, 1), then the equation is (x1 , x2 , x3 ).n = d, i.e. ax1 + bx2 + cx3 = d. (2.10.1) The case that N = (0, 0, 1) lies within this plane is then precisely that (0, 0, 1).n = d, i.e. c = d. The intersection of any such plane with S 2 is a circle in S 2 . All such circles arise in this way. Now plug in the values (x1 , x2 , x3 ) given by the formula (2.2.2) into the formula (2.10.1), to give a formula for the image of this circle under π. 2.10 Exercises π −1 (z) 1 s r Figure 3: Antipodal points In the case c = d you should end up with an equation of a line in the plane. (I obtained ax + by = d.) This is the case of the last question. In the case c 6= d, you should get the equation of a√ circle. By completing the square, you find b 1−d2 a , d−c ). The radius should be |d−c| . that the centre is ( d−c 2.5. The points 0 and ∞ in C∞ , when viewed as points in S 2 by applying π −1 , are the south and north poles respectively. They are therefore antipodal points. Given any other point z ∈ C∞ , i.e. z ∈ C that is nonzero, show that −1/z̄ ∈ C corresponds to the antipodal point. More precisely, show that π −1 (z) and π −1 (−1/z̄) are antipodal. If you first convince yourself that z, 0 and −1/z̄ are co-linear in C, then you can work in a cross-section as in Figure 3. 2.6. In Example 2.17 we mapped the unit disc to the upper half plane with a Möbius transformation. Viewed as a transformation of the Riemann sphere, seen as a 2-sphere in R3 , what is this transformation? What is the square of this transformation? That is, which Möbius transformation is it, and what does it look like as a transformation of the Riemann sphere seen as a 2-sphere in R3 ? 2.7. Where does the map 1+z 1−z send the unit disc? What is the inverse of this map? What is it as a transformation of the Riemann sphere S 2 ? f (z) = 2.8. Show that the quarter disc Q̂ := {z ∈ C : |z| < 1 and =(z) > 0 and 0}
is conformally equivalent to the unit disc D.
2.9. Show that the slit plane
S := C [0, ∞)
is conformally equivalent to the unit disc D.
26
2.10
Exercises
2.10. Construct explicitly the Möbius transformation that was constructed geometrically in Example
2.24.
2.11. Construct explicitly the unique Möbius transformation f that sends D to the half space H0 ,
while sending 0 to 12 and sending −1 to 0. By composing with the map z 7→ z 2 and then
z 7→ z − 41 , show that the map
z
K(z) :=
(1 − z)2
is a conformal map from D to the slit plane C (−∞, − 41 ].
This function is known as the Koebe function, and we will revisit it once we know a little more
theory. Amongst all the conformal maps from D to C(−∞, − 41 ], it is the unique one satisfying
the normalisation K(0) = 0 and K 0 (0) = 1.
27
3
Review of material from MA244 Analysis 3 – second portion
3.1
Power series
VIDEO: Power Series
In Section 4.2 of Analysis 3 you learned about power series, i.e. expressions of the form

X
an z n ,
n=0
where an is a complex-valued sequence.
Theorem 3.1 (Theorem 4.13 from Analysis 3). Given a complex-valued sequence (an ), define the
R :=
Then the power series
1
∈ [0, ∞].
lim sup |an |1/n

X
an z n
n=0
converges for all |z| < R and diverges for all |z| > R.
Nothing is claimed in this theorem about z for which |z| = R. That question can be somewhat
delicate. The cases considered in the theorem follow easily from the root test.
Power series can be differentiated term-by-term within their radius of convergence:
Theorem 3.2 (Theorem 4.15 from Analysis 3). If the radius of convergence R of the power series
f (z) =

X
an z n
n=0
can define a single valued choice of arg(z), although different people may make a choice of arg(z)
that differs by a fixed constant multiple of 2π. The ray here is known as a branch cut.
A particularly useful resolution of this 2π ambiguity will be given in Lemma 4.1.
The (complex) logarithm is defined for z 6= 0 by
log(z) = log |z| + i arg(z).
Here the logarithm on the right-hand side is taking real values, so it is uniquely defined. But the
argument is only defined up to the addition of an integer multiple of 2π, so log(z) is only defined
up to the addition of an integer multiple of 2πi. As for the argument arg(z), we are sometimes
content with this state of affairs, and sometimes we ask arg(z) to take its principal value in (−π, π],
in which case log(z) is a well defined function that extends the usual logarithm. Unfortunately it is
2
even smooth, i.e. infinity real differentiable
31
3.4
Complex integration
discontinuous across the negative real axis {x < 0} ⊂ C, and we sometimes make a branch cut by removing the half-line. The complex logarithm inherits many of the useful properties of the real logarithm. For example, it is the inverse of the exponential function in the sense that elog z = elog |z|+i arg(z) = |z|ei arg(z) = z, and log(ez ) = log |ez | + i arg(ez ) = log ex + i arg(eiy ) = x + iy = z, (3.3.3) where the latter computation is carried out modulo 2πi. By (3.3.2) we have the familiar identity log(zw) = log |zw| + i arg(zw) = log |z| + log |w| + i(arg(z) + arg(w)) = log z + log w, (3.3.4) again modulo 2πi. Beware that identities that are claimed modulo 2πi should not be expected to work if we take the principal values of the functions log or arg. As you saw in Analysis 3, the function log(z) allows us to define what it means to raise a complex number to a complex power, albeit with complications arising from log(z) only being defined up to a multiple of 2πi. 3.4 Complex integration VIDEO: Complex integration Recall that in Analysis 3 you defined what it means to integrate a suitable (e.g. continuous) function f : [a, b] → C. You defined Z b Z b Z b f (t)dt := 0 we could write Z Z f (z)dz := f (z)dz, ∂Br (a) γ where γ : [0, 2π] → C is defined by γ(θ) = a + reiθ . Alternatively, if A := {z ∈ C : R1 < |z| < R2 } is an annulus, then we have two boundary components that are parametrised by curves γ1 , γ2 : [0, 2π] → C defined by γ2 (θ) = R2 eiθ and γ1 (θ) = R1 e−iθ , and we analogously define Z Z Z f (z)dz := f (z)dz + f (z)dz. ∂A γ1 γ2 In both Definitions 3.7 and 3.9, we say that the curve is closed if γ(a) = γ(b). Informally this means that it closes up into a loop. The curve is a simple closed curve if whenever a ≤ x < y ≤ b and γ(x) = γ(y), we have x = a and y = b. 3.5 Anti-derivatives, and a baby version of Cauchy’s theorem VIDEO: Anti-derivatives; baby Cauchy’s theorem Later, in Section 5, we will turn our attention to a theorem that is at the heart of complex analysis, namely Cauchy’s theorem. Loosely speaking it will tell us that in certain situations when we integrate holomorphic functions around closed curves we obtain zero. In this section we see a baby version of this theory in which our holomorphic function f is the derivative of some other holomorphic function F. 34 3.5 Anti-derivatives; baby Cauchy’s theorem Lemma 3.10. Suppose Ω ⊂ C is open. Suppose further that f : Ω → C and F : Ω → C are holomorphic with F 0 (z) = f (z). If γ is a piecewise C 1 closed curve in Ω. Then Z f (z)dz = 0. γ This lemma follows immediately from the following type of fundamental theorem of calculus. Lemma 3.11. Suppose that F : Ω → C is holomorphic, with F 0 continuous. Suppose further that γ : [a, b] → Ω is a piecewise C 1 curve. Then Z F 0 (z) dz = F (γ(b)) − F (γ(a)). (3.5.1) γ In particular, if γ is a closed curve (i.e. γ(b) = γ(a)) then we have R γ F 0 (z) dz = 0. Proof. By the definition of contour integration and the chain rule of Lemma 1.5, we have Z Z b Z b d 0 0 0 F (z) dz = F (γ(t)) γ (t) dt = F (γ(t)) dt = F (γ(b)) − F (γ(a)). γ a a dt Note that to be able to apply the usual fundamental theorem of calculus in the last equality, we need some regularity on the integrand such as continuity, which is why we are assuming that F 0 is continuous. The result assumes that the derivative F 0 of the holomorphic function F is continuous. Later we will see that this is always true, but we can’t assume that now or our arguments will be circular. Corollary 3.12. Suppose n ∈ Z does not equal −1. Then for γ : [a, b] → C {0} any piecewise C 1 closed curve, we have Z z n dz = 0. γ Proof. If we define F (z) := z n+1 /(n + 1), then F 0 (z) = z n , so the result follows from Lemma 3.10. This corollary fails in a very important way if n = −1! In Q. 3.8 you will compute: Example 3.13. For r > 0 and k ∈ Z let γ : [0, 2π] → C be the closed C 1 curve γ(θ) = reikθ that
travels anticlockwise k times around the circle of radius r. Then
Z
dz
= 2πi k.
γ z
35
3.6
Exercises
P
n
3.1. Suppose Alice has P
a power series ∞
n=0 an z with radius of convergence R1 > 0, and Bob

has a power series n=0 bn z n with radius of convergence R2 ≥ R1 . Suppose we are told that
these power series give the same function on the ball BR1 (0) where they are both converging.
Prove that an = bn for every n ∈ N0 := {0, 1, 2, . . .}, i.e., the power series and their radii of
convergence agree.
P
1
k
3.2. Write the function z 7→ 1−z
as a power series ∞
k=0 ak z and give its radius of convergence.
3.3. By using the previous question and a theorem from Section 3.1, write down the power series of
1
the function z 7→ (1−z)
2 and its radius of convergence.
We will use this exercise when we take a closer look at the Koebe function.
P
1
k
3.4. Suppose w ∈ C {0}. Write the function z 7→ w−z
as a power series ∞
k=0 ak z and give its
We will use this fact when we prove Taylor’s theorem.
P
k
3.5. Suppose that n ∈ N, and that the power series ∞
k=n ak z , which omits the first n terms of a
general power series, has radius of convergence R > 0 and thus defines a holomorphic function
f : BR (0) → C. Prove that there exists a holomorphic function g : BR (0) → C such that
f (z) = z n g(z) for all z ∈ BR (0),
and that g can be written as a power series with radius of convergence R.
We’ll use this fact when we study the zeros of holomorphic functions.
3.6. Consider the function f : C {0} → C defined by
f (z) =
sin z
.
z
Prove that we can extend f to a function on the whole of C (by defining f (0) to be a suitable
value in C) that is entire, i.e. holomorphic on the whole of C.
The ‘singularity’ of f at 0 in this example will be known as a ‘removable singularity’.
3.7.
(a) For R > 0, Compute
1
2i
Z
z̄ dz,
∂BR (0)
and show that it agrees with the area of the ball BR (0).
(b) For R := {z ∈ C : 0,
compute
Z
1
z̄ dz,
2i ∂R
and show that it agrees with the area of R.
This is not a fluke. We are seeing a couple of instances of a general fact that could be derived
from an appropriate form of Stokes’ theorem.
3.6
Exercises
3.8. For r > 0 and k ∈ Z let γ : [0, 2π] → C be the closed C 1 curve γ(θ) = reikθ that travels
anticlockwise k times around the circle of radius r. Prove that
Z
dz
= 2πi k.
γ z
37
4
Winding numbers
4.1
Winding numbers of continuous closed paths
VIDEO: Winding numbers of continuous closed paths
Watch the video for an instant explanation-by-pictures of what the winding number is!
In Section 3.3 we have discussed the function arg(z), and the issue that it is only defined modulo an
integer multiple of 2π. The following lemma will tell us that if we decide on a choice of arg(z) at one
point z, and then move along a continuous path/curve that stays away from 0, then this determines a
unique continuously varying choice of the argument along this path.
Lemma 4.1 (Lifting lemma). Suppose γ : [a, b] → C {0} is continuous, and fix θ0 ∈ R such that
γ(a) = |γ(a)|eiθ0 . Then there exists a unique continuous function θ : [a, b] → R such that θ(a) = θ0
and γ(t) = |γ(t)|eiθ(t) for all t ∈ [a, b].
For example, for a curve γ : [0, 2π] → C given by γ(t) = eit , if we choose θ0 = 0 rather than any
other value in 2πZ then θ(t) = t. In particular, even though the start and end points are the same, the
argument differs by 2π.
Those of you who have already studied Introduction to Topology will know how to prove this lemma.
There is an obvious ‘covering map’ R 7→ R/(2πZ) given by θ 7→ θ + 2πZ, and we are taking a lift of
the function arg ◦ γ : [a, b] → R/(2πZ).
For the Introduction to Topology course in 2020/21, Section 8.3 talks about the homotopy lifting
property, and its special case, Definition 8.6, the path lifting property. In Section 10, Proposition
10.1, you saw that covering maps satisfy the homotopy lifting property.
For those who have not taken the course Introduction to Topology, here is a self-contained proof.
The video could be useful in order to understand this proof!
Proof. First observe that if γ avoids a slit {−reiθ0 : r > 0} on the opposite side of the starting point,
then the existence of θ(t) is clear: We just ask θ(t) to take values in the interval (θ0 − π, θ0 + π).
In the general case, we are free to replace γ(t) by the curve γ̃(t) := γ(t)/|γ(t)| without changing the
argument. Since γ̃ is a continuous function from a closed interval, it is uniformly continuous, and by
dividing up [a, b] into a large enough number of equal intervals, we can be sure that γ̃ does not move
too far when restricted to each of these sub-intervals. More precisely, by taking n ∈ N large enough,
we can be sure that for each k ∈ {0, 1, . . . , n − 1} and each t ∈ [ck , ck+1 ], where ck := a + nk (b − a),
we have |γ̃(t) − γ̃(ck )| < 1. The idea then is to do the lifting on each of these sub-intervals in turn. Indeed, the restriction of γ̃ to [c0 , c1 ] must avoid a slit {−reiθ0 : r > 0} on the opposite side of the starting point, so by the
38
4.2
Nearby closed paths have the same winding number
comment at the start of the proof we can find our function θ(t) at least for t ∈ [c0 , c1 ], with θ(c0 ) = θ0 .
At this point we can use θ(c1 ) as a new starting argument analogous to θ0 , and do our lifting on the
next interval [c1 , c2 ], where γ̃(t) avoids the new opposite slit {−reiθ(c1 ) : r > 0}. This extends
θ(t) to the interval [c0 , c2 ]. By repeating this process a total of n times, we obtain a lift to the whole
interval [c0 , cn−1 ] = [a, b].
To establish uniqueness of θ(t), suppose we have a second continuous lift θ̂(t) also with θ̂(a) = θ0 .
Then t 7→ θ(t) − θ̂(t) is a continuous function that vanishes at t = a, and takes values in 2πZ because
both θ(t) and θ̂(t) represent the argument, modulo 2π. Thus θ(t) − θ̂(t) = 0 for all t ∈ [a, b], as
required.
Lemma 4.1 allows us to unambigously define the change in argument as we move along a continuous
path.
Definition 4.2. Suppose γ : [a, b] → C {0} is continuous, and let θ : [a, b] → R be a function
arising in Lemma 4.1. We define
](γ) := θ(b) − θ(a).
The function θ was only defined up to a constant multiple of 2π that was determined by θ0 . However,
when we subtract θ(a) from θ(b) this unknown multiple of 2π will disappear.
Definition 4.3. Suppose γ : [a, b] → C {0} is a closed continuous path. Then we define the index
or winding number of γ around 0 to be
I(γ, 0) :=
1
](γ) ∈ Z.

More generally, if w ∈ C and γ : [a, b] → C {w} is a closed continuous path then we define the
index of γ around w to be
1
I(γ, w) :=
](γw ),

where γw : [a, b] → C {0} is the path γ translated to send w to the origin, i.e. γw (t) := γ(t) − w.
Example 4.4. For n ∈ Z, consider the curve γ : [0, 2π] → C defined by γ(θ) = reinθ , for some
r > 0, which winds around the origin n times in an anticlockwise direction. Then
I(γ, 0) = n.
Remark 4.5. In the case that γ maps into a region of C on which we can make a choice of arg(z),
for example if γ : [a, b] → C − {reiθ : r > 0} is a continuous closed path (see Section 3.3 for a
discussion of branch cuts) then we can choose θ(t) = arg(γ(t)), and we see that I(γ, 0) = 0. The
branch cut prevents γ from winding around the origin.
4.2
Nearby closed paths have the same winding number
39
4.3
Winding number and simply connected domains
VIDEO: Nearby closed paths have the same winding number
In this section we prove that if we have a closed path γ : [a, b] → C {0}, then a small-enough
perturbation of γ will wind round 0 the same number of times as γ itself.
In the video this will be completely self evident!
However, a little care is required. If γ goes very close to 0 then we must ensure that we perturb very
Lemma 4.6. Suppose γ : [a, b] → C Bε (0) is a continuous closed path, and γ̃ : [a, b] → C {0} is
a continuous closed path with |γ(t) − γ̃(t)| < ε for every t ∈ [a, b]. Then I(γ, 0) = I(γ̃, 0). Note that if γ : [a, b] → C {0} is a continuous closed path then we can always find some ε > 0 such
that the hypotheses of the lemma are satisfied. The smaller that is, the less we can perturb.
Proof. Let θ(t) and θ̃(t) be lifts of the arguments of γ(t) and γ̃(t) respectively, as given by Lemma
4.1. Define a continuous function α : [a, b] → R by α(t) := θ̃(t) − θ(t). By definition of winding
number, we have
1
1
[θ̃(b) − θ̃(a)] −
[θ(b) − θ(a)]
I(γ̃, 0) − I(γ, 0) =

(4.2.1)
1
=
(α(b) − α(a)) .

Because the winding number is always an integer, the difference α(b) − α(a) must be an integer
multiple of 2π, and our task is to show that it is zero.
If this were not the case, then α(t) would either increase or decrease by at least 2π as t varied from
a to b. By continuity of α, we could be sure that there would exist some t0 ∈ [a, b] with α(t0 ) = π2 ,
modulo 2π, and we will use this to derive a contradiction.
By dividing the hypothesis |γ(t) − γ̃(t)| < ε by |γ(t)|, which is at least as large as ε, we find that 1− which forces γ̃(t) γ(t) γ̃(t) < 1, γ(t) to have positive real part. Draw a picture or watch the video! But iπ γ̃(t) γ(t) = | γ̃(t) |eiα(t) , γ(t) and at t = t0 this will be purely imaginary because eiα(t0 ) = e 2 = i, giving a contradiction. 4.3 Winding number and simply connected domains 40 4.4 The winding number as an integral VIDEO: Winding number and simply connected domains In Definition 2.26 we recalled what it meant for an open subset Ω ⊂ C to be simply connected. The most useful consequence of being simply connected for us will be: Theorem 4.7. If an open set Ω ⊂ C is simply connected then for every w ∈ C Ω and every continuous closed path γ : [a, b] → Ω, we have I(γ, w) = 0. In the video we will draw a picture that makes this seem completely obvious. Proof. In order to simply notation, we translate Ω and γ by −w in order to reduce to the case that w = 0. Note that then 0 ∈ / Ω. Because Ω is simply connected, there exists a homotopy of γ to a constant path. More precisely, there exists a continuous map h : [0, 1] × [a, b] → Ω such that for all t ∈ [a, b] and all s ∈ [0, 1] we have h(0, t) = γ(t), h(1, t) = γ(a), h(s, a) = γ(a), and h(s, b) = γ(a). (4.3.1) In particular, for each s ∈ [0, 1], we have a simple closed curve γs : [a, b] → Ω defined by γs (t) := h(s, t). The theorem will be proved if we can show that the index I(γs , 0) is the same for each s ∈ [0, 1], because I(γ0 , 0) = I(γ, 0) and I(γ1 , 0) = 0 since γ1 is a constant path. Because h is continuous, and [0, 1] × [a, b] is closed, the image of h will be closed in Ω ⊂ C {0}, and in particular we can find ε > 0 so that this image does not intersect Bε (0).
Because h is continuous on its closed domain, it is also uniformly continuous. In particular, we can
pick δ > 0 so that whenever t ∈ [a, b] and s1 , s2 ∈ [0, 1] with |s1 − s2 | < δ, we must have |γs1 (t) − γs2 (t)| = |h(s1 , t) − h(s2 , t)| < ε. By Lemma 4.6, we then have I(γs1 , 0) = I(γs2 , 0). This implies that s 7→ I(γs , 0) is locally constant, and then constant for all s ∈ [0, 1]. 4.4 The winding number as an integral VIDEO: The winding number as an integral In the last minute or two of the video I refer to something that is ‘coming in the next section’. But I subsequently decided to move that material to Corollary 3.12 back in Section 3.5. This makes the ordering more logical. If we have a simple closed path that is not just continuous but also piecewise C 1 , then we can characterise the winding number as an integral: 41 4.4 The winding number as an integral Lemma 4.8. If w ∈ C and γ : [a, b] → C {w} is a closed piecewise C 1 curve, then Z 1 dz I(γ, w) = . 2πi γ z − w This expression for the winding number is often used as the definition. It is often easier to use in rigorous proofs, but the definition we gave is more immediately visual, more general, and is useful when considering homotopies. In Example 3.13 and Q. 3.8 we showed that for r > 0, n ∈ Z, and γ : [0, 2π] → C defined by
γ(θ) = reinθ , which winds around the origin n times in an anticlockwise direction, we have
Z
1
1
dz = n.
2πi γ z
Thus by Example 4.4 the claimed formula at least works for I(γ, 0).
Proof of Lemma 4.8. By translation, we may assume that w = 0. We give the proof assuming that γ
is C 1 . The modifications to handle piecewise C 1 curves are straightforward. We must control
Z
Z b 0
dz
γ (t)
=
dt.
γ z
a γ(t)
Taking any function θ(t) from the Lifting lemma 4.1, so γ(t) = |γ(t)|eiθ(t) , we notice that because γ
is C 1 and keeps away from 0, the function θ(t) is also C 1 and we can compute
γ 0 (t) = eiθ(t)
and so
d
|γ(t)| + |γ(t)|iθ0 (t)eiθ(t) ,
dt
γ 0 (t)
d
=
log |γ(t)| + iθ0 (t).
γ(t)
dt
Integrating gives

Z
Z b
dz
d
0
=
log |γ(t)| + iθ (t) dt = 0 + i[θ(b) − θ(a)] = i](γ) = 2πiI(γ, 0)
dt
γ z
a
because γ is closed.
Given our definition of the winding number I(γ, w) as the number of times a curve γ winds around a
point w, it is intuitively obvious that it will be constant as we vary w continuously without touching
the image of γ. We formulate and prove a precise version of this assertion in the case that γ is
piecewise C 1 .
Lemma 4.9. Suppose γ : [a, b] → C is a piecewise C 1 closed curve. Then on each connected
component of C γ([a, b]), the function w 7→ I(γ, w) is constant.
42
4.4
The winding number as an integral
Note that as the continuous image of a closed set, we know that γ([a, b]) is closed, and so C γ([a, b])
is open.
Proof. It suffices to prove that the map w 7→ I(γ, w) is holomorphic, with derivative zero, for w not
in the image of γ (so that I(γ, w) is defined). That then implies that on each component of Cγ([a, b])
the function is constant, as required. To see this last implication, one can recall that a holomorphic
function with zero derivative has all its partial derivatives equal to zero, so it is locally constant and
hence constant on each connected component.
From the definition of

∂ w̄
we have

1
I(γ, w) =
∂ w̄
2πi
Z
γ

∂ w̄

1
z−w

dz = 0.
Here we are using the fact that we can differentiate under the integral sign because the partial derivatives of the integrand with respect to the real and imaginary components of w are uniformly continuous
for z = γ(t), t ∈ [a, b], and w in a small neighbourhood of its starting point.
Differentiation under the integral sign was covered in Analysis 3, and I am not planning to examine
you on it in this context.
On the other hand
1

I(γ, w) =
∂w
2πi
Z
γ

∂w

1
z−w

by Corollary 3.12.
43
1
dz =
2πi
Z
γ
1
dz = 0
(z − w)2
4.5
Exercises
4.1. Given a continuous path γ : [a, b] → C {0}, we write −γ for the continuous path [a, b] 7→
C {0} defined by t 7→ γ(a + b − t), which reverses the direction.
(a) Verify that ](−γ) = −](γ).
(b) Now suppose additionally that γ is closed. What is I(−γ, w) in terms of I(γ, w)?
4.2. Suppose that Ω ⊂ C contains the closure of the ball Br (a) of radius r > 0 centred at a ∈ Ω.
Prove that for all z0 ∈ Br (a) we have
I(∂Br (a), z0 ) = 1
by showing that we may as well take z0 = a and computing.
Recall that ∂Br (a) refers to a curve γ passing once around ∂Br (a) in an anticlockwise direction. Please use the integral formulation of index, i.e. Lemma 4.8, to make it easier to formulate
a rigorous argument.
4.3. Prove that if γ : [a, b] → C is a closed piecewise C 1 curve then the set of points w ∈ Cγ([a, b])
for which I(γ, w) 6= 0 is bounded.
1
4.4. In Q. 3.4, hopefully you wrote (for w ∈ C {0}) the function z 7→ w−z
as a power series
P∞
−k−1 k
z valid for |z| < |w|. By replacing z by 1/z and setting w = 1/z0 , where k=0 w 1 z0 ∈ Br (0) for some r > 0, obtain an expansion for z−z
that is valid for |z| > |z0 | and
0
converges uniformly for z in ∂Br . By integrating around ∂Br (0), prove that
I(∂Br (0), z0 ) = 1
thus reproving Q. 4.2.
5
Cauchy’s Theorem
Baron Augustin-Louis Cauchy (1789 – 1857).
5.1
Preamble and connection with Analysis 3
VIDEO: Cauchy’s theorem: Analysis 3 reminder
In Analysis 3 (Theorem 4.29 in 2019/20) you heard about the iconic theorem of Cauchy from which a
spectacular amount of wonderful theory gushes forth. To restate the particular form of the result that
you saw, we need the notion of simply connected from Definition 2.26.
Theorem 5.1 (Cauchy’s theorem on simply connected domains). Suppose Ω ⊂ C is open and simply
connected. Suppose further that f : Ω → C is holomorphic and γ is a piecewise C 1 closed curve in
Ω. Then
Z
f (z)dz = 0.
γ
Cauchy’s theorem has a huge number of applications. An example application that you saw in Analysis 3 (and that we’ll review shortly) is that a function f : Ω → C that is holomorphic is necessarily
infinitely differentiable. Amazing. This is nothing like what happens for real differentiable functions.
Just think of the function f : R → R that is zero for x < 0 and equal to x2 for x ≥ 0. You also saw, in Analysis 3, a heuristic proof of Theorem 5.1 that, strictly speaking, requires a few extra hypotheses. Indeed, you saw a proof somewhat along the lines that Cauchy originally gave. That proof requires extra regularity for the curve γ (it needs to be simple and have nonvanishing derivative) and extra regularity for the holomorphic function f (it needs to have continuous derivative in order to apply Stokes’ theorem in its standard form). The required extra regularity for f will always hold, but in order to prove this we will need Cauchy’s theorem itself! In this section we would like to sketch the layout of the theory required to give a rigorous proof of a slightly restricted form of this theorem. Indeed, in Theorem 5.7 we will prove it in the special case of so-called star-shaped domains. In due course we will also see a much more general form of Cauchy’s theorem that includes Theorem 5.1 as a special case. Before doing that, we recall the following special case of Example 3.13 and Q. 3.8, which shows that Cauchy’s theorem fails on C {0}, so the requirement that Ω is simply connected cannot simply be dropped. Example 5.2. Consider the holomorphic function f (z) = z1 on Ω := C{0}, and for r > 0 let
γ : [0, 2π] → C be the simple closed C 1 curve γ(θ) = reiθ that travels anticlockwise around the circle
Z
f (z)dz = 2πi.
γ
45
5.2
Goursat’s theorem – Cauchy’s theorem on triangles
Although you have already done a more complicated computation in Q. 3.8, let’s redo it in this special
case. Note that f (γ(θ)) = f (reiθ ) = r−1 e−iθ , and γ 0 (θ) = ireiθ , and so
Z
Z 2π
Z 2π
−1 −iθ

r e ire dθ = i
dθ = 2πi.
f (z)dz =
0
γ
5.2
0
Goursat’s theorem – Cauchy’s theorem on triangles
VIDEO: Goursat’s theorem: Cauchy’s theorem on triangles
Édouard Jean-Baptiste Goursat (1858-1936). Known for his contribution to the current rigorous
theory for Cauchy’s theorem and his spectacular moustache.
A first situation in which one rigorously proves Cauchy’s theorem for general holomorphic f is when
one heavily restricts the curves γ one allows and integrates around the boundary of a triangle. This
will then later be used as a tool in order to prove more general results such as Theorem 5.1. This
special case is named after Goursat, who came long after Cauchy.
Theorem 5.3 (Gour…

attachment

Tags:
holomorphic function

uniform convergence

limit function

Riemann Mapping Theorem

Hurwitzs theorem

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