aarjaneiro/planetmath huggingface dataset:
generalized Cauchy integral formula

generalized Cauchy integral formula

Theorem

.

Let

U

⊂

C

be a domain with

C

1

boundary. Let

f

:

U

→

C

be a

C

1

function

that is

C

1

up to the boundary. Then for

z

∈

U

,

f

⁢

(

z

)

=

1

2

⁢

π

⁢

i

⁢

∫

∂

⁡

U

f

⁢

(

w

)

w

-

z

⁢

𝑑

w

-

1

2

⁢

π

⁢

i

⁢

∫

U

∂

⁡

f

∂

⁡

z

¯

⁢

(

w

)

w

-

z

⁢

𝑑

w

¯

∧

d

⁢

w

.

Note that

C

1

up to the boundary means that the function and the

derivative

extend to be

continuous
functions

on the

closure

of

U

.

The

theorem

follows from Stokes’ theorem. When

f

is

holomorphic

,
then the second term is zero and this is the classical

Cauchy integral formula

.

References

1

Lars Hörmander.
,
North-Holland Publishing Company, New York, New York, 1973.

2

Steven G. Krantz.
,
AMS Chelsea Publishing, Providence, Rhode Island, 1992.

Title

generalized Cauchy integral formula

Canonical name

GeneralizedCauchyIntegralFormula

Date of creation

2013-03-22 17:46:41

Last modified on

2013-03-22 17:46:41

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

6

Author

jirka (4157)

Entry type

Theorem

Classification

msc 30E20

Synonym

generalized Cauchy formula

Generated on Fri Feb 9 20:20:39 2018 by

LaTeXML
holomorphic function associated with continuous function

holomorphic function associated with continuous function

Theorem

.

If

f

⁢

(

z

)

is

continuous

on a (finite)

contour

γ

of the

complex plane

, then the contour integral

g

(

z

)

=

:

∫

γ

f

⁢

(

t

)

t

-

z

d

t

,

(1)

defines a

function

z

↦

g

⁢

(

z

)

which is

holomorphic

in any domain

D

not containing points of

γ

.  Moreover, the

derivative

has the

expression

g

′

⁢

(

z

)

=

∫

γ

f

⁢

(

t

)

(

t

-

z

)

2

⁢

𝑑

t

.

(2)

Proof.

The right hand side of (2) is defined since its

integrand

is continuous.  On has to show that it equals

lim

Δ

⁢

z

→

0

⁡

g

⁢

(

z

+

Δ

⁢

z

)

-

g

⁢

(

z

)

Δ

⁢

z

.

Let

z

1

=

:

z

+

Δ

z

∉

γ

,

Δ

⁢

z

≠

0

.  We may write first

g

⁢

(

z

1

)

-

g

⁢

(

z

)

z

1

-

z

=

1

Δ

⁢

z

⁢

∫

γ

f

⁢

(

t

)

⁢

[

1

t

-

z

1

-

1

t

-

z

]

⁢

𝑑

t

=

∫

γ

f

⁢

(

t

)

(

t

-

z

1

)

⁢

(

t

-

z

)

⁢

𝑑

t

,

whence

E

=

:

g

⁢

(

z

1

)

-

g

⁢

(

z

)

z

1

-

z

-

∫

γ

f

⁢

(

t

)

(

t

-

z

)

2

=

Δ

z

⋅

∫

γ

f

⁢

(

t

)

(

t

-

z

1

)

⁢

(

t

-

z

)

2

d

t

.

Because

f

is continuous in the compact set

γ

, there is a

positive

constant

M

such that

|

f

⁢

(

t

)

|

<

M

∀

t

∈

γ

.

As well, we have a positive constant

d

such that

|

t

-

z

|

≧

d

∀

t

∈

γ

.

When we choose

|

Δ

⁢

z

|

<

d

2

,  it follows that

|

t

-

z

1

|

=

|

(

t

-

z

)

-

Δ

⁢

z

|

≧

|

t

-

z

|

-

|

Δ

⁢

z

|

>

d

-

d

2

=

d

2

.

Consequently,

|

f

⁢

(

t

)

(

t

-

z

1

)

⁢

(

t

-

z

)

2

|

=

|

f

⁢

(

t

)

|

|

t

-

z

1

|

⁢

|

t

-

z

|

2

<

M

d

2

⋅

d

2

=

2

⁢

M

d

3

and, by the

estimating theorem of contour integral

,

|

E

|

=

|

Δ

⁢

z

|

⋅

|

∫

γ

f

⁢

(

t

)

(

t

-

z

1

)

⁢

(

t

-

z

)

2

⁢

𝑑

t

|

<

|

Δ

⁢

z

|

⋅

2

⁢

M

d

3

⋅

k

,

where

k

is the length of the contour.  The last expression tends to zero as

Δ

⁢

z

→

0

.  This settles the proof.

Remark 1.

By

induction

, one can prove the following generalisation of (2):

g

(

n

)

(

z

)

=

n

!

∫

γ

f

⁢

(

t

)

(

t

-

z

)

n

+

1

d

t

(

n

=

0

,

1

,

2

,

…

)

(3)

Remark 2.

The contour

γ

may be .  If it especially is a circle, then (1) defines a

holomorphic function

inside

γ

and another outside it.

Title

holomorphic function associated with continuous function

Canonical

name

HolomorphicFunctionAssociatedWithContinuousFunction

Date of creation

2013-03-22 19:14:29

Last modified on

2013-03-22 19:14:29

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 30E20

Classification

msc 30D20

Related topic

DifferentiationUnderIntegralSign

Related topic

CauchyIntegralFormula

Generated on Fri Feb 9 20:20:41 2018 by

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logarithmic integral

logarithmic integral

The European or Eulerian version of

logarithmic integral

(in Latin

logarithmus integralis

) is defined as

Li

⁡

x

:=

∫

2

x

d

⁢

t

ln

⁡

t

,

(1)

and the American version is

li

⁡

x

:=

∫

0

x

d

⁢

t

ln

⁡

t

,

(2)

The

integrand

1

ln

⁡

t

has a singularity

t

=

1

,
and for

x

>

1

the latter

definition

is interpreted as
the

Cauchy principal value

li

⁡

x

=

lim

ε

→

0

+

⁡

(

∫

0

1

-

ε

d

⁢

t

ln

⁡

t

+

∫

1

+

ε

x

d

⁢

t

ln

⁡

t

)

.

The connection between (1) and (2) is

Li

⁡

x

=

li

⁡

x

-

li

⁡

2

.

The logarithmic integral appears in some physical problems
and in a formulation of the

prime number theorem

(

Li

⁡

x

gives
a slightly better approximation for the

prime counting function

than

li

⁡

x

).

One has the

asymptotic series

expansion

Li

⁡

x

=

x

ln

⁡

x

⁢

∑

n

=

0

∞

n

!

(

ln

⁡

x

)

n

.

The definition of the logarithmic integral may be extended to the whole

complex plane

, and one gets the

analytic function

Li

⁡

z

having
the

branch point

z

=

1

and the

derivative

1

log

⁡

z

.

Title

logarithmic integral

Canonical

name

LogarithmicIntegral

Date of creation

2013-03-22 17:03:05

Last modified on

2013-03-22 17:03:05

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Definition

Classification

msc 30E20

Classification

msc 33E20

Classification

msc 26A36

Synonym

Li

Related topic

SineIntegral

Related topic

PrimeNumberTheorem

Related topic

PrimeCountingFunction

Related topic

LaTeXSymbolForCauchyPrincipalValue

Related topic

ConvergenceOfIntegrals

Defines

logarithmic integral

Defines

logarithmus integralis

Defines

Eulerian logarithmic integral

Generated on Fri Feb 9 20:20:43 2018 by

LaTeXML
Möbius circle transformation theorem

Möbius circle transformation theorem

Möbius

transformations

always transform circles into circles.

Title

Möbius circle transformation theorem

Canonical name

MobiusCircleTransformationTheorem

Date of creation

2013-03-22 13:35:47

Last modified on

2013-03-22 13:35:47

Owner

Johan (1032)

Last modified by

Johan (1032)

Numerical id

4

Author

Johan (1032)

Entry type

Theorem

Classification

msc 30E20

Related topic

UnitDiskUpperHalfPlaneConformalEquivalenceTheorem

Generated on Fri Feb 9 20:20:45 2018 by

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Möbius transformation cross-ratio preservation theorem

Möbius transformation cross-ratio preservation theorem

A Möbius

transformation

f

:

z

↦

w

preserves the cross-ratios, i.e.

(

w

1

-

w

2

)

⁢

(

w

3

-

w

4

)

(

w

1

-

w

4

)

⁢

(

w

3

-

w

2

)

=

(

z

1

-

z

2

)

⁢

(

z

3

-

z

4

)

(

z

1

-

z

4

)

⁢

(

z

3

-

z

2

)

Conversely

, given two

quadruplets

which have the same

cross-ratio

, there
exists a Möbius transformation which maps one quadruplet to the other.

A

consequence

of this result is that the cross-ratio of

(

a

,

b

,

c

,

d

)

is the
value at

a

of the Möbius transformation that takes

b

,

c

,

d

, to

1

,

0

,

∞

respectively.

Title

Möbius transformation cross-ratio preservation

theorem

Canonical name

MobiusTransformationCrossratioPreservationTheorem

Date of creation

2013-03-22 13:35:50

Last modified on

2013-03-22 13:35:50

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 30E20

Related topic

CrossRatio

Generated on Fri Feb 9 20:20:47 2018 by

LaTeXML
order of factors in infinite product

order of factors in infinite product

Theorem

.

If the infinite product

∏

ν

=

1

∞

(

1

+

c

ν

)

=

(

1

+

c

1

)

⁢

(

1

+

c

2

)

⁢

⋯

of

complex numbers

1

+

c

ν

is

absolutely convergent

(

http://planetmath.org/AbsoluteConvergenceOfInfiniteProduct

), then its

value

, i.e.

lim

n

→

∞

⁡

∏

ν

=

1

n

(

1

+

c

ν

)

, does not depend on the is zero.

Title

order of factors in infinite product

Canonical name

OrderOfFactorsInInfiniteProduct

Date of creation

2013-03-22 14:37:27

Last modified on

2013-03-22 14:37:27

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 30E20

Related topic

AbsoluteConvergenceOfInfiniteProductAndSeries

Related topic

ConvergenceOfComplexTermSeries

Related topic

SumOfSeriesDependsOnOrder

Defines

value of an infinite product

Generated on Fri Feb 9 20:20:49 2018 by

LaTeXML
proof of absolute convergence implies convergence for an infinite product

proof of absolute convergence implies convergence for an infinite product

This comes at once from the

link between infinite products and sums

and the

absolute convergence theorem

for infinite sums.

Title

proof of

absolute convergence

implies

convergence

for an infinite product

Canonical name

ProofOfAbsoluteConvergenceImpliesConvergenceForAnInfiniteProduct

Date of creation

2013-03-22 13:41:47

Last modified on

2013-03-22 13:41:47

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

4

Author

paolini (1187)

Entry type

Proof

Classification

msc 30E20

Generated on Fri Feb 9 20:20:50 2018 by

LaTeXML
proof of argument principle

proof of argument principle

Since

f

is

meromorphic

,

f

′

is meromorphic, and hence

f

′

/

f

is meromorphic. The singularities of

f

′

/

f

can only occur at the zeros and the poles of

f

.

I claim that all singularities of

f

′

/

f

are

simple poles

. Furthermore, if

f

has a zero at some point

p

, then the

residue

of the pole at

p

is

positive

and equals the

multiplicity

of the zero of

f

at

p

. If

f

has a pole at some point

p

, then the residue of the pole at

p

is negative and equals minus the multiplicity of the pole of

f

at

p

.

To prove these assertions, write

f

⁢

(

x

)

=

(

x

-

p

)

n

⁢

g

⁢

(

x

)

with

g

⁢

(

p

)

≠

0

. Then

f

′

⁢

(

x

)

f

⁢

(

x

)

=

n

x

-

p

+

g

′

⁢

(

x

)

g

⁢

(

x

)

Since

g

⁢

(

p

)

≠

0

, the only singularity of

f

′

/

f

at

p

comes from the first summand. Since

n

is either the order of the zero of

f

at

p

if

f

has a zero at

p

or minus the order of the pole of

f

at

p

if

f

has a pole at

p

, the assertion is proven.

By the

Cauchy residue theorem

, the integral

1

2

⁢

π

⁢

i

⁢

∫

C

f

′

⁢

(

z

)

f

⁢

(

z

)

⁢

𝑑

z

equals the sum of the residues of

f

′

/

f

. Combining this fact with the

characterization

of the poles of

f

′

/

f

and their residues given above, one deduces that this integral equals the number of zeros of

f

minus the number of poles of

f

, counted with multiplicity.

Title

proof of argument principle

Canonical

name

ProofOfArgumentPrinciple

Date of creation

2013-03-22 14:34:32

Last modified on

2013-03-22 14:34:32

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

11

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 30E20

Synonym

Cauchy’s

argument principle

Generated on Fri Feb 9 20:20:52 2018 by

LaTeXML
proof of Cauchy integral formula

proof of Cauchy integral formula

Let

D

=

{

z

∈

ℂ

:

∥

z

-

z

0

∥

<

R

}

be a disk in the

complex plane

,

S

⊂

D

a finite subset, and

U

⊂

ℂ

an
open domain that contains the closed disk

D

¯

. Suppose that

•

f

:

U

\

S

→

ℂ

is

holomorphic

, and that

•

f

⁢

(

z

)

is

bounded

near all

z

∈

D

\

S

.

Hence, by a straightforward compactness argument we also have that

f

⁢

(

z

)

is bounded on

D

¯

\

S

, and hence bounded on

D

\

S

.

Let

z

∈

D

\

S

be given, and set

g

⁢

(

ζ

)

=

f

⁢

(

ζ

)

-

f

⁢

(

z

)

ζ

-

z

,

ζ

∈

D

\

S

′

,

where

S

′

=

S

∪

{

z

}

. Note that

g

⁢

(

ζ

)

is holomorphic and
bounded on

D

\

S

′

.
The second assertion is true, because

g

⁢

(

ζ

)

→

f

′

⁢

(

z

)

,

as

⁢

ζ

→

z

.

Therefore, by the

Cauchy integral theorem

∮

C

g

⁢

(

ζ

)

⁢

𝑑

ζ

=

0

,

where

C

is the counterclockwise circular

contour

parameterized by

ζ

=

z

0

+

R

⁢

e

i

⁢

t

,

0

≤

t

≤

2

⁢

π

.

Hence,

∮

C

f

⁢

(

ζ

)

ζ

-

z

⁢

𝑑

ζ

=

∮

C

f

⁢

(

z

)

ζ

-

z

⁢

𝑑

ζ

.

(1)

𝐋𝐞𝐦𝐦𝐚

If

z

∈

ℂ

is such that

∥

z

∥

≠

1

, then

∮

∥

ζ

∥

=

1

d

⁢

ζ

ζ

-

z

=

{

0

if

⁢

∥

z

∥

>

1

2

⁢

π

⁢

i

if

⁢

∥

z

∥

<

1

The proof is a fun exercise in elementary integral calculus, an
application of the half-angle trigonometric substitutions.

Thanks to the Lemma, the right hand side of (

1

) evaluates to

2

⁢

π

⁢

i

⁢

f

⁢

(

z

)

.

Dividing through by

2

⁢

π

⁢

i

, we obtain

f

⁢

(

z

)

=

1

2

⁢

π

⁢

i

⁢

∮

C

f

⁢

(

ζ

)

ζ

-

z

⁢

𝑑

ζ

,

z

∈

D

,

as desired.

Since a circle is a

compact set

, the defining limit for the

derivative

d

d

⁢

z

⁢

f

⁢

(

ζ

)

ζ

-

z

=

f

⁢

(

ζ

)

(

ζ

-

z

)

2

,

z

∈

D

converges uniformly

for

ζ

∈

∂

⁡

D

. Thanks to the

uniform
convergence

, the order of the derivative and the integral operations
can be interchanged. In this way we obtain the second formula:

f

′

⁢

(

z

)

=

1

2

⁢

π

⁢

i

⁢

d

d

⁢

z

⁢

∮

C

f

⁢

(

ζ

)

ζ

-

z

⁢

𝑑

ζ

=

1

2

⁢

π

⁢

i

⁢

∮

C

f

⁢

(

ζ

)

(

ζ

-

z

)

2

⁢

𝑑

ζ

,

z

∈

D

.

Title

proof of Cauchy integral formula

Canonical name

ProofOfCauchyIntegralFormula

Date of creation

2013-03-22 12:47:23

Last modified on

2013-03-22 12:47:23

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

15

Author

rmilson (146)

Entry type

Proof

Classification

msc 30E20

Generated on Fri Feb 9 20:20:54 2018 by

LaTeXML
proof of Cauchy residue theorem

proof of Cauchy residue theorem

Being

f

holomorphic

by

Cauchy-Riemann equations

the

differential form

f

⁢

(

z

)

⁢

d

⁢

z

is closed. So by the lemma about

closed differential forms

on a simple connected domain we know that the integral

∫

C

f

⁢

(

z

)

⁢

𝑑

z

is equal to

∫

C

′

f

⁢

(

z

)

⁢

𝑑

z

if

C

′

is any curve which is homotopic to

C

.
In particular we can consider a curve

C

′

which turns around the points

a

j

along small circles and join these small circles with segments. Since the curve

C

′

follows each segment two times with opposite

orientation

it is enough to sum
the integrals of

f

around the small circles.

So letting

z

=

a

j

+

ρ

⁢

e

i

⁢

θ

be a parameterization of the curve around the point

a

j

, we have

d

⁢

z

=

ρ

⁢

i

⁢

e

i

⁢

θ

⁢

d

⁢

θ

and hence

∫

C

f

⁢

(

z

)

⁢

𝑑

z

=

∫

C

′

f

⁢

(

z

)

⁢

𝑑

z

=

∑

j

η

⁢

(

C

,

a

j

)

⁢

∫

∂

⁡

B

ρ

⁢

(

a

j

)

f

⁢

(

z

)

⁢

𝑑

z

=

∑

j

η

⁢

(

C

,

a

j

)

⁢

∫

0

2

⁢

π

f

⁢

(

a

j

+

ρ

⁢

e

i

⁢

θ

)

⁢

ρ

⁢

i

⁢

e

i

⁢

θ

⁢

𝑑

θ

where

ρ

>

0

is chosen so small that the balls

B

ρ

⁢

(

a

j

)

are all disjoint and all contained in the domain

U

. So by linearity, it is enough to prove that
for all

j

i

⁢

∫

0

2

⁢

π

f

⁢

(

a

j

+

e

i

⁢

θ

)

⁢

ρ

⁢

e

i

⁢

θ

⁢

𝑑

θ

=

2

⁢

π

⁢

i

⁢

Res

⁢

(

f

,

a

j

)

.

Let now

j

be fixed and
consider now the

Laurent series

for

f

in

a

j

:

f

⁢

(

z

)

=

∑

k

∈

ℤ

c

k

⁢

(

z

-

a

j

)

k

so that

Res

⁢

(

f

,

a

j

)

=

c

-

1

. We have

∫

0

2

⁢

π

f

⁢

(

a

j

+

e

i

⁢

θ

)

⁢

ρ

⁢

e

i

⁢

θ

⁢

𝑑

θ

=

∑

k

∫

0

2

⁢

π

c

k

⁢

(

ρ

⁢

e

i

⁢

θ

)

k

⁢

ρ

⁢

e

i

⁢

θ

⁢

𝑑

θ

=

ρ

k

+

1

⁢

∑

k

c

k

⁢

∫

0

2

⁢

π

e

i

⁢

(

k

+

1

)

⁢

θ

⁢

𝑑

θ

.

Notice now that if

k

=

-

1

we have

ρ

k

+

1

⁢

c

k

⁢

∫

0

2

⁢

π

e

i

⁢

(

k

+

1

)

⁢

θ

⁢

𝑑

θ

=

c

-

1

⁢

∫

0

2

⁢

π

𝑑

θ

=

2

⁢

π

⁢

c

-

1

=

2

⁢

π

⁢

Res

⁢

(

f

,

a

j

)

while for

k

≠

-

1

we have

∫

0

2

⁢

π

e

i

⁢

(

k

+

1

)

⁢

θ

⁢

𝑑

θ

=

[

e

i

⁢

(

k

+

1

)

⁢

θ

i

⁢

(

k

+

1

)

]

0

2

⁢

π

=

0

.

Hence the result follows.

Title

proof of Cauchy residue theorem

Canonical name

ProofOfCauchyResidueTheorem

Date of creation

2013-03-22 13:42:04

Last modified on

2013-03-22 13:42:04

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

7

Author

paolini (1187)

Entry type

Proof

Classification

msc 30E20

Generated on Fri Feb 9 20:20:57 2018 by

LaTeXML
proof of closed curve theorem

proof of closed curve theorem

Let

f

⁢

(

x

+

i

⁢

y

)

=

u

⁢

(

x

,

y

)

+

i

⁢

v

⁢

(

x

,

y

)

.

Hence we have

∫

C

f

⁢

(

z

)

⁢

𝑑

z

=

∫

C

ω

+

i

⁢

∫

C

η

where

ω

and

η

are the

differential forms

ω

=

u

⁢

(

x

,

y

)

⁢

d

⁢

x

-

v

⁢

(

x

,

y

)

⁢

d

⁢

y

,

η

=

v

⁢

(

x

,

y

)

⁢

d

⁢

x

+

u

⁢

(

x

,

y

)

⁢

d

⁢

y

.

Notice that by

Cauchy-Riemann equations

ω

and

η

are

closed differential forms

. Hence by the lemma on closed differential forms on a simply connected domain we get

∫

C

1

ω

=

∫

C

2

ω

,

∫

C

1

η

=

∫

C

2

η

.

and hence

∫

C

1

f

⁢

(

z

)

⁢

𝑑

z

=

∫

C

2

f

⁢

(

z

)

⁢

𝑑

z

Title

proof of closed curve theorem

Canonical name

ProofOfClosedCurveTheorem

Date of creation

2013-03-22 13:33:34

Last modified on

2013-03-22 13:33:34

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

6

Author

paolini (1187)

Entry type

Proof

Classification

msc 30E20

Generated on Fri Feb 9 20:20:59 2018 by

LaTeXML
proof of conformal Möbius circle map theorem

proof of conformal Möbius circle map theorem

Let

f

be a conformal map from the

unit disk

Δ

onto itself.
Let

a

=

f

⁢

(

0

)

. Let

g

a

⁢

(

z

)

=

z

-

a

1

-

a

¯

⁢

z

.
Then

g

a

∘

f

is a conformal map from

Δ

onto itself,
with

g

a

∘

f

⁢

(

0

)

=

0

.
Therefore, by Schwarz’s Lemma for all

z

∈

Δ

|

g

a

∘

f

⁢

(

z

)

|

≤

|

z

|

.

Because

f

is a conformal map onto

Δ

,

f

-

1

is also a conformal map of

Δ

onto itself.

(

g

a

∘

f

)

-

1

⁢

(

0

)

=

0

so that by Schwarz’s Lemma

|

(

g

a

∘

f

)

-

1

⁢

(

w

)

|

≤

|

w

|

for all

w

∈

Δ

.
Writing

w

=

g

a

∘

f

⁢

(

z

)

this becomes

|

z

|

≤

|

g

a

∘

f

⁢

(

z

)

|

.

Therefore, for all

z

∈

Δ

|

g

a

∘

f

⁢

(

z

)

|

=

|

z

|

.
By Schwarz’s Lemma,

g

a

∘

f

is a

rotation

.
Write

g

a

∘

f

⁢

(

z

)

=

e

i

⁢

θ

⁢

z

, or

f

⁢

(

z

)

=

e

i

⁢

θ

⁢

g

a

-

1

.

Therefore,

f

is a Möbius

Transformation

.

Title

proof of conformal Möbius circle map

theorem

Canonical name

ProofOfConformalMobiusCircleMapTheorem

Date of creation

2013-03-22 13:36:45

Last modified on

2013-03-22 13:36:45

Owner

brianbirgen (2180)

Last modified by

brianbirgen (2180)

Numerical id

5

Author

brianbirgen (2180)

Entry type

Proof

Classification

msc 30E20

Related topic

SchwarzLemma

Related topic

MobiusTransformation

Related topic

AutomorphismsOfUnitDisk

Generated on Fri Feb 9 20:21:01 2018 by

LaTeXML
proof of convergence condition of infinite product

proof of convergence condition of infinite product

proof of theorem of

convergence

of infinite productFernando Sanz Gamiz

Proof.

Let

p

n

=

∏

i

=

1

n

u

i

. We have to study the convergence of the

sequence

{

p

n

}

. The sequence

{

p

n

}

converges

to a not null limit iff

{

log

⁡

p

n

}

(

log

is restricted to its

principal branch

) converges
to a finite limit. By the

Cauchy criterion

, this happens iff for
every

ϵ

′

>

0

there exist

N

such that

|

log

⁡

p

n

+

k

-

log

⁡

p

n

|

<

ϵ

′

for all

n

>

N

and all

k

=

1

,

2

,

…

, i.e, iff

|

log

⁡

p

n

+

k

p

n

|

=

|

log

⁡

u

n

+

1

⁢

u

n

+

2

⁢

⋯

⁢

u

n

+

k

|

<

ϵ

′

;

as

log

⁡

(

z

)

is an injective function and

continuous at

z

=

1

and

log

⁡

(

1

)

=

0

this will happen iff for every

ϵ

>

0

|

u

n

+

1

⁢

u

n

+

2

⁢

⋯

⁢

u

n

+

k

-

1

|

<

ϵ

for

n

greater

than

N

and

k

=

1

,

2

,

…

∎

Title

proof of convergence condition of infinite product

Canonical name

ProofOfConvergenceConditionOfInfiniteProduct

Date of creation

2013-03-22 17:22:27

Last modified on

2013-03-22 17:22:27

Owner

fernsanz (8869)

Last modified by

fernsanz (8869)

Numerical id

5

Author

fernsanz (8869)

Entry type

Proof

Classification

msc 30E20

Generated on Fri Feb 9 20:21:03 2018 by

LaTeXML
proof of converse of Möbius transformation cross-ratio preservation theorem

proof of converse of Möbius transformation cross-ratio preservation theorem

Suppose that

a

,

b

,

c

,

d

are distinct. Consider the transform

μ

defined as

μ

⁢

(

z

)

=

(

b

-

d

)

⁢

(

c

-

d

)

(

c

-

b

)

⁢

(

z

-

d

)

-

b

-

d

c

-

b

.

Simple calculation reveals that

μ

⁢

(

b

)

=

1

,

μ

⁢

(

c

)

=

0

, and

μ

⁢

(

d

)

=

∞

.
Furthermore,

μ

⁢

(

a

)

equals the

cross-ratio

of

a

,

b

,

c

,

d

.

Suppose we have two tetrads with a common cross-ratio

λ

. Then, as above, we may

construct

a transform

μ

1

which maps the first tetrad to

(

λ

,

1

,

0

,

∞

)

and a
transform

μ

2

which maps the first tetrad to

(

λ

,

1

,

0

,

∞

)

. Then

μ

2

-

1

∘

μ

1

maps the former tetrad to the latter and, by the group property, it is also a
Möbius

transformation

.

Title

proof of

converse

of Möbius transformation cross-ratio preservation

theorem

Canonical name

ProofOfConverseOfMobiusTransformationCrossratioPreservationTheorem

Date of creation

2013-03-22 17:01:51

Last modified on

2013-03-22 17:01:51

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 30E20

Generated on Fri Feb 9 20:21:04 2018 by

LaTeXML