Prove that suppose that the boundary of a simply
connected region contains a line segment y as a
one-sided free boundary are. Then the function
f(x) which maps onto the unit disk can be
extended to a function which is analytic and one to
one on Quy. The image of y is an are y'on the
unit circle.
Answers
Answer:
1.3 Fractional Linear Transformations
Finally, as a specific example of conformal maps consider the fractional linear
transformations. A fractional linear transformation or FLT is just a function
of the form
f(z) =
az + b
cz + d
,
where a, b, c, d ∈ C are constants such that ad−bc 6= 0. We can extend such a
function f to the extended complex plane or Riemann sphere C
∗ = C∪{∞} by
defining f(∞) = a/c if c 6= 0 and ∞ otherwise, and by defining f(−d/c) = ∞
if c 6= 0.
Such a function f has no critical points and hence is conformal everywhere
(except −d/c if c 6= 0). FLT’s have other interesting properties. In particular,
if we extend the definition of “circles” in the Riemann sphere to include
lines, which we consider to be “circles passing through ∞”, then FLT’s take
circles to circles. More significantly for this course, we will see later than
any bijective conformal map from the unit disk to itself must be an FLT;
combined with the Riemann Mapping Theorem, this allows us to classify
the set of conformal self-maps of any simply connected open subset of the
complex plane.
2 Week 2
Recall that if f : D → C is a function and γ : [a, b] → C is a path whose
image is contained in D, then the path integral of f along γ is defined to be:
Z
γ
f dz =
Z b
a
f(γ(t))γ
0
(t) dt.
If γ is a loop and f is analytic, it doesn’t take many examples before one
begins to notice a pattern: the path integral of an analytic function around
a loop is related to the poles of the function, if any, inside the loop. This
simplest case of this is Cauchy’s Theorem: if f is analytic on a domain
D and extends continuously to ∂D, then H
∂D
f dz = 0, if ∂D is oriented
appropriately (see below).
It’s important to note that Cauchy’s theorem applies even if the boundary
of D is not a single loop: if ∂D has more than one component, just add up
the path integrals along each component. However, each component m
Step-by-step explanation: